C H A P T E R 7 Functions of Several Variables

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© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. C H A P T E R 7 Functions of Several Variabl... es Section 7.1 The Three-Dimensional Coordinate System.....................................417 Section 7.2 Surfaces in Space................................................................................423 Section 7.3 Functions of Several Variables ..........................................................428 Section 7.4 Partial Derivatives ..............................................................................434 Section 7.5 Extrema of Functions of Two Variables ...........................................443 Quiz Yourself .............................................................................................................453 Section 7.6 Lagrange Multipliers ..........................................................................456 Section 7.7 Least Squares Regression Analysis ...................................................467 Section 7.8 Double Integrals and Area in a Plane ................................................471 Section 7.9 Applications of Double Integrals.......................................................478 Review Exercises ........................................................................................................484 Test Yourself .............................................................................................................494 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 417 C H A P T E R 7 Functions of Several Variables Section 7.1 The Three-Dimensional Coordinate System 1. 2. Skills Warm Up 1. (5, 1), (3, 5) ( )2 ( )2 3 5 5 1 4 16 20 2 5 d = − + − = + = = 2. (2, 3), (−1, −1) ( )2 ( )2 1 2 1 3 9 16 25 5 d = − − + − − = + = = 3. (−5, 4), (−5, −4) ( ( )) ( ) 2 2 d = −5 − −5 + −4 − 4 = 64 = 8 4. (−3, 6), (−3, −2) ( ( )) ( ) 2 2 d = −3 − −3 + −2 − 6 = 64 = 8 5. (2, 5), (6, 9) Midpoint 2 6, 5 9 (4, 7) 2 2 ⎛ + + ⎞ = ⎜ ⎟ = ⎝ ⎠ 6. (−1, −2), (3, 2) Midpoint 1 3, 2 2 (1, 0) 2 2 ⎛ − + − + ⎞ = ⎜ ⎟ = ⎝ ⎠ 7. (−6, 0), (6, 6) Midpoint 6 6, 0 6 (0, 3) 2 2 ⎛ − + + ⎞ = ⎜ ⎟ = ⎝ ⎠ 8. (−4, 3), (2, −1) 4 2 3 ( 1) ( ) Midpoint , 1, 1 2 2 ⎛ − + + − ⎞ = ⎜ ⎟ = − ⎝ ⎠ 9. ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 : 2, 3, 2 2 3 2 2 3 4 c r x y x y = − + − = − + − = 10. 4 ( 2), 0 8 (1, 4) 2 2 C ⎛ + − + ⎞ = ⎜ ⎟ = ⎝ ⎠ 1 ( 2 4)2 (8 0)2 2 1 36 64 2 1 100 2 5 r = − − + − = + = = ( ) ( ) ( ) ( ) 2 2 2 2 2 1 4 5 1 4 25 x y x y − + − = − + − = x y z (2, 1, 3) (3, −2, 5) (−1, 2, 1) ( , 4, −2) 32 −2 2 4 4 2 x y z (1, 3, 1) (5, −2, −2) (−2, 4, −3) (−5, −2, 2) −2 −4 −4 4 4 2 2 4 418 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 3. 4. 5. x = −3, y = 4, z = 5: (−3, 4, 5) 6. x = 7, y = −2, z = −1: (7, −2, −1) 7. y = z = 0, x = 10: (10, 0, 0) 8. x = 0, y = 3, z = 2: (0, 3, 2) 9. The z-coordinate is 0. 10. The y-coordinate is 0. 11. (4, 1, 5), (8, 2, 6) ( )2 ( )2 ( )2 d = 8 − 4 + 2 − 1 + 6 − 5 = 18 = 3 2 12. (−4, −1, 1), (2, −1, 5) ( )2 ( )2 ( )2 2 4 1 1 5 1 52 2 13 d = + + − + + − = = 13. (−1, −5, 7), (−3, 4, −4) ( )2 ( )2 ( )2 d = −3 + 1 + 4 + 5 + −4 − 7 = 206 14. (8, −2, 2), (8, −2, 4) ( )2 ( )2 ( )2 d = 8 − 8 + −2 + 2 + 4 − 2 = 2 15. (6, −4, 2), (−2, 1, 3) 6 ( 2) 4 1 2 3 Midpoint , , 2 2 2 2, 3, 5 2 2 ⎛ + − − + + ⎞ =⎜ ⎟ ⎝ ⎠ = ⎛ − ⎞ ⎜ ⎟ ⎝ ⎠ 16. (4, 0, −6), (8, 8, 20) Midpoint 4 8, 0 8, 6 20 (6, 4, 7) 2 2 2 ⎛ + + − + ⎞ = ⎜ ⎟ = ⎝ ⎠ 17. (−5, −2, 5), (6, 3, −7) 5 6 2 3 5 ( 7) Midpoint , , 2 2 2 1, 1, 1 2 2 ⎛ − + − + + − ⎞ =⎜ ⎟ ⎝ ⎠ = ⎛ − ⎞ ⎜ ⎟ ⎝ ⎠ 18. (0, −2, 5), (4, 2, 7) Midpoint 0 4, 2 2, 5 7 (2, 0, 6) 2 2 2 ⎛ + − + + ⎞ = ⎜ ⎟ = ⎝ ⎠ 19. ( ) ( ) 2 1 1 2, 1, 3 , , 2 2 2 ⎛ x + − y + z + ⎞ − =⎜ ⎟ ⎝ ⎠ 2 2 2 4 2 6 x x x − = = − = 1 1 2 2 1 3 y y y + − = − = + = − 3 1 2 6 1 5 z z z + = = + = (x, y, z) = (6, −3, 5) 20. ( ) ( ) 0 2 1 1, 0, 0 , , 2 2 2 ⎛ x + y + − z + ⎞ =⎜ ⎟ ⎝ ⎠ 1 2 = x 0 2 2 y − = 0 1 2 z + = 0 = y − 2 0 = z + 1 x = 2 y = 2 z = −1 (x, y, z) = (2, 2, −1) 21. 3, 1, 2 2, 0, 3 2 2 2 2 ⎛ ⎞ = ⎛ x + y + z + ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 3 2 2 2 x + = 1 2 = y 2 3 2 z + = 3 = x + 2 4 = z + 3 x = 1 y = 2 z = 1 (x, y, z) = (1, 2, 1) 22. (0, 1, 1) 3, 3, 0 2 2 2 = ⎛ x + y + z + ⎞ ⎜ ⎟ ⎝ ⎠ 0 3 2 x + = 1 3 2 y + = 1 2 = z 0 = x + 3 2 = y + 3 x = −3 y = −1 z = 2 (x, y, z) = (−3, − 1, 2) x y z (4, 0, 5) (0, 4, −5) (−2, , 0) 12 (− , 3, 1) 12 −2 −4 −2 2 4 2 4 2 4 x y z (1, 3, 4) (−1, −3, −2) (2, −1, 1) (−3, 0, −1) −2 −4 −4 4 2 4 2 4 Section 7.1 The Three-Dimensional Coordinate System 419 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 23. Let A = (0, 0, 0), B = (2, 2, 1), and C = (2, −4, 4). Then you have ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 0 2 0 1 0 3 2 0 4 0 4 0 6 2 2 4 2 4 1 3 5. d AB d AC d BC = − + − + − = = − + − − + − = = − + − − + − = The triangle is a right triangle because ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 3 6 45 3 5 . d AB d AC d BC + = + = = = 24. Let A = (5, 3, 4), B = (7, 1, 3), and C = (3, 5, 3).Then you have ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 7 5 1 3 3 4 3 3 5 5 3 3 4 3 3 7 5 1 3 3 4 2. d AB d AC d BC = − + − + − = = − + − + − = = − + − + − = Because d(AB) = d(AC), the triangle is isosceles. The triangle is not a right triangle because ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 3 3 18 4 2 32 . d AB d AC d BC + = + = ≠ = = 25. Let A = (−1, 0, −2), B = (−1, 5, 2), and C = (−3, −1, 1). Then you have ( ) ( ) ( ) ( ) 2 2 2 1 1 0 5 2 2 41 d AB = ⎡⎣− − − ⎤⎦ + − + − − = ( ) ( ) ( ) ( ) 2 2 2 1 3 0 1 2 1 14 d AC = ⎡⎣− − − ⎤⎦ + ⎡⎣ − − ⎤⎦ + − − = ( ) ( ) ( ) ( ) 2 2 2 1 3 5 1 2 1 41. d BC = ⎡⎣− − − ⎤⎦ + ⎡⎣ − − ⎤⎦ + − = Because d(AB) = d(BC), the triangle is isosceles. The triangle is not a right triangle because ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 41 41 82 14 14 . d AB d BC d AC + = + = ≠ = = 26. Let A = (5, 0, 0), B = (0, 2, 0), and C = (0, 0, −3). Then you have ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 0 5 2 0 0 0 29 0 5 0 0 3 0 34 0 0 0 2 3 0 13. d AB d AC d BC = − + − + − = = − + − + − − = = − + − + − − = The triangle is not a right triangle because ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 29 13 42 34 34 . d AB d BC d AC + = + = ≠ = = The triangle is neither right nor isosceles. 27. Each z-coordinate is decreased by 5 units: (0, 0, −5), (2, 2, −4), (2, −4, −1) 28. Each y-coordinate is increased by 3 units: (5, 6, 4), (7, 4, 3), (3, 8, 3) 29. x2 + (y − 2)2 + (z − 2)2 = 4 30. ( )2 ( )2 ( )2 x − 2 + y − 3 + z − 1 = 9 31. The midpoint of the diameter is the center. 2 1 1 3 3 ( 1) 3 Center , , , 2, 1 2 2 2 2 ⎛ + + + − ⎞ ⎛ ⎞ = ⎜ ⎟ = ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ The radius is the distance between the center and either endpoint. ( ) ( ) 2 Radius 2 3 1 2 2 3 1 2 2 1 1 4 4 21 2 = ⎛ − ⎞ + − + − ⎜ ⎟ ⎝ ⎠ = + + = ( ) ( ) 2 3 2 2 1 2 21 2 4 ⎛⎜ x − ⎞⎟ + y − + z − = ⎝ ⎠ 420 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 32. The midpoint of the diameter is the center. Center 1 0, 2 3, 1 3 1, 1, 2 2 2 2 22 ⎛ − + − + + ⎞ ⎛ ⎞ = ⎜ ⎟ = ⎜− ⎟ ⎝ ⎠ ⎝ ⎠ The radius is the distance from the center to either endpoint. ( ) 2 2 Radius 1 0 1 3 2 3 2 1 25 1 30 2 2 4 4 2 = ⎛− − ⎞ + ⎛ − ⎞ + − = + + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ( ) 2 2 1 1 2 2 15 2 2 2 ⎛⎜ x + ⎞⎟ + ⎛⎜ y − ⎞⎟ + z − = ⎝ ⎠ ⎝ ⎠ 33. ( )2 ( )2 ( )2 x − 3 + y + 2 + z + 3 = 16 34. ( )2 ( )2 ( )2 x − 4 + y + 1 + z − 1 = 25 35. The midpoint of the diameter is the center. Center 2 0, 0 6, 0 0 (1, 3, 0) 2 2 2 ⎛ + + + ⎞ = ⎜ ⎟ = ⎝ ⎠ The radius is the distance from the center to either endpoint. ( )2 ( )2 ( )2 Radius = 1 − 2 + 3 − 0 + 0 − 0 = 10 (x − 1)2 + (y − 3)2 + z2 = 10 36. The midpoint of the diameter is the center. Center 1 0, 0 5, 0 0 1, 5, 0 2 2 2 2 2 ⎛ + + + ⎞ ⎛ ⎞ = ⎜ ⎟ = ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ The radius is the distance from the center to either endpoint. ( ) 2 2 Radius 1 1 0 5 0 0 2 2 2 1 25 26 4 4 2 = ⎛ − ⎞ + ⎛ − ⎞ + − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = + = 2 2 1 5 2 13 2 2 2 ⎛⎜ x − ⎞⎟ + ⎛⎜ y − ⎞⎟ + z = ⎝ ⎠ ⎝ ⎠ 37. The distance from (−4, 3, 2) to the xy-plane is the radius r = 2. ( )2 ( )2 ( )2 x + 4 + y − 3 + z − 2 = 4 38. The distance from (1, 2, 0) to the yz-plane is the radius r = 1. (x − 1)2 + (y − 2)2 + z2 = 1 39. ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 25 25 4 4 5 25 2 4 5 5 0 0 0 x x y z x y z x x y z − + + + = + + − = − + − + − = Center: (5 ) 2, 0, 0 Radius: 5 2 40. ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 8 0 8 16 16 0 4 0 16 x y z y x y y z x y z + + − = + − + + = − + − + − = Center: (0, 4, 0) Radius: 4 41. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 4 2 8 4 0 4 4 2 1 8 16 4 4 1 16 2 1 4 25 x y z x y z x x y y z z x y z + + + − + − = + + + − + + + + = + + + + + − + + = Center: (−2, 1, −4) Radius: 5 42. ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 4 6 4 0 4 4 6 9 4 4 9 0 2 3 9 x y z y z x y y z z x y z + + − + + = + − + + + + = − + + − + − + + = Center: (0, 2, −3) Radius: 3 Section 7.1 The Three-Dimensional Coordinate System 421 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 43. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 4 12 8 3 0 2 1 6 9 4 4 3 1 9 4 2 1 3 2 25 2 x y z x y z x x y y z z x y z + + − − − + = − + + − + + − + = − + + + − + − + − = Center: (1, 3, 2) Radius: 5 5 2 2 2 = 44. ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 11 4 11 4 9 4 4 4 4 8 16 11 0 2 4 0 2 1 4 4 1 4 1 2 x y z x y x y z x y x x y y z x y z + + − + + = + + − + + = − + + + + + = − + + − + + + = Center: (1, −2, 0) Radius: 32 45. ( )2 ( )2 ( )2 x − 1 + y − 3 + z − 2 = 25 To find the xy-trace, let z = 0. ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 1 3 0 2 25 1 3 21 x y x y − + − + − = − + − = 46. ( )2 ( )2 ( )2 x + 1 + y + 2 + z − 2 = 16 To find the xy-trace, let z = 0. ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 1 2 0 2 16 1 2 12 x y x y + + + + − = + + + = 47. x2 + y2 + z2 − 6x − 10y + 6z + 30 = 0 To find the xy-trace, let z = 0. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 0 6 10 60 30 0 6 9 10 25 30 9 25 3 5 4 x y x y x x y y x y + + − − + + = − + + − + = − + + − + − = 48. x2 + y2 + z2 − 4y + 2z − 60 = 0 To find the xy-trace, let z = 0. ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 0 4 20 60 0 4 60 4 4 60 4 2 64 x y y x y y x y y x y + + − + − = + − = + − + = + + − = 49. x2 + (y + 3)2 + z2 = 25 To find the yz-trace, let x = 0. ( ) ( ) 2 2 2 2 2 0 3 25 3 25 y z y z + + + = + + = 4 2 4 2 4 6 2 y x z y x −2 −4 −6 2 −6 4 z y x 6 4 2 2 4 6 4 8 z y x −8 8 −12 12 4 4 z y x 2 −4 −6 4 6 4 4 2 −2 −4 2 −6 z 422 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 50. (x + 2)2 + (y − 3)2 + z2 = 9 To find the yz-trace, let x = 0. ( ) ( ) ( ) 2 2 2 2 2 0 2 3 9 3 5 y z y z + + − + = − + = 51. x2 + y2 + z2 − 4x − 4y − 6z − 12 = 0 To find the yz-trace, let x = 0. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 0 40 4 6 12 0 4 4 6 9 12 4 9 2 3 25 y z y z y y z z y z + + − − − − = − + + − + = + + − + − = 52. x2 + y2 + z2 − 6x − 10y + 6z + 30 = 0 To find the yz-trace, let x = 0. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 0 60 10 6 30 0 10 25 6 9 30 25 9 5 3 4 y z y z y y z z y z + + − − + + = − + + + + = − + + − + + = 53. x2 + y2 + z2 = 25 (a) To find the trace, let z = 3. 2 2 2 2 2 3 25 16 x y x y + + = + = (b) To find the trace, let x = 4. 2 2 2 2 2 4 25 9 y z y z + + = + = 54. x2 + y2 + z2 = 169 (a) To find the trace, let x = 5. 2 2 2 2 2 5 169 144 y z y z + + = + = (b) To find the trace, let y = 12. 2 2 2 2 2 12 169 25 x z x z + + = + = y x 4 −4 −4 −6 6 2 6 8 4 6 z y x 4 4 6 z y x (0, 5, −3) 4 2 −2 −4 4 6 8 2 −4 −6 z x y 6 4 2 6 z x y 4 8 8 4 4 8 z y x 25 20 10 5 10 15 15 z y x 9 6 −6 −9 3 6 9 12 9 6 −6 −9 z Section 7.2 Surfaces in Space 423 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 55. x2 + y2 + z2 − 4x − 6y + 9 = 0 (a) To find the trace, let x = 2. (b) To find the trace, let y = 3. ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 42 6 9 0 6 9 9 4 8 9 3 2 y z y y y z y z + + − − + = − + + = − − + + − + = ( ) ( ) ( ) 2 2 2 2 2 2 2 2 3 4 6 3 9 0 4 4 9 18 9 4 2 2 x z x x x z x z + + − − + = − + + = − + − + − + = 56. x2 + y2 + z2 − 8x − 6z + 16 = 0 (a) To find the trace, let x = 4. (b) To find the trace, let z = 3. ( ) ( ) ( ) 2 2 2 2 2 2 2 4 84 6 16 0 6 9 16 32 16 9 3 9 y z z y z z y z + + − − + = + − + = − + − + + − = ( ) ( ) ( ) 2 2 2 2 2 2 2 3 8 63 16 0 8 16 9 18 16 16 4 9 x y x x x y x y + + − − + = − + + = − + − + − + = 57. ( )2 2 2 2 2 2 2 165 2 6806.25 x y z x y z + + = + + = 58. (a) (3, 3, 3) (b) (4, 4, 8) Section 7.2 Surfaces in Space Skills Warm Up 1. 3x + 4y = 12 Let x = 0 to find the y-intercept. 3(0) 4 12 3 y y + = = y-intercept: (0, 3) Let y = 0 to find the x-intercept. 3 4(0) 12 4 x x + = = x-intercept: (4, 0) 2. 6x + y = −8 Let x = 0 to find the y-intercept. 6(0) 8 8 y y + = − = − y-intercept: (0, −8) Let y = 0 to find the x-intercept. 6 0 8 4 3 x x + = − = − x-intercept: 4, 0 3 ⎛− ⎞ ⎜ ⎟ ⎝ ⎠ x y 8 4 8 4 8 z x y 8 4 8 8 4 z y x 8 6 2 4 8 6 4 z y x 8 6 2 4 8 6 4 z 424 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. 4x + 2y + 6z = 12 To find the x-intercept, let y = 0 and z = 0. 4x = 12 ⇒ x = 3 To find the y-intercept, let x = 0 and z = 0. 2y = 12 ⇒ y = 6 To find the z-intercept, let x = 0 and y = 0. 6z = 12 ⇒ z = 2 2. 3x + 6y + 2z = 6 To find the x-intercept, let y = 0 and z = 0. 3x = 6 ⇒ x = 2 To find the y-intercept, let x = 0 and z = 0. 6y = 6 ⇒ y = 1 To find the z-intercept, let x = 0 and y = 0. 2z = 6 ⇒ z = 3 3. 3x + 3y + 5z = 15 To find the x-intercept, let y = 0 and z = 0. 3x = 15 ⇒ x = 5 To find the y-intercept, let x = 0 and z = 0. 3y = 15 ⇒ y = 5 To find the z-intercept, let x = 0 and y = 0. 5z = 15 ⇒ z = 3 4. x + y + z = 3 To find the x-intercept, let y = 0 and z = 0. x = 3 To find the y-intercept, let x = 0 and z = 0. y = 3 To find the z-intercept, let x = 0 and y = 0. z = 3 5. 2x − y + 3z = 4 To find the x-intercept, let y = 0 and z = 0. 2x = 4 ⇒ x = 2 To find the y-intercept, let x = 0 and z = 0. −y = 4 ⇒ y = −4 To find the z-intercept, let x = 0 and y = 0. 43 3z = 4 ⇒ z = 6. 2x − y + z = 4 To find the x-intercept, let y = 0 and z = 0. 2x = 4 ⇒ x = 2 To find the y-intercept, let x = 0 and z = 0. −y = 4 ⇒ y = −4 To find the z-intercept, let x = 0 and y = 0. z = 4 Skills Warm Up —continued— 3. −2x + y = −2 Let x = 0 to find the y-intercept. 2(0) 2 2 y y − + = − = − y-intercept: (0, −2) Let y = 0 to find the x-intercept. 2 0 2 1 x x − + = − = x-intercept: (1, 0) 4. −x − y = 5 Let x = 0 to find the y-intercept. 0 5 5 y y − − = = − y-intercept: (0, −5) Let y = 0 to find the x-intercept. 0 5 5 x x − − = = − x-intercept: (−5, 0) 5. 2 2 2 2 2 2 1 4 16x 16y 16z 4 x y z + + = + + = 6. 2 2 2 2 2 2 9 9 9 36 4 x y z x y z + + = + + = x y 4 6 4 z (0, 0, 2) (3, 0, 0) (0, 6, 0) x 2 2 3 y 3 3 (0, 0, 3) (2, 0, 0) (0, 1, 0) z x y 5 3 5 z (5, 0, 0) (0, 5, 0) (0, 0, 3) x y (3, 0, 0) (0, 3, 0) (0, 0, 3) 3 3 3 z x y 4 3 1 1 3 2 −2 −2 −1 −4 (2, 0, 0) (0, −4, 0) 4 3 0, 0, ( ( z x 1 y −4 3 4 1 2 (0, 0, 4) (2, 0, 0) (0, −4, 0 ) z Section 7.2 Surfaces in Space 425 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 7. z = 8 Because the coefficients of x and y are zero, the only intercept is the z-intercept of 8. The plane is parallel to the xy-plane. 8. x = 5 Because the coefficients of y and z are zero, the only intercept is the x-intercept. The plane is parallel to the yz-plane. 9. y + z = 5 Because the coefficient of x is zero, there is no x-intercept. To find the y-intercept, let z = 0. y = 5 To find the z-intercept, let y = 0. z = 5 The plane is parallel to the x-axis. 10. x + 2y = 4 Because the coefficient of z is zero, there is no z-intercept. To find the x-intercept, let y = 0. x = 4 To find the y-intercept, let x = 0. 2y = 4 ⇒ y = 2 The plane is parallel to the z-axis. 11. x + z = 6 To find the x-intercept, let z = 0, so x = 6. Because the coefficient of y is zero, there is no y-intercept. To find the z-intercept, let x = 0, so z = 6. The plane is parallel to the y-axis. 12. x − 3z = 3 Because the coefficient of y is zero, there is no y-intercept. To find the x-intercept, let z = 0. x = 3 To find the z-intercept, let x = 0. −3z = 3 ⇒ z = −1 The plane is parallel to the y-axis. 13. For the first plane, 5x − 3y + z = 4, a1 = 5, b1 = −3, and c1 = 1. For the second plane, x + 4y + 7z = 1, a2 = 1, b2 = 4, and c2 = 7.So you have 1 2 1 2 1 2 (5)(1) ( 3)(4) (1)(7) 5 12 7 0. a a + b b + c c = + − + = − + = The planes are perpendicular. 14. For the first plane, 3x + y − 4z = 3, a1 = 3, b1 = 1, and c1 = −4. For the second plane, −9x − 3y + 12z = 4, a2 = −9, b2 = −3, and c2 = 12. So you have a2 = −3a1, b2 = −3b1, and c2 = −3c1. The planes are parallel. 15. For the first plane, x − 5y − z = 1, a1 = 1, b1 = −5, and c1 = −1. For the second plane, 5x − 25y − 5z = −3, a2 = 5, b2 = −25, and c2 = −5. So you have a2 = 5a1, b2 = 5b1, and c2 = 5c1. The planes are parallel. 16. For the first plane, x + 3y + 2z = 6, a1 = 1, b1 = 3, and c1 = 2. For the second plane, 4x − 12y + 8z = 24, a2 = 4, b2 = −12, and c2 = 8. The planes are not parallel because a2 = 4a1 and b2 ≠ 4b1. The planes are not perpendicular because 1 2 1 2 1 2 (1)(4) (3)( 12) (2)(8) 4 36 16 16 0. a a + b b + c c = + − + = − + = ≠ 17. For the first plane, x + 2y = 3, a1 = 1, b1 = 2, and c1 = 0. For the second plane, 4x + 8y = 5, a2 = 4, b2 = 8, and c2 = 0.So you have a2 = 4a1, b2 = 4b1, and c2 = 4c1. The planes are parallel. y z x 4 2 6 4 6 10 2 2 4 6 (0, 0, 8) x y 6 −6 −6 −8 6 8 10 6 z (5, 0, 0) x y 6 6 (0, 0, 5) (0, 5, 0) 6 z x 4 3 y 4 (4, 0, 0) (0, 2, 0) z x y z (0, 0, 6) (6, 0, 0) 2 4 6 2 4 2 6 x y (3, 0, 0) (0, 0, −1) 4 1 1 2 1 2 3 4 z 426 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 18. For the first plane, x + 3y + z = 7, a1 = 1, b1 = 3, and c1 = 1.For the second plane, x − 5z = 0, a2 = 1, b2 = 0, and c2 = −5.The planes are not parallel because a1 = a2 and b1 ≠ b2. The planes are not perpendicular because 1 2 1 2 1 2 (1)(1) (3)(0) (1)( 5) 1 5 4 0. a a + b b + c c = + + − = − = − ≠ 19. For the first plane, 2x + y = 3, a1 = 2, b1 = 1, and c1 = 0. For the second plane, 3x − 5z = 0, a2 = 3, b2 = 0, and c2 = −5.The planes are not parallel because 3a1 = 2a2 and 3b1 ≠ 2b2. The planes are not perpendicular because 1 2 1 2 1 2 (2)(3) (1)(0) (0)( 5) 6 0. a a + b b + c c = + + − = ≠ 20. For the first plane, 2x − z = 1, a1 = 2, b1 = 0, and c1 = −1. For the second plane, 4x + 8y = 5, a2 = 4, b2 = 1, and c2 = 8. So you have 1 2 1 2 1 2 (2)(4) (0)(1) ( 1)(8) 8 8 0. a a + b b + c c = + + − = − = The planes are perpendicular. 21. For the first plane, x = 3, a1 = 1, b1 = 0, and c1 = 0. For the second plane, z = −1, a2 = 0, b2 = 0, and c2 = 1. So you have 1 2 1 2 1 2 (1)(0) (0)(0) (0)(1) 0. a a + b b + c c = + + = The planes are perpendicular. 22. For the first plane, x = −2, a1 = 1, b1 = 0, and c1 = 0. For the second plane, y = 4, a2 = 0, b2 = 1, and c2 = 0.So you have a1a2 + b1b2 + c1c2 = (1)(0) + (0)(1) + (0)(0) = 0. The planes are perpendicular. 23. 2 2 2 1 9 16 9 x + y + z = is an ellipsoid. Matches graph (c). 24. 15x2 − 4y2 + 15z2 = −4 is a hyperboloid of two sheets. Matches graph (e). 25. 4x2 − y2 + 4z2 = 4 is a hyperboloid of one sheet. Matches graph (f ). 26. y2 = 4x2 + 9z2 is an elliptic cone. Matches graph (b). 27. 4x2 − 4y + z2 = 0 is an elliptic paraboloid. Matches graph (d). 28. 4x2 − y2 + 4z = 0 is a hyperbolic paraboloid. Matches graph (a). 29. z = x2 − y2 (a) Trace in xy-plane (z = 0): 0 x2 y2 x y = − ± = Lines (b) Trace in plane x = 3: z = 9 − y2 Parabola (c) Trace in xz-plane (y = 0): z = x2 Parabola The graph is a hyperbolic parabola. 30. y = x2 + z2 (a) Trace in xy-plane (z = 0): y = x2 Parabola (b) Trace in plane y = 1: x2 + z2 = 1 Circle (c) Trace in yz-plane (x = 0): y = z2 Parabola The graph is an elliptic paraboloid. Section 7.2 Surfaces in Space 427 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 31. 2 2 2 1 4 x + y + z = (a) Trace in xy-plane (z = 0): 2 2 1 4 x + y = Ellipse (b) Trace in xz-plane (y = 0): 2 2 1 4 x + z = Ellipse (c) Trace in yz-plane (x = 0): y2 + z2 = 1 Circle The graph is an ellipsoid. 32. y2 + z2 − x2 = 1 (a) Trace in xy-plane (z = 0): y2 − x2 = 1 Hyperbola (b) Trace in xz-plane (y = 0): z2 − x2 = 1 Hyperbola (c) Trace in yz-plane (x = 0): y2 + z2 = 1 Circle The graph is an elliptic cone. 33. 2 2 2 1 9 16 z − x − y = (a) Trace in xz-plane (y = 0): 2 2 1 9 z − x = Hyperbola (b) Trace in plane x = 2: 2 2 2 2 13 16 9 1 13 9 208 9 z y z y − = − = Hyperbola (c) Trace in plane z = 4: 2 2 2 2 15 9 16 1 135 240 x y x y + = + = Ellipse The graph is a hyperboloid of two sheets. 34. 2 2 2 0 4 y + z − x = (a) Trace in plane y = −1: 2 2 1 4 z − x = Hyperbola (b) Trace in plane z = 4: x2 − y2 = 1 Hyperbola (c) Trace in yz-plane (x = 0): 2 2 0 4 y + z = Point The graph is an elliptic cone. 35. The graph of 2 2 2 1 4 x + y + z = is an ellipsoid. 36. 2 2 2 4 z = x + y Standard form: 2 2 2 0 4 x + y − z = The graph is an elliptic cone. 37. 25x2 + 25y2 − z2 = 5 Standard form: 2 2 2 1 1 5 1 5 5 x + y − z = The graph is a hyperboloid of one sheet. 38. z = 4x2 + y2 Standard form: 2 2 1 4 z = x + y The graph is an elliptic paraboloid. 428 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 39. x2 − y2 + z = 0 Standard form: z = y2 − x2 The graph is a hyperbolic paraboloid. 40. The graph of 2 2 2 1 4 z − x − y = is a hyperboloid of two sheets. 41. x2 − y + z2 = 0 Standard form: y = x2 + z2 The graph is an elliptic paraboloid. 42. 9x2 + 4y2 − 8z2 = 72 Standard form: 2 2 2 1 8 18 9 x + y − z = The graph is a hyperboloid of one sheet. 43. z2 = 9x2 + y2 Standard form: 2 2 2 0 9 9 x + y − z = The graph is an elliptic cone. 44. The graph of 2 2 2 1 9 16 16 x + y + z = is an ellipsoid. 45. 2x2 − y2 + 2z2 = −4 Standard form: 2 2 2 1 2 4 2 − x + y − z = The graph is a hyperboloid of two sheets. 46. 4y = x2 + z2 Standard form: 2 2 4 4 y = x + z The graph is an elliptic paraboloid. 47. 3z = −y2 + x2 Standard form: 2 2 3 3 z = x − y The graph is a hyperbolic paraboloid. 48. z2 = 2x2 + 2y2 Standard form: 2 2 2 0 1 2 1 2 x + y − z = The graph is an elliptic cone. 49. 2 2 2 2 2 2 1 3963 3963 3950 x + y + z = 50. (a) You are viewing the paraboloid from the x-axis: (20, 0, 0) (b) You are viewing the paraboloid from above, but not on the z-axis: (10, 10, 20) (c) You are viewing the paraboloid from the z-axis: (0, 0, 20) (d) You are viewing the paraboloid from the y-axis: (0, 20, 0) 51. z = 0.62x − 0.41y + 0.38 (a) The approximated values of z are very close to the actual values. (b) According to the model, increases in expenditures of recreation types y and z will correspond to an increase in expenditures of recreation type x. Section 7.3 Functions of Several Variables Skills Warm Up 1. f (x) = 5 − 2x, x = −3 f (−3) = 5 − 2(−3) = 11 2. f (x) = −x2 + 4x + 5, x = −3 ( ) ( )2 ( ) f −3 = − −3 + 4 −3 + 5 = −16 Year 2004 2005 2006 2007 2008 2009 x 33.1 34.9 37.4 40.6 43.0 41.8 y 13.2 13.8 14.9 15.0 15.4 14.5 z (actual) 15.5 16.3 17.8 19.5 20.5 20.7 z (model) 15.5 16.4 17.5 19.4 20.7 20.4 Section 7.3 Functions of Several Variables 429 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. f (x, y) x y = (a) (3, 2) 3 2 f = (b) ( 1, 4) 1 4 f − = − (c) (30, 5) 30 6 5 f = = (d) f (5, y) 5 y = (e) ( , 2) 2 f x = x (f ) f (5, t) 5 t = 2. f (x, y) = 4 − x2 − 4y2 (a) f (0, 0) = 4 − 02 − 4(0)2 = 4 (b) f (0, 1) = 4 − 02 − 4(1)2 = 0 (c) f (2, 3) = 4 − 22 − 4(3)2 = −36 (d) f (1, y) = 4 − 12 − 4y2 = 3 − 4y2 (e) f (x, 0) = 4 − x2 − 4(0)2 = 4 − x2 (f ) f (t, 1) = 4 − t2 − 4(1)2 = −t2 3. f (x, y) = xey (a) f (5, 0) = 5e0 = 5 (b) f (3, 2) = 3e2 (c) f (2, 1) 2e 1 2 e − = − = (d) f (5, y) = 5ey (e) f (x, 2) = xe2 (f ) f (t, t) = tet 4. g(x, y) = ln x + y (a) g(2, 3) = ln 2 + 3 = ln 5 (b) g(5, 6) = ln 5 + 6 = ln 11 (c) g(e, 0) = ln e + 0 = ln e = 1 (d) g(0, 1) = ln 0 + 1 = ln 1 = 0 (e) g(2, −3) = ln 2 + (−3) = ln 1 = 0 (f ) g(e, e) = ln e + e = ln(2e) = 1 + ln 2 5. h(x, y, z) xy z = (a) ( ) ( )( ) 2 3 2 2, 3, 9 9 3 h = = (b) ( ) ( )( ) 1 0 1, 0, 1 0 1 h = = 6. f (x, y, z) = x + y + z (a) f (0, 5, 4) = 0 + 5 + 4 = 9 = 3 (b) f (6, 8, −3) = 6 + 8 + (−3) = 11 7. V(r, h) = π r2h (a) ( ) ( )2 ( ) V 3, 10 = π 3 10 = 90π (b) ( ) ( )2 ( ) V 5, 2 = π 5 2 = 50π Skills Warm Up —continued— 3. y = 4x2 − 3x + 4, x = −3 ( )2 ( ) y = 4 −3 − 3 −3 + 4 = 49 = 7 4. y = 3 34 − 4x + 2x2 , x = −3 ( ) ( ) y = 3 34 − 4 −3 + 2 −3 2 = 3 64 = 4 5. f (x) = 5x2 + 3x − 2 Domain: (−∞, ∞) 6. ( ) 1 2 2 3 g x x x = − + Domain: (−∞, −3) ∪ (−3, 0) ∪ (0, ∞) 7. h(y) = y − 5 Domain: [5, ∞) 8. f (y) = y2 − 5 Domain: (−∞, − 5⎤⎦ ∪ ⎡⎣ 5, ∞) 9. ( )0.65 476 ≈ 55.0104 10. ( )0.35 251 ≈ 6.9165 430 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 8. ( , ) 500 1 12 n F r n = ⎛ + r ⎞ ⎜ ⎟ ⎝ ⎠ (a) ( ) 60 0.09, 60 500 1 0.09 782.84 12 F = ⎛⎜ + ⎞⎟ ≈ ⎝ ⎠ (b) ( ) 240 0.14, 240 500 1 0.14 8090.14 12 F = ⎛⎜ + ⎞⎟ ≈ ⎝ ⎠ 9. ( ) 12 , , 1 1 1 12 12 t A P r t P r r ⎡⎛ ⎞ ⎤⎛ ⎞ = ⎢⎜ + ⎟ − ⎥⎜ + ⎟ ⎢⎣⎝ ⎠ ⎥⎦⎝ ⎠ (a) ( ) 120 100, 0.10, 10 100 1 0.10 1 1 12 $20,655.20 12 0.10 A ⎡⎛ ⎞ ⎤⎛ ⎞ = ⎢⎜ + ⎟ − ⎥⎜ + ⎟ = ⎢⎣⎝ ⎠ ⎥⎦⎝ ⎠ (b) ( ) 480 275, 0.0925, 40 275 1 0.0925 1 1 12 $1,397,672.67 12 0.0925 A ⎡⎛ ⎞ ⎤⎛ ⎞ = ⎢⎜ + ⎟ − ⎥⎜ + ⎟ = ⎢⎣⎝ ⎠ ⎥⎦⎝ ⎠ 10. A(P, r, t) = Pert (a) A(500, 0.10, 5) = 500e(0.10)(5) = 500e0.5 ≈ 824.36 (b) A(1500, 0.12, 20) = 1500e(0.12)(20) = 1500e2.4 ≈ 16,534.76 11. ( , ) (2 3) y x f x y = ∫ t − dt (a) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 1 1, 2 2 3 3 2 2 0 f t dt t t = − = ⎡ − ⎤ ⎣ ⎦ = − − − = ∫ (b) ( ) ( ) ( ) ( ) 4 1 2 4 1 1, 4 2 3 3 4 2 6 f t dt t t = − = ⎡ − ⎤ ⎣ ⎦ = − − = ∫ 12. ( , ) 1 y x g x y dt t = ∫ (a) ( ) 1 1 4 4 g 4, 1 1 dt ln t ln 1 ln 4 ln 4 t = ∫ = ⎡⎣ ⎤⎦ = − = − (b) ( ) 3 6 3 6 6, 3 1 ln ln 3 ln 6 ln 1 2 ln 2 g dt t t = = ⎡⎣ ⎤⎦ = − = = − ∫ 13. f (x, y) = x2 − 2y (a) f (x + Δx, y) = (x + Δx)2 − 2y = x2 + 2x Δx + (Δx)2 − 2y (b) ( , ) ( , ) 2 2( ) ( 2 2 ) 2 2 2 2 2 2 2, 0 f x y y f x y x y y x y x y y x y y y y y y y + Δ − ⎡⎣ − + Δ ⎤⎦ − − − − Δ − + Δ = = = − = − Δ ≠ Δ Δ Δ Δ 14. f (x, y) = 3xy + y2 (a) f (x + Δx, y) = 3(x + Δx)y + y2 (b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 , , 3 3 3 3 2 3 3 2 3 2 , 0 f x y y f x y x y y y y xy y y y xy x y y y y y xy y y x y y y y x y y y y + Δ − ⎡ + Δ + + Δ ⎤ − + = ⎣ ⎦ Δ Δ + Δ + + Δ + Δ − − = Δ Δ + Δ + Δ = = + + Δ Δ ≠ Δ Section 7.3 Functions of Several Variables 431 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 15. f (x, y) = 16 − x2 − y2 The domain is the set of all points inside and on the circle x2 + y2 = 16 because 16 − x2 − y2 ≥ 0. The range is [0, 4]. 16. z = 4 − x2 − y2 The domain is the set of all points inside and on the circle x2 + y2 = 4 because 4 − x2 − y2 ≥ 0. The range is [0, 2]. 17. f (x, y) = x2 + y2 The domain is the set of all points in the xy-plane. The range is [0, ∞). 18. f (x, y) = x2 + y2 − 1 The domain is the set of all points in the xy-plane. The range is [−1, ∞). 19. f (x, y) = ex y The domain is the set of all points above or below the x-axis because y ≠ 0. The range is (0, ∞). 20. f (x, y) = ye1 x The domain is the set of all points in the xy-plane above or below the y-axis because x ≠ 0. The range is (−∞, ∞). 21. g(x, y) = ln(4 − x − y) The domain is the half-plane below the line y = − x + 4 because 4 − x − y > 0.The range is (−∞, ∞). 22. f (x, y) = ln(x + y) The domain is the half plane above the line y = −x because x + y > 0. The range is (−∞, ∞). 23. z = 9 − 3x2 − y2 The domain is the set of all points inside and on the ellipse 3x2 + y2 = 9 because 9 − 3x2 − y2 ≥ 0. The range is [0, 3]. 24. z = 4 − x2 − 4y2 The domain is the set of all points inside or on the ellipse x2 + 4y2 = 4 because 4 − x2 − 4y2 ≥ 0. The range is [0, 4]. 25. z y x = The domain is the set of all points in the xy-plane above or below the y-axis because x ≠ 0. The range is (−∞, ∞). 26. f (x, y) x y = The domain is the set of all points in the xy-plane above or below the x-axis because y ≠ 0. The range is (−∞, ∞). 27. f (x, y) 1 xy = The domain is the set of all points in the xy-plane except those on the x-axis and y-axis because x ≠ y ≠ 0. The range is all z ≠ 0, or (−∞, 0) ∪ (0, ∞). 28. g(x, y) 1 x y = − The domain is the set of all points in the xy-plane except those on the line y = x because x ≠ y. The range is all z ≠ 0, or (−∞, 0) ∪ (0, ∞). 29. h(x, y) = x y The domain is the set of all points in the xy-plane such that y ≥ 0. The range is (−∞, ∞). 30. f (x, y) = xy The domain is the set of all points in the xy-plane that lie in the first quadrant and the third quadrant, as well as the x-axis and y-axis because xy ≥ 0.The range is [0, ∞). 31. ( ) 2 , 2 4 f x y = x + y The contour map consists of ellipses 2 2 . 4 x + y = C Matches (b). 32. ( ) f x, y = e1− x2 + y2 The contour map consists of curves e1− x2 + y2 = C, or 1 − x2 + y2 = ln C, which are hyperbolas. Matches (d). 432 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 33. f (x, y) = e1− x2 − y2 The contour map consists of curves e1− x2 − y2 = C, or 1 − x2 − y2 = ln C ⇒ x2 + y2 = 1 − ln C, circles. Matches (a). 34. f (x, y) = ln y − x2 The contour map consists of curves ln y − x2 = C, or y − x2 = eC ⇒ y = x2 + eC , which are parabolas. Matches (c). 35. 1, 0, 2, 4, c c c c = − = = = 1 , 0 , 2 , 4 , x y x y x y x y − = + = + = + = + 1 2 4 y x y x y x y x = − − = − = − + = − + The level curves are parallel lines. 36. 0, 2, 4, 6, 8, 10, c c c c c c = = = = = = 0 6 2 3 , 2 6 2 3 , 4 6 2 3 , 6 6 2 3 , 8 6 2 3 , 10 6 2 3 , x y x y x y x y x y x y = − − = − − = − − = − − = − − = − − 2 3 6 2 3 4 2 3 2 2 3 0 2 3 2 2 3 4 x y x y x y x y x y x y + = + = + = + = + = − + = − The level curves are parallel lines. 37. c = 0, 0 = 25 − x2 − y2 , x2 + y2 = 25 c = 1, 1 = 25 − x2 − y2 , x2 + y2 = 24 c = 2, 2 = 25 − x2 − y2 , x2 + y2 = 21 c = 3, 3 = 25 − x2 − y2 , x2 + y2 = 16 c = 4, 4 = 25 − x2 − y2 , x2 + y2 = 9 c = 5, 5 = 25 − x2 − y2 , x2 + y2 = 0 The level curves are circles. 38. c = 0, 0 = x2 + y2 c = 2, 2 = x2 + y2 c = 4, 4 = x2 + y2 c = 6, 6 = x2 + y2 c = 8, 8 = x2 + y2 The level curves are circles. 39. 1, 2, 3, 4, 5, 6, c c c c c c = ± = ± = ± = ± = ± = ± 1 2 3 4 5 6 xy xy xy xy xy xy = ± = ± = ± = ± = ± = ± The level curves are hyperbolas. 40. 1 2 1 3 1 4 1, 2, 3, 4, , , , c c c c c c c = = = = = = = 1 2 1 3 1 4 1 , 0 2 , ln2 3 , ln3 4 , ln4 , ln2 , ln3 , ln4 xy xy xy xy xy xy xy e xy e xy e xy e xy e xy e xy e xy = = = = = = = = = − = = − = = − = The level curves are hyperbolas. −1 5 4 3 2 1 −1 1 2 3 4 5 x c = −1 c = 0 c = 2 c = 4 y x −2 3 c = 0 c = 2 c = 4 c = 6 c = 8 c = 10 y −2 −1 1 2 −2 1 2 c = 0 c = 1 c = 2 c = 5 c = 4 c = 3 x y 1 1 x c = 8 c = 6 c = 4 c = 2 c = 0 y x y c = 1 c = 2 c = 3 c = 4 c = 5 c = 6 c = −6 c = −5 c = −4 c = −3 c = −1 c = −2 −1 1 1 −1 1 −1 1 x c = 2 c = 3 c = 4 c = 12 c = 13 c = 14 y Section 7.3 Functions of Several Variables 433 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 41. 1, 2 c = 2 2 1 , 2 x x y = + (x − 1)2 + y2 = 1 1, 2 c = − 2 2 1 , 2 x x y − = + (x + 1)2 + y2 = 1 c = 1, 2 2 1 x , x y = + 2 1 2 1 2 4 ⎛⎜ x − ⎞⎟ + y = ⎝ ⎠ c = −1, 2 2 1 x , x y − = + 2 1 2 1 2 4 ⎛⎜ x + ⎞⎟ + y = ⎝ ⎠ 3, 2 c = 2 2 3 , 2 x x y = + 2 1 2 1 3 9 ⎛⎜ x − ⎞⎟ + y = ⎝ ⎠ 3, 2 c = − 2 2 3 , 2 x x y − = + 2 1 2 1 3 9 ⎛⎜ x + ⎞⎟ + y = ⎝ ⎠ c = 2, 2 2 2 x , x y = + 2 1 2 1 4 16 ⎛⎜ x − ⎞⎟ + y = ⎝ ⎠ c = −2, 2 2 2 x , x y − = + 2 1 2 1 4 16 ⎛⎜ x + ⎞⎟ + y = ⎝ ⎠ The level curves are circles. 42. c = 0, ln(x − y) = 0, x − y = 1 1 2c = , ( ) 1 2 ln x − y = , x − y = e1 2 1 2c = − , ( ) 1 2 ln x − y = − , x − y = e−1 2 c = 1, ln(x − y) = 1, x − y = e c = −1, ln(x − y) = −1, x − y = e−1 32 , c = ( ) 32 ln x − y = , x − y = e3 2 32 , c = − ( ) 32 ln x − y = − , x − y = e−3 2 c = 2, ln(x − y) = 2, x − y = e2 c = −2, ln(x − y) = −2, x − y = e−2 The level curves are lines. 43. f (x, y) = 100x0.75 y0.25 ( ) ( )0.75( )0.25 1500, 1000 100 1500 1000 135,540 units f = ≈ 44. From Example 4, f (x, y) = Cxa y1−a. (2 , 2 ) (2 ) (2 )1 2 21 1 2 21 1 2 1 2 ( , ) f x y C x a y a C a xa a y a a aCxa y a Cxa y a f x y = − = − − = − − = − = 45. ( ) ( ) ( ) ( ) ( ) ( 2 ) ( 2 ) P x1, x 2 = 50 x1 + x 2 − C1 x1 − C2 x 2 = 50 x1 + x 2 − 0.02x1 + 4x1 + 500 − 0.05x 2 + 4x2 + 275 (a) ( ) ( ) ( )2 ( ) ( )2 ( ) P 250, 150 = 50 250 + 150 − ⎡⎣0.02 250 + 4 250 + 500⎤⎦ − ⎡⎣0.05 150 + 4 150 + 275⎤⎦ = $15,250 (b) ( ) ( ) ( )2 ( ) ( )2 ( ) P 300, 200 = 50 300 + 200 − ⎡⎣0.02 300 + 4 300 + 500⎤⎦ − ⎡⎣0.05 200 + 4 200 + 275⎤⎦ = $18,425 (c) ( ) ( ) ( )2 ( ) ( )2 ( ) P 600, 400 = 50 600 + 400 − ⎡⎣0.02 600 + 4 600 + 500⎤⎦ − ⎡⎣0.05 400 + 4 400 + 275⎤⎦ = $30,025 46. w(x, y) 1 x y = − (a) (15, 10) 1 hr 1 hr 12 min 15 10 5 W = = = − (b) (12, 9) 1 hr 1 hr 20 min 12 9 3 W = = = − (c) (12, 6) 1 hr 1 hr 10 min 12 6 6 W = = = − (d) (4, 2) 1 hr 1 hr 30 min 4 2 2 W = = = − 47. ( ) ( ) 10 1 0.101 , 2000 1 R V I R I ⎡ + − ⎤ = ⎢ ⎥ ⎣ + ⎦ −2 2 x 3 2 3 2 2 −2 c = −2 c = −1 c = 2 c = 1 c = 1 2 1 2 c = c = − c y x 6 −4 −6 c = 2 c = 32 c = 12 c = 1 c = 0 c = 1 2 − c = 3 2 − c = −1 c = −2 y I R 0 0.03 0.05 0 $5187.48 $3859.98 $3184.67 0.28 $4008.46 $2982.67 $2460.85 0.35 $3754.27 $2793.53 $2304.80 434 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 48. A(r, t) = 5000ert 49. (a) C, highest pressure (b) A, lowest pressure (c) B, highest wind velocity 50. (a) The purple and blue areas represent the lowest levels of ozone. (b) No, the level curves are uneven and sporadically spaced. 51. z = 0.379x − 0.135y − 3.45 (a) 0.379(20) 0.135(10) 3.45 $2.78 earnings per share z = − − = (b) The x-variable Explanations will vary. Sample answer: The x-variable has a greater influence on the earnings per share because the absolute value of its coefficient is larger than the absolute value of the coefficient of the y-term. 52. z = 0.175x + 0.772y − 275 (a) 0.175(1000) 0.772(500) 275 $286 million z = + − = (b) The y-variable Explanations will vary. Sample answer: The y-variable has a greater influence on shareholders’ equity because the absolute value of its coefficient is greater than that of the coefficient of the x-term. 53. ( ) 12 12 1 1 1 12 t P r M r ⎛ ⋅ ⎞ ⎜ ⎟ = ⎝ ⎠ ⎡ ⎤ − ⎢ ⎥ ⎢⎣ + ⎥⎦ (a) ( ) ( ) 12(20) 120,000 0.08 12 $1003.73 1 1 1 0.0812 M ⎡ ⎤ ⎢ ⎥ = ⎣ ⎦ = ⎡ ⎤ − ⎢ ⎥ ⎢⎣ + ⎥⎦ $1003.73 × 240 payments = $240,895.20 (b) ( ) ( ) 12(30) 120,000 0.07 12 $798.36 1 1 1 0.0712 M ⎡ ⎤ ⎢ ⎥ = ⎣ ⎦ = ⎡ ⎤ − ⎢ ⎥ ⎢⎣ + ⎥⎦ $798.36 × 360 payments = $287,409.60 (c) ( ) ( ) 12(15) 120,000 0.07 12 $1078.59 1 1 1 0.0712 M ⎡ ⎤ ⎢ ⎥ = ⎣ ⎦ = ⎡ ⎤ − ⎢ ⎥ ⎢⎣ + ⎥⎦ $1078.59 × 180 payments = $194,146.20 Choices will vary, as well as explanations. Section 7.4 Partial Derivatives Skills Warm Up 1. ( ) ( ) ( ) ( ) 2 2 1 2 2 3 1 3 2 2 3 f x x f x x x x x − = + ′ = + = + 2. ( ) ( ) ( ) ( ) ( ) ( ) 2 3 2 2 2 2 3 3 3 2 6 3 g x x g x x x x x = − ′ = − − = − − 3. ( ) ( ) ( ) ( ) ( ) 2 1 2 1 2 2 1 1 2 1 2 1 t t t t g t te gt te e e t + + + + = ′ = + = + Number of years Rate 5 10 15 20 0.02 $5525.85 $6107.01 $6749.29 $7459.12 0.04 $6107.01 $7459.12 $9110.59 $11,127.70 0.06 $6749.29 $9110.59 $12,298.02 $16,600.58 0.08 $7459.12 $11,127.70 $16,600.58 $24,765.16 Section 7.4 Partial Derivatives 435 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. 3 5 z xz y ∂ = ∂ ∂ = ∂ 2. 2 2 z x xz y ∂ = ∂ ∂ = − ∂ 3. ( ) ( ) , 3 , 12 x y f x y f x y y = = − 4. ( ) ( ) , 1 , 6 x y f x y f x y y = = Skills Warm Up —continued— 4. f (x) = e2x 1 − e2x ( ) ( ) ( )( ) ( ) ( ) ( ( )) ( ) 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 11 2 1 2 2 2 1 21 2 3 1 1 1 x x x x x x x x x x x x x x x x x f x e e e e e e e e e e e e e e e e e ′ = ⎛ ⎞ − − − + − ⎜ ⎟ ⎝ ⎠ − = + − = − + − = − − − − 5. ( ) ( ) ( ) ln 3 2 2 3 2 fx x f x x = − ′ − = − 6. ( ) ( ) ( ) ( ) ( ) ( ) 3 3 1 2 2 3 2 2 ln 6 1 1 6 3 6 6 2 3 2 2 6 u t t t u t t t t t t t t t − = − ′ = ⎛ ⎞ − − ⎜ ⎟ − ⎝ ⎠ − = − 7. ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( ) ( ) 2 2 2 2 4 2 3 3 5 4 1 4 1 10 5 2 4 1 4 4 1 4 1 10 40 4 1 10 4 1 g x x x x x x x g x x x x x x x x = − − − − ′ = − − − = − = − − 8. ( ) ( ) ( ) 3 2 2 2 9 x f x x + = − ( ) ( ) ( )( ) ( ) ( )( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 3 2 2 4 2 2 3 2 2 2 2 2 3 2 3 2 3 9 3 2 2 2 9 2 9 3 2 9 4 2 2 3 9 4 2 2 8 27 9 9 9 x x x x x f x x x x x x x x x x x x x x x x − + − + − ′ = − + − − + + ⎡ − − + ⎤ + + + = = ⎣ ⎦ = − − − − 9. f (x) = x2ex−2 f ′(x) = x2ex−2 + ex−2 (2x) f ′(2) = (2)2e2−2 + e2−2 (2(2)) = 4 + 4 = 8 10. g(x) = x x2 − x + 2 ( ) ( ) ( ) 1 2 2 2 2 2 2 1 2 2 1 2 2 2 2 2 g x x x x x x x x x x x x x ′ = ⎛ ⎞ − + − − + − + = − + − + ⎜ ⎟ ⎝ ⎠ − + ( ) 2 2 2 2 2 2 2 2 2 2 3 2 7 2 2 2 2 2 g ′ − = + − + = + = − + 436 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 5. ( ) ( ) 2 2 , 1 , x y f x y y f x y xy x y − = = − = − 6. 1 1 2 2 2 z y x z xy x y y − ∂ = ∂ ∂ = = ∂ 7. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 2 2 2 1 2 2 2 , 1 2 2 , 1 2 2 x y f x y x y x x x y f x y x y y y x y − − = + = + = + = + 8. ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 x , x y y xy x y y x f x y x y x y + − − = = + + ( ) ( )( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 y , x y x xy y xx y f x y x y x y + − − = = + + 9. 2 2 2 2 2 y y z xe xz x e y ∂ = ∂ ∂ = ∂ 10. x y x y ( 1) x y x y z xe e x e xz xe y + + + + ∂ = + = + ∂ ∂ = ∂ 11. ( ) ( ) ( ) ( ) 2 2 2 2 , 2 , 2 x y x x y y h x y xe h x y ye − + − + = − = − 12. ( ) ( ) 2 , 1 , x y x x y y g x y e y g x y x e y = = − 13. ( ) ( ) 2 2 2 2 ln ln ln 1 1 2 1 1 2 z x y x y x y x y z y x x y x y x y z x y x y x y x y + = = + − − − ∂ = − =− ∂ + − − ∂ = + = ∂ + − − 14. ( ) ( ) 2 2 2 2 , 2 , 2 x y g x y x x y g x y y x y = + = + 15. fx (x, y) = 6x + y, fx (2, 1) = 13 fy (x, y) = x − 2y, fy (2, 1) = 0 16. fx (x, y) = 2x − 3y, fx (1, −1) = 5 fy (x, y) = −3x + 2y, fy (1, −1) = −5 17. ( , ) 3 3xy , fx x y = ye fx (0, 4) = 12 ( , ) 3 3xy , fy x y = xe fy (0, 4) = 0 18. ( , ) x 2 , fx x y = e y fx (0, 2) = 4 ( , ) 2 x , fy x y = e y fy (0, 2) = 4 19. ( ) ( ) ( ) ( ) ( ) 2 2 2 1 x , , x y y xy y f x y x y x y − − = =− − − (2, 2) 1 4 fx − = − ( ) ( ) ( ) ( ) ( ) 2 2 2 1 y , , x y x xy x f x y x y x y − − − = = − − (2, 2) 1 4 fy − = 20. ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 2 2 2 2 1 2 2 2 2 3 2 2 2 2 3 2 2 2 3 2 4 4 1 2 2 4 4 4 x , , x y y xy x y x x y x x y y f x y x y x y x y + − ⎛ ⎞ + − ⎜ ⎟ + − = ⎝ ⎠ = = + + + fx (1, 0) = 0 ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 2 2 2 2 1 2 2 2 2 3 2 2 2 2 3 2 2 2 3 2 4 4 1 2 2 4 4 4 y , , x y x xy x y y x y x xy x f x y x y x y x y + − ⎛ ⎞ + − ⎜ ⎟ + − = ⎝ ⎠ = = + + + fy (1, 0) = 4 21. ( , ) 3 , 3 5 fx x y x y = + fx (1, 0) = 1 ( , ) 5 , 3 5 fy x y x y = + (1, 0) 5 3 fy = 22. ( , ) ln( )1 2 1 ln 1 ln 2 2 f x y = xy = x + y ( , ) 1 , 2 fx x y x = ( 1, 1) 1 2 fx − − = − ( , ) 1 , 2 fy x y y = ( 1, 1) 1 2 fy − − = − Section 7.4 Partial Derivatives 437 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 23. (a) z y x ∂ = ∂ z (1, 2, 2) 2 x ∂ = ∂ (b) z x y ∂ = ∂ z (1, 2, 2) 1 y ∂ = ∂ 24. (a) 25 2 2 z x x x y ∂ = − ∂ − − (3, 0, 4) 3 4 z x ∂ = − ∂ (b) 25 2 2 z y y x y ∂ = − ∂ − − z (3, 0, 4) 0 y ∂ = ∂ 25. (a) z 2x x ∂ = − ∂ z (1, 1, 2) 2 x ∂ = − ∂ (b) z 2y y ∂ = − ∂ z (1, 1, 2) 2 y ∂ = − ∂ 26. (a) z 2x x ∂ = ∂ z ( 2, 1, 3) 4 x ∂ − = − ∂ (b) z 2y y ∂ = − ∂ z ( 2, 1, 3) 2 y ∂ − = − ∂ 27. wx = y2z4 , wy = 2xyz4 , wz = 4xy2z3 28. wx = 3x2 yz2 , wy = x3z2 , wz = 2x3yz 29. ( )( ) ( ) ( )2 ( )2 0 2 1 2 x x y z z w x y x y + − = =− + + ( )( ) ( ) ( )2 ( )2 0 2 1 2 y x y z z w x y x y + − = =− + + 2 wz x y = + 30. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 1 1 0 1 x y z x y z y xy y yz w x y z x y z x y z x xy x xz w x y z x y z x y z xy xy w x y z x y z + + − + = = + + + + + + − + = = + + + + + + − = =− + + + + 31. wx = 2z2 + 3yz, wx (1, −1, 2) = 2 wy = 3xz − 12yz, wy (1, −1, 2) = 30 wz = 4xz + 3xy − 6y2 , wz (1, −1, 2) = −1 32. wx = 6xy − 5yz, wx (3, 4, −2) = 112 wy = 3x2 − 5xz + 10z2 , wy (3, 4, −2) = 97 wz = −5xy + 20yz, wz (3, 4, −2) = −220 33. 2 2 2 , x w x x y z = + + (2, 1, 2) 2 3 wx − = 2 2 2 , y w y x y z = + + (2, 1, 2) 1 3 wy − = − 2 2 2 , z w z x y z = + + (2, 1, 2) 2 3 wz − = 34. 2 2 2 3 , 3 2 x w x x y z = + − (1, 2, 1) 3 5 5 wx − = 2 2 2 , 3 2 y w y x y z = + − (1, 2, 1) 2 5 5 wy − = − 2 2 2 2 , 3 2 z w z x y z − = + − (1, 2, 1) 2 5 5 wz − − = 35. 4 3 2 2x2 , wx = xy z e ( ) 1 2wx , −1, 2 = −8 e 3 2 2 2x2 , wy = y z e ( ) 1 2wy , −1, 2 = 12 e 2 3 2x2 , wz = y ze ( ) 1 2wz , −1, 2 = −4 e 36. 2 z , wx = ye wx (2, 1, 0) = 1 2 z , wy = xe wy (2, 1, 0) = 2 2 2 z , wz = xyze wz (2, 1, 0) = 0 37. 3 5 , 5 2 3 wx x y z = + − (4, 1, 1) 1 5 wx − = 2 3 6 , 5 2 3 y w y x y z = + − (4, 1, 1) 6 25 wy − = 3 3 , 5 2 3 wz x y z = − + − (4, 1, 1) 3 25 wz − = − 38. ln 2 2 2 1 ln( 2 2 2 ) 2 w = x + y + z = x + y + z 2 2 2 , x w x x y z = + + (3, 0, 4) 3 25 wx = 2 2 2 , y w y x y z = + + wy (3, 0, 4) = 0 2 2 2 , z w z x y z = + + (3, 0, 4) 4 25 wz = 39. ( ) ( ) , 2 4 4 0 , 4 2 16 0 x y f x y x y f x y x y = + − = ⇒ = + + = ⇒ 4 8 8 4 2 16 6 24 4 6 x y x y y y x − − = − + = − − = − = = − Solution: (−6, 4) 438 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 40. ( , ) 9 2 12 0 3 2 4 fx x y = x − y = ⇒ y = x fy (x, y) = −12x + 3y2 = 0 ⇒ ( ) 2 2 4 3 12 3 3 0 4 12 27 0 16 3 9 64 0 16 x x x x x x ⎛ ⎞ − + ⎜ ⎟ = ⎝ ⎠ − + = − = x = 0 or 4 3 3 3 x = y = 0 or 4 3 9 3 y = Solutions: ( ) 4 3 3 4 3 9 0, 0 , , 3 3 ⎛ ⎞ ⎜⎜ ⎟⎟ ⎝ ⎠ 41. ( ) ( ) 2 2 2 2 , 1 0 1 , 1 0 1 x y f x y y x y x f x y x y x y = − + = ⇒ = ⎫⎪⎪⎬⎪ = − + = ⇒ = ⎪⎭ x = y = 1 Solution: (1, 1) 42. ( ) ( ) 2 2 2 2 , 2 0 0 1 , 2 0 0 1 x y f x y x x x y f x y y y x y = = ⇒ = + + = = ⇒ = + + Solution: (0, 0) 43. 3 2 8 z x xz y y ∂ = ∂ ∂ = − ∂ 2 2 2 2 2 2 6 0 0 8 z x x z x y z y x z y ∂ = ∂ ∂ = ∂ ∂ ∂ = ∂ ∂ ∂ = − ∂ 44. 4 4 5 z x xz y y ∂ = ∂ ∂ = ∂ 2 2 2 2 3 2 2 4 0 20 0 z x z x y z y y z y x ∂ = ∂ ∂ = ∂ ∂ ∂ = ∂ ∂ = ∂ ∂ 45. 2 2 2 6 z x y xz x y y ∂ = − ∂ ∂ = − + ∂ 2 2 2 2 2 2 2 2 2 6 z x z x y z y x z y ∂ = ∂ ∂ = − ∂ ∂ ∂ = − ∂ ∂ ∂ = ∂ 46. 2 2 4 3 8 z y xz y xy y ∂ = − ∂ ∂ = − ∂ 2 2 2 2 2 2 0 8 8 6 8 z x z y x y z x y x z y x y ∂ = ∂ ∂ = − ∂ ∂ ∂ = − ∂ ∂ ∂ = − ∂ 47. ( ) ( ) ( ) ( ) ( ) ( ) 4 3 2 3 3 4 3 2 4 3 2 2 2 4 3 2 3 3 2 12 36 3 2 3 3 2 6 18 3 2 z x y x x x y xz x y y y x y y ∂ = − = − ∂ ∂ = − − = − − ∂ ( )( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )( ) 2 2 3 4 3 3 4 3 2 2 4 3 4 3 2 2 3 4 3 2 32 4 3 2 2 2 4 3 2 4 3 4 3 4 3 2 2 2 4 3 3 36 2 3 2 12 3 2 108 108 3 2 11 2 36 2 3 2 6 432 3 2 18 2 3 2 6 3 2 36 36 3 2 3 8 18 2 3 2 12 432 z x x y x x y x x x y x y x z x x y y xy x y y y z y x y y x y y y x y x y y z y x y x y x ∂ ⎡ ⎤ = ⎣ − ⎦ + − = − − ∂ ∂ ⎡ ⎤ = ⎣ − − ⎦ = − − ∂ ∂ ∂ ⎡ ⎤ = − ⎣ − − ⎦ + − − = − − − ∂ ∂ ⎡ ⎤ = − ⎣ − ⎦ = − ∂ ∂ x3y2 (3x4 − 2y3) Section 7.4 Partial Derivatives 439 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 48. ( ) ( ) ( ) ( ) 2 2 1 2 2 2 2 2 1 2 2 2 19 2 2 9 19 2 2 9 z x y x x x x y z x y y y y x y − − ∂ = − − − = − ∂ − − ∂ = − − − = − ∂ − − ( ) ( ) ( ) ( ) 2 2 2 2 1 2 2 2 2 2 2 2 2 3 2 9 1 1 9 2 2 9 9 9 x y x x y x z y x x y x y − − − ⎛ ⎞ − − − − ∂ ⎜ ⎟ − = − ⎝ ⎠ = ∂ − − − − ( ) ( ) ( ) ( ) 2 2 2 2 1 2 2 2 2 2 2 3 2 9 0 1 9 2 2 9 9 x y x x y y z xy x y x y x y − − − ⎛ ⎞ − − − − ∂ ⎜ ⎟ = − ⎝ ⎠ = − ∂ ∂ − − − − ( ) ( ) ( ) ( ) 2 2 2 2 1 2 2 2 2 2 2 3 2 9 0 1 9 2 2 9 9 x y y x y x z xy y x x y x y − − − ⎛ ⎞ − − − − ∂ ⎜ ⎟ = − ⎝ ⎠ = − ∂ ∂ − − − − ( ) ( ) ( ) ( ) 2 2 2 2 1 2 2 2 2 2 2 2 2 3 2 9 1 1 9 2 2 9 9 9 x y y x y y z x y x y x y − − − ⎛ ⎞ − − − − ∂ ⎜ ⎟ − = − ⎝ ⎠ = ∂ − − − − 49. ( ) ( )( ) ( ) ( )( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 2 2 2 2 4 2 z xy x x y y x y x xy x y z xy y x y x x y y xy xy ∂ − − + = = ∂ ∂ − − − + = =− ∂ ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) 2 2 2 2 2 42 3 2 2 2 2 2 2 2 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 24 3 2 2 4 4 2 2 2 4 2 2 2 2 4 2 2 2 4 4 z x y x x y xy y x xy x z x y y x y x x y x y x y x y z xy x x y y x y y x x y x y z xy y x y xy x y xy y ∂ − + = =− ∂ ∂ − + − = =− ∂ ∂ ∂ − + − = − = − ∂ ∂ ∂ − + = − = ∂ 50. ( )( ) ( ) ( ) ( ) ( ) 2 2 2 z x y 1 x 1 y x x y x y z x y x y ∂ + − = = ∂ + + ∂ = − ∂ + ( ) ( ) () ( )( )() ( ) ( ) ( ) () ( )( )( ) ( ) ( ) ( ) 2 2 3 2 2 4 3 2 2 4 3 2 2 3 2 1 2 1 1 2 1 2 z y x x y z x y y x y x y x y x y x y z x y x x y x y y x x y x y z x y x y ∂ = − ∂ + ∂ + − + − = = ∂ ∂ + + ∂ + − + − = − = ∂ ∂ + + ∂ = ∂ + 51. fx (x, y) = 4x3 − 6xy2 , fy (x, y) = −6x2 y + 2y fxx (x, y) = 12x2 − 6y2 , fxx (1, 0) = 12 fxy (x, y) = −12xy, fxy (1, 0) = 0 fyx (x, y) = −12xy, fyx (1, 0) = 0 fyy (x, y) = −6x2 + 2, fyy (1, 0) = −4 52. fx (x, y) = 3x2 + 2y3, fy (x, y) = 6xy2 − 3 fxx (x, y) = 6x, fxx (3, 2) = 18 fxy (x, y) = 6y2 , fxy (3, 2) = 24 fyx (x, y) = 6y2 , fyx (3, 2) = 24 fyy (x, y) = 12xy, fyy (3, 2) = 72 53. ( ) , 2 3 x2 , fx x y = xy e ( ) , 3 2 x2 fy x y = y e ( , ) x2 (4 2 3 2 3), fxx x y = e x y + y fxx (1, −1) = −6e ( ) , 6 2 x2 , fxy x y = xy e fxy (1, −1) = 6e ( ) , 6 2 x2 , fyx x y = xy e fyx (1, −1) = 6e ( ) 2 , 6 x , fyy x y = ye fyy (1, −1) = −6e 440 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 54. ( , ) 2 y , fx x y = xe ( ) , 2 y fy x y = x e ( , ) 2 y , fxx x y = e fxx (−1, 0) = 2 ( , ) 2 y , fxy x y = xe fxy (−1, 0) = −2 ( , ) 2 y , fyx x y = xe fyx (−1, 0) = −2 ( , ) 2 y , fyy x y = x e fyy (−1, 0) = 1 55. wx = 2x − 3y, wy = −3x + 4z, wz = 4y + 3z2 wxx = 2, wyx = −3, wzx = 0 wxy = −3, wyy = 0, wzy = 4 wxz = 0, wyz = 4, wzz = 6z 56. wx = 2xy3 + 2yz, wy = 3x2 y2 + 2xz − 3z, wz = 2xy − 3y, wxx = 2y3, wyx = 6xy2 + 2z, wzx = 2y, wxy = 6xy2 + 2z, wyy = 6x2 y, wzy = 2x − 3, wxz = 2y, wyz = 2x − 3, wzz = 0, 57. ( )( ) () ( )2 ( )2 4 4 1 4 x x y z xz yz w x y x y + − = = + + ( )( ) () ( )2 ( )2 0 4 1 4 y x y xz xz w x y x y + − − = = + + 4 z w x x y = + ( ) () ( ) 3 3 4 2 1 8 xx w yz x y yz x y = ⎡− + − ⎤ = − ⎣ ⎦ + ( ) () ( ) ( ) ( ) ( ) 3 2 3 4 xy 4 2 1 4 z x y w yz x y x y z x y = ⎡− + − ⎤ + + − = − ⎣ ⎦ + ( )2 4 xz w y x y = + ( ) () ( ) ( ) ( ) ( ) 3 2 3 4 yx 4 2 1 4 z x y w xz x y x y z x y = − ⎡− + − ⎤ + + − − = − ⎣ ⎦ + ( ) () ( ) 3 3 4 2 1 8 yy w xz x y xz x y = − ⎡− + − ⎤ = ⎣ ⎦ + ( )2 4 yz w x x y = − + ( )( ) () ( )2 ( )2 4 4 1 4 zx x y x y w x y x y + − = = + + ( )( ) ( ) ( ) 2 2 4 1 1 4 zy w x y x x y = ⎡ − + − ⎤ = − ⎣ ⎦ + wzz = 0 Section 7.4 Partial Derivatives 441 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 58. ( ) ( ) ( ) ( ) 2 2 2 1 x x y z y xy w y yz x y z x y z + + − + = = + + + + ( ) ( ) ( ) ( ) 2 2 2 1 y x y z x xy x xz w x y z x y z + + − + = = + + + + ( )( ) ( ) ( )2 ( )2 0 1 z x y z xy xy w x y z x y z + + − − = = + + + + ( )( )( ) ( ) ( ) ( ) 2 2 3 3 2 xx 2 1 y yz w y yz x y z x y z − + = + ⎡ − + + ⎤ = − ⎣ ⎦ + + ( )( )( ) ( ) ( ) ( ) ( ) 2 2 3 2 3 2 1 2 2 xy w y yz x y z x y z y z xy xz yz z x y z = + ⎡ − + + − ⎤ + + + − + = + + + ⎣ ⎦ + + ( )( )( ) ( ) ( ) ( ) ( ) 2 2 3 2 3 xz 2 1 w y yz x y z x y z y xy yz y x y z = + ⎡ − + + − ⎤ + + + − = − − ⎣ ⎦ + + ( )( )( ) ( ) ( ) ( ) ( ) 2 2 3 2 3 2 1 2 2 yx w x xz x y z x y z x z xy xz yz z x y z = + ⎡ − + + − ⎤ + + + − + = + + + ⎣ ⎦ + + ( )( )( ) ( ) ( ) ( ) 2 2 3 3 2 yy 2 1 x xz w x xz x y z x y z − − + = + ⎡ − + + ⎤ = ⎣ ⎦ + + ( )( )( ) ( ) ( ) ( ) ( ) 2 2 3 2 3 yz 2 1 w x xz x y z x y z x xy x xz x y z = + ⎡ − + + − ⎤ + + + − = − − ⎣ ⎦ + + ( ) ( )( ) ( ) ( ) ( ) ( ) 2 3 2 3 zx 2 1 xy y yz w xy x y z x y z y x y z = − ⎡ − + + − ⎤ + + + − − = − − ⎣ ⎦ + + ( ) ( )( ) ( ) ( ) ( ) ( ) 2 3 2 3 zy 2 1 xy x xz w xy x y z x y z x x y z = − ⎡ − + + − ⎤ + + + − − = − − ⎣ ⎦ + + ( ) ( )( ) ( ) ( ) 3 3 2 1 2 zz w xy x y z xy x y z = − ⎡ − + + − ⎤ = ⎣ ⎦ + + 59. (a) C 5y 149, x xy ∂ = + ∂ C(120, 160) 154.77 x ∂ ≈ ∂ C 5x 189, y xy ∂ = + ∂ C(120, 160) 193.33 y ∂ ≈ ∂ (b) Increasing the production of racing bikes increases the cost at a higher rate than increasing the production of mountain bikes. This is determined by comparing the marginal cost for mountain bikes C $154.77 x ∂ = ∂ to that for racing bikes C $193.33 y ∂ = ∂ at the production level (120, 160). 60. (a) 1 2 1 R 200 8x 8x x ∂ = − − ∂ When x1 = 4 and x2 = 12, 1 R 72. x ∂ = ∂ (b) 1 2 2 R 200 8x 8x x ∂ = − − ∂ When x1 = 4 and x2 = 12, 1 2 R 72. x ∂ = ∂ 61. (a) 0.3 f 140x 0.3y0.3 140 y x x ∂ = − = ⎛⎜ ⎞⎟ ∂ ⎝ ⎠ When x = 1000 and y = 500, 0.3 0.3 140 500 140 1 113.72. 1000 2 f x ∂ ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ = ⎜ ⎟ ≈ ∂ ⎝ ⎠ ⎝ ⎠ (b) 0.7 f 60x0.7 y 0.7 60 x y y ∂ − ⎛ ⎞ = = ⎜ ⎟ ∂ ⎝ ⎠ When x = 1000 and y = 500, ( ) 0.7 60 1000 60 2 0.7 97.47. 500 f y ∂ ⎛ ⎞ = ⎜ ⎟ = ≈ ∂ ⎝ ⎠ 442 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 62. (a) 0.25 f 75x 0.25 y0.25 75 y x x ∂ = − = ⎛⎜ ⎞⎟ ∂ ⎝ ⎠ When x = 1000 and y = 500, 0.25 0.25 75 500 75 1 63.07. 1000 2 f x ∂ ⎛ ⎞ ⎛ ⎞ = ⎜ ⎟ = ⎜ ⎟ ≈ ∂ ⎝ ⎠ ⎝ ⎠ (b) 0.75 f 25x0.75 y 0.75 25 x y y ∂ − ⎛ ⎞ = = ⎜ ⎟ ∂ ⎝ ⎠ When x = 1000 and y = 500, ( ) 0.75 25 1000 25 2 0.25 42.04. 500 f y ∂ ⎛ ⎞ = ⎜ ⎟ = ≈ ∂ ⎝ ⎠ 63. Complementary because 1 2 5 0 2 x p ∂ = − < ∂ and 2 1 3 0. 2 x p ∂ = − < ∂ 64. Substitute because 1 2 x 1.8 0 p ∂ = > ∂ and 2 1 x 0.75 0. p ∂ = > ∂ 65. z = 0.62x − 0.41y + 0.38 (a) z 0.62; x ∂ = ∂ z 0.41 y ∂ = − ∂ (b) For every increase of $1 billion in expenditures on amusement parks and campgrounds, the expenditures for spectator sports will increase by $0.62 billion. For every increase of $1 billion in expenditures on live entertainment (excluding sports), the expenditures for spectator sports will decrease by $0.41 billion. 66. z = 0.175x − 0.772y − 275 (a) z 0.175; x ∂ = ∂ z 0.772 y ∂ = − ∂ (b) For every increase of $1 million in sales for Skechers, the shareholder’s equity for Skechers will increase by $0.175 million. For every increase of $1 million in total assets for Skechers, the shareholder’s equity will decrease by $0.772 million. 67. ( ) ( ) ( ) 2 2 100 100 12, 10 100 10 10 100 12 12, 10 12 10 M C M C IQ C IQ M C IQ IQ = − = = = − = = − For a child who has a current mental age of 12 years and a chronological age of 10 years, the IQ is increasing at a rate of 10 IQ points for every increase of 1 year in the child’s mental age. For a child who has a current mental age of 12 years and a chronological age of 10 years, the IQ is decreasing at a rate of 12 IQ points for every increase of 1 year in the child’s chronological age. 68. (a) fx (4, 1) < 0 (b) fy (4, 1) > 0 (c) fx (−1, −2) < 0 (d) fy (−1, −2) < 0 69. ( ) ( ) 10 1 0.101 , 1000 1 R V I R I ⎡ + − ⎤ = ⎢ ⎥ ⎣ + ⎦ ( ) ( ) ( ) ( ) ( ) ( ) 9 10 2 11 1 0.10 1 1 0.10 1 1 0.10 1 , 10,000 10,000 1 1 1 I R R R V I R I I I ⎡ + − ⎤ ⎡ + − ⎤ ⎡⎣ + − ⎤⎦ = ⎢ ⎥ ⎢− ⎥ = − ⎣ + ⎦ ⎢⎣ + ⎥⎦ + VI (0.03, 0.28) ≈ −14,478.99 ( ) ( ) ( ) ( ) ( ) 9 9 10 1 0.10 1 0.10 1 0.10 1 , 10,000 1000 1 1 1 0.03, 0.28 1391.17 R R R R V I R I I I V ⎡ + − ⎤ ⎡ ⎤ ⎡⎣ + − ⎤⎦ = ⎢ ⎥ ⎢− ⎥ = − ⎣ + ⎦ ⎣ + ⎦ + ≈ − The rate of inflation has the greater negative influence on the growth of the investment because −14,478.99 > −1391.17 . 70. Since both first partials are negative, an increase in the charge for food and housing or tuition will cause a decrease in the number of applicants. Section 7.5 Extrema of Functions of Two Variables 443 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 71. (a) Ux = −10x + y (b) Uy = x − 6y (c) When x = 2 and y = 3, Ux = −17 and Uy = −16. The person should consume one more unit of good y, since the rate of decrease of satisfaction is less for y. The slope of U in the x-direction is 0 when y = 10x and negative when y < 10x. The slope of U in the y-direction is 0 when x = 6y and negative when x < 6y. 72. Answers will vary. Section 7.5 Extrema of Functions of Two Variables Skills Warm Up 1. 5 15 3 2 5 x x y = ⎧⎨ ⎩ − = 5 15 3 x x = = Substitute in the other equation. 3(3) 2 5 2 4 2 y y y − = − = − = The solution is (3, 2). 2. 1 2 3 5 19 y x y ⎧ = ⎪⎨ ⎩⎪− + = 1 2 3 6 y y = = Substitute in the other equation. 5(6) 19 11 11 x x x − + = − = − = The solution is (11, 6). 3. 5 3 x y x y + = ⎧⎨ ⎩ − = − Adding the two equations gives 2x = 2, so x = 1. Substitute. 1 5 4 y y + = = The solution is (1, 4). 4. 8 2 4 x y x y + = ⎧⎨ ⎩ − = Adding the two equations gives 3x = 12, so x = 4. Substitute. 4 8 4 y y + = = The solution is (4, 4). 5. 2 8 3 4 7 x y x y − = ⎧⎨ ⎩ − = Equation 1 Equation 2 Multiply Equation 1 by −4: −8x + 4y = −32 Add the new equation to Equation 2: −5x = −25 Simplify: x = 5 Substitute 5 for x in Equation 1: 2(5) − y = 8 Solve for y: y = 2 The solution is (5, 2). 6. 2 4 14 3 7 x y x y − = ⎧⎨ ⎩ + = Equation 1 Equation 2 Multiply equation 2 by 4: 12x + 4y = 28 Add new equation to Equation 1: 14x = 42 Simplify: x = 3 Substitute 3 for x in Equation 2: 3(3) + y = 7 Simplify: y = −2 The solution is (3, −2). 7. 2 0 2 0 x x yx y ⎧ + = ⎪⎨ ⎩⎪ + = Equation 1 Equation 2 Factor Equation 1: x(x + 1) = 0 Solve equation 1 for x: x = −1 or x = 0 Substitute −1 for x in Equation 2: 2y(−1) + y = 0 Solve for y: y = 0 Substitute 0 for x in Equation 2: 2y(0) + y = 0 Solve for y: y = 0 The solutions are (−1, 0) and (0, 0). y x z 444 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Skills Warm Up —continued— 8. 3 2 6 0 2 0 y y xy x ⎧ + = ⎪⎨ ⎩⎪ + + = Equation 1 Equation 2 Factor Equation 1: y(3y + 6) = 0 Solve for y: 0 3 6 0 2 y y y = + = = − Substitute 0 for y in Equation 2: x(0) + x + 2 = 0 Solve for x: x = −2 Substitute −2 for y in Equation 2: x(−2) + x + 2 = 0 Solve for x: x = 2 The solutions are (−2, 0) and (2, −2). 9. z = 4x3 − 3y2 z 12x2 , x ∂ = ∂ z 6y y ∂ = − ∂ 2 2 z 24x, x ∂ = ∂ 2 2 z 6 y ∂ = − ∂ 2 z 0, x y ∂ = ∂ ∂ 2 z 0 y x ∂ = ∂ ∂ 10. z = 2x5 − y3 z 10x4 , x ∂ = ∂ z 3y2 y ∂ = − ∂ 2 3 2 z 40x , x ∂ = ∂ 2 2 z 6y y ∂ = − ∂ 2 z 0, x y ∂ = ∂ ∂ 2 z 0 y x ∂ = ∂ ∂ 11. z = x4 − xy + 2y 4 3 4 3 2 2 z x y x xy x xy x ∂ = − = − ∂ 2 2 2 2 z x xy x xy y ∂ = − + = − + ∂ 2 2 2 2 2 2 2 12 12 4 4 4 z x y xy x xy x x xy x x ∂ = − + = + ∂ 2 2 4 2 2 4 2 z x xy xy y y xy y y ∂ = − + = ∂ 2 4 4 z x xy x y x xy xy ∂ = − = − ∂ ∂ 2 4 4 z x xy y y x xy xy ∂ = − = − ∂ ∂ 12. z = 2x2 − 3xy + y2 z 4x 3y, x ∂ = − ∂ z 3x 2y y ∂ = − + ∂ 2 2 z 4, x ∂ = ∂ 2 z 2 y ∂ = ∂ 2 z 3, x y ∂ = − ∂ ∂ 2 z 3 y x ∂ = − ∂ ∂ 13. z = yexy2 z y(y2 )exy2 y3exy2 x ∂ = = ∂ ( ) ( ) z y 2xy exy2 exy2 1 2xy2exy2 exy2 y ∂ = + = + ∂ ( ) 2 2 2 3 2 5 2 z y y exy y exy x ∂ = = ∂ ( ) ( ) 2 2 2 2 2 2 2 3 2 2 2 2 4 2 4 6 xy xy xy xy xy z xy xye e xy xye y x y e xye ∂ = + + ∂ = + 2 2 2 z 2xy4exy 3y2exy x y ∂ = + ∂ ∂ ( ) 2 2 2 2 2 z y3 2xy exy exy 3y2 2xy4exy 3y2exy y x ∂ = + = + ∂ ∂ 14. z = xexy z xyexy exy exy (xy 1) x ∂ = + = + ∂ z x2exy y ∂ = ∂ ( ) ( ) ( ) 2 2 z exy y xy 1 yexy yexy xy 2 x ∂ = + + = + ∂ 2 3 2 z x exy y ∂ = ∂ ( ) ( ) ( ) 2 z exy x xy 1 xexy xexy xy 2 x y ∂ = + + = + ∂ ∂ ( ) ( ) 2 z x2 yexy exy 2x xexy xy 2 y x ∂ = + = + ∂ ∂ Section 7.5 Extrema of Functions of Two Variables 445 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. f (x, y) = x2 − y2 + 4x − 8y − 11 The first partial derivatives of f , fx (x, y) = 2x + 4 and fy (x, y) = −2y − 8, are zero at the critical point (−2, −4). Because fxx (x, y) = 2, fyy (x, y) = −2, and fxy (x, y) = 0, it follows that fxx (−2, −4) > 0 and ( ) ( ) ( ) 2 fxx −2, −4 fyy −2, −4 − ⎡⎣ fxy −2, −4 ⎤⎦ = −4 < 0. So, (−2, −4, 1) is a saddle point. There are no relative extrema. 2. f (x, y) = x2 + y2 + 2x − 6y + 6 The first partial derivatives of f , fx (x, y) = 2(x + 1) and fy (x, y) = 2(y − 3), are zero at the critical point (−1, 3). Because fxx (x, y) = 2, fyy (x, y) = 2, and fxy (x, y) = 0, it follows that fxx (−1, 3) > 0 and ( ) ( ) ( ) 2 fxx −1, 3 fyy −1, 3 − ⎡⎣ fxy −1, 3 ⎤⎦ = 4 > 0. So, (−1, 3, −4) is a relative minimum. 3. f (x, y) = x2 + y2 + 1 The first partial derivatives of f, ( ) 2 2 , 1 x f x y x x y = + + and ( ) 2 2 , , 1 y f x y y x y = + + are zero at the critical point (0, 0). Because ( ) 2 2 2 3 2 ( , ) 1 , 1 xx f x y y x y + = + + ( ) ( ) 2 2 2 3 2 , 1 , 1 yy f x y x x y + = + + and ( ) ( )2 2 3 2 , , 1 xy f x y xy x y − = + + it follows that fxx (0, 0) = 1 > 0 and ( ) ( ) ( ) 2 fxx 0, 0 fyy 0, 0 − ⎡⎣ fxy 0, 0 ⎤⎦ = 1 > 0. So, (0, 0, 1) is a relative minimum. 4. f (x, y) = 25 − (x − 2)2 − y2 The first partial derivatives of f, ( ) ( )2 2 , 2 25 2 x f x y x x y − = − − − and ( ) ( )2 2 , , 25 2 y f x y y x y − = − − − are zero at the critical point (2, 0). Because ( ) ( ) 2 2 3 2 2 , 25 , 25 2 xx f x y y x y − = ⎡ − − − ⎤ ⎣ ⎦ ( ) ( ) ( ) 2 2 3 2 2 2 25 , , 25 2 yy x f x y x y − − = ⎡ − − − ⎤ ⎣ ⎦ and ( ) ( ) ( ) 2 3 2 2 2 , , 25 2 xy x y f x y x y − = ⎡ − − − ⎤ ⎣ ⎦ it follows that (2, 0) 1 0 5 fxx = − < and ( ) ( ) ( ) 2, 0 2, 0 2, 0 2 1 0. 25 fxx fyy − ⎡⎣ fxy ⎤⎦ = > So, (2, 0, 5) is a relative maximum. 5. ( ) ( )2 ( )2 f x, y = x − 1 + y − 3 The first partial derivatives of f , fx (x, y) = 2(x − 1) and fy (x, y) = 2(y − 3), are zero at the critical point (1, 3). Because fxx (x, y) = 2, fyy (x, y) = 2, and fxy (x, y) = 0, it follows that fxx (1, 3) > 0 and ( ) ( ) ( ) 2 fxx 1, 3 fyy 1, 3 − ⎡⎣ fxy 1, 3 ⎤⎦ = 4 > 0. So, (1, 3, 0) is a relative minimum. 6. ( ) ( )2 ( )2 f x, y = 9 − x − 3 − y + 2 The first partial derivatives of f , fx (x, y) = −2(x − 3) and fy (x, y) = −2(y + 2), are zero at the critical point (3, −2). Because fxx (x, y) = −2, fyy (x, y) = −2, and fxy (x, y) = 0, it follows that fxx (3, −2) < 0 and ( ) ( ) ( ) 2 fxx 3, −2 fyy 3, −2 − ⎡⎣ fxy 3, −2 ⎤⎦ = 4 > 0. So, (3, −2, 9) is a relative maximum. 446 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 7. f (x, y) = 2x2 + 2xy + y2 + 2x − 3 The first partial derivatives of f, fx (x, y) = 4x + 2y + 2 and fy (x, y) = 2x + 2y, are zero at the critical point (−1, 1). Because fxx (x, y) = 4, fyy (x, y) = 2, and fxy (x, y) = 2, it follows that fxx (−1, 1) > 0 and ( ) ( ) ( ) 2 fxx −1, 1 fyy −1, 1 − ⎡⎣ fxy −1, 0 ⎤⎦ = 4 > 0. So, (−1, 1, −4) is a relative minimum. 8. f (x, y) = −x2 − 5y2 + 8x − 10y − 13 The first partial derivatives of f, fx (x, y) = −2x + 8 and fy (x, y) = −10y − 10, are zero at the critical point (4, −1). Because fxx (x, y) = −2, fyy (x, y) = −10, and fxy (x, y) = 0, it follows that fxx (4, −1) < 0 and ( ) ( ) ( ) 2 fxx 4, −1 fyy 4, −1 − ⎡⎣ fxy 4, −1 ⎤⎦ = 20 > 0. So, (4, −1, 8) is a relative maximum. 9. f (x, y) = −5x2 + 4xy − y2 + 16x + 10 The first partial derivatives of f, fx (x, y) = −10x + 4y + 16 and fy (x, y) = 4x − 2y, are zero at the critical point (8, 16). Because fxx (x, y) = −10, fyy (x, y) = −2, and fxy (x, y) = 4, it follows that fxx (8, 16) < 0 and fxx (8, 16) fyy (8, 16) − ⎡⎣ fxy (8, 16)⎤⎦ 2 = 4 > 0. So, (8, 16, 74) is a relative maximum. 10. f (x, y) = x2 + 6xy + 10y2 − 4y + 4 The first partial derivatives of f, fx (x, y) = 2x + 6y = 2(x + 3y) and fy (x, y) = 6x + 20y − 4 = 2(3x + 10y − 2), are zero at the critical point (−6, 2). Because fxx (x, y) = 2, fyy (x, y) = 20, and fxy (x, y) = 6, it follows that fxx (−6, 2) > 0 and ( ) ( ) ( ) 2 fxx −6, 2 fyy −6, 2 − ⎡⎣ fxy −6, 2 ⎤⎦ = 4 > 0. So, (−6, 2, 0) is a relative minimum. 11. f (x, y) = 3x2 + 2y2 − 6x − 4y + 16 The first partial derivatives of f, fx (x, y) = 6x − 6 = 6(x − 1) and fy (x, y) = 4y − 4 = 4(y − 1), are zero at the critical point (1, 1). Because fxx (x, y) = 6, fyy (x, y) = 4, and fxy (x, y) = 0, it follows that fxx (1, 1) > 0 and ( ) ( ) ( ) 2 fxx 1, 1 fyy 1, 1 − ⎡⎣ fxy 1, 1 ⎤⎦ = 24 > 0. So, (1, 1, 11) is a relative minimum. 12. f (x, y) = −3x2 − 2y2 + 3x − 4y + 5 The first partial derivatives of f, fx (x, y) = −6x + 3 and fy (x, y) = −4y − 4, are zero at the critical point (1 ) 2, −1 .Because fxx (x, y) = −6, fyy (x, y) = −4, and fxy (x, y) = 0, it follows that (1 ) 2fxx , −1 < 0 and ( ) ( ) ( ) 2 1 1 1 2 2 2 fxx , −1 fyy , −1 − ⎣⎡ fxy , −1 ⎦⎤ = 24 > 0. So, (1 31) 2 4 , −1, is a relative maximum. 13. f (x, y) = −x3 + 4xy − 2y2 + 1 The first partial derivatives of f, fx (x, y) = −3x2 + 4y and fy (x, y) = 4x − 4y, are zero at the critical points (0, 0) and (4 4) 3 3 , . Because fxx (x, y) = −6x, fyy (x, y) = −4, and fxy (x, y) = 4, it follows that fxx (0, 0) = 0 and ( ) ( ) ( ) 2 fxx 0, 0 fyy 0, 0 − ⎣⎡ fxy 0, 0 ⎦⎤ = −16 < 0. So, (0, 0, 1) is a saddle point. Because fxx (x, y) = −6x, fyy (x, y) = −4, and fxy (x, y) = 4, it follows that (4 4) 3 3 fxx , < 0 and ( ) ( ) ( ) 2 4 4 4 4 4 4 3 3 3 3 3 3 fxx , fyy , − ⎡⎣ fxy , ⎤⎦ = 16 > 0. So, (4 4 59 ) 3 3 27 , , is a relative maximum. 14. f (x, y) = x2 − 3xy − y2 The first partial derivatives of f , fx (x, y) = 2x − 3y and fy (x, y) = −3x − 2y, are zero at the critical point (0, 0). Because fxx (x, y) = 2, fyy (x, y) = −2, and fxy (x, y) = −3, it follows that fxx (0, 0) > 0 and ( ) ( ) ( ) 2 fxx 0, 0 fyy 0, 0 − ⎣⎡ fxy 0, 0 ⎦⎤ = −13 < 0. So, (0, 0, 0) is a saddle point. Section 7.5 Extrema of Functions of Two Variables 447 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 15. ( ) 1 f x, y = 2 xy The first partial derivatives of f, ( ) 1 2 fx x, y = y and ( ) 1 2 fy x, y = x, are zero at the critical point (0, 0). Because fxx (x, y) = 0, fyy (x, y) = 0, and ( ) 1 2 fxy x, y = , it follows that fxx (0, 0) = 0 and ( ) ( ) ( ) 2 1 4 fxx 0, 0 fyy 0, 0 − ⎡⎣ fxy 0, 0 ⎤⎦ = − < 0. So, (0, 0, 0) is a saddle point. 16. f (x, y) = x + y + 2xy − x2 − y2 The first partial derivatives of f, fx (x, y) = 1 + 2y − 2x and fy (x, y) = 1 + 2x − 2y, are not zero at a specific point. Because the partial derivatives exist for all real values of x and y and are never both zero, f does not have any relative extrema. 17. ( ) ( ) f x, y = x + y e1− x2 − y2 The first partial derivatives of f, ( , ) ( 2 2 2 1) 1 x2 y2 fx x y x xy e = − − + − − and ( , ) ( 2 2 2 1) 1 x2 y2 , fy x y y xy e = − − + − − are zero at the critical points (1 1 ) 2 2 , and ( 1 1 ) 2 2 − , − . Because ( , ) (4 3 4 2 6 2 ) 1 x2 y2 , fxx x y x x y x y e = + − − − − ( , ) (4 3 4 2 6 2 ) 1 x2 y2 , fyy x y y xy y x e = + − − − − and ( , ) (4 2 4 2 2 2 ) 1 x2 y2 , fxy x y x y xy y x e = + − − − − it follows that (1 1 ) 1 2 2 2fxx , = −3e < 0, ( ) ( ) ( ) 2 1 1 1 1 1 1 2 2 2 2 2 2 fxx , fyy , − ⎡⎣ fxy , ⎤⎦ = 0, ( 1 1 ) 1 2 2 2 fxx − , − = 3e > 0, and ( ) ( ) ( ) 2 1 1 1 1 1 1 2 2 2 2 2 2 fxx − , − fyy − , − − ⎡⎣ fxy − , − ⎤⎦ = 0. So, (1 1 1 2 ) 2 2 , , e is a relative maximum and ( 1 1 1 2 ) 2 2 − , − , −e is a relative minimum. 18. ( ) ( ) 2 2 , 3 x y f x y e− + = The first partial derivatives of f , ( ) ( 2 2) , 6 x y fx x y xe− + = − and ( ) ( ) 2 2 , 6 , x y fy x y ye− + = − are zero at the critical point (0, 0).Because ( ) ( )( ) 2 2 , 6 2 2 1 , x y fxx x y e x − + = − ( ) ( )( ) 2 2 , 6 2 2 1 , x y fyy x y e y − + = − and ( ) ( 2 2) , 12 , x y fxy x y xye− + = it follows that fxx (0, 0) = −6 < 0 and ( ) ( ) ( ) 2 fxx 0, 0 fyy 0, 0 − ⎣⎡ fxy 0, 0 ⎦⎤ = 36 > 0. So, (0, 0, 3) is a relative maximum. 19. fxx > 0 and ( ) ( )( ) 2 9 4 62 0 fxx fyy − fxy = − = Insufficient information 20. fxx < 0 and ( ) ( )( ) 2 3 8 22 0 fxx fyy − fxy = − − − > f has a relative maximum at (x0 , y0 ). 21. fxx < 0 and ( ) ( )( ) 2 9 6 102 0 fxx fyy − fxy = − − < f has a saddle point at (x0 , y0 ). 22. fxx > 0 and ( ) ( )( ) 2 25 8 102 0 fxx fyy − fxy = − > f has a relative minimum at (x0 , y0 ). 23. fxx > 0 and ( ) ( )( ) 2 5 5 32 0 fxx fyy − fxy = − > f has a relative minimum at (x0 , y0 ). 24. fxx > 0 and ( ) ( )( ) 2 8 7 92 0 fxx fyy − fxy = − < f has a saddle point at (x0 , y0 ). 25. ( ) ( )2 f x, y = xy The first partial derivatives of f , fx (x, y) = 2xy2 and fy (x, y) = 2x2 y, are zero at the critical points (a, 0) and (0, b) where a and b are any real numbers. Because fxx (x, y) = 2y2 , fyy (x, y) = 2x2 , and fxy (x, y) = 4xy, it follows that ( ) ( ) ( ) 2 fxx a, 0 fyy a, 0 − ⎡⎣ fxy a, 0 ⎤⎦ = 0 and ( ) ( ) ( ) 2 fxx 0, b fyy 0, b − ⎣⎡ fxy 0, b ⎦⎤ = 0 and the Second-Derivative Test fails. Note that ( ) ( )2 f x, y = xy is nonnegative for all (a, 0, 0) and (0, b, 0) where a and b are real numbers. So, (a, 0, 0) and (0, b, 0) are relative minima. 448 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 26. f (x, y) = x2 + y2 The first partial derivatives of f, ( ) 2 2 x , f x y x x y = + and ( ) 2 2 y , , f x y y x y = + are undefined at the point (0, 0). Because ( ) ( ) 2 2 2 3 2 xx , , f x y y x y = + ( ) ( ) 2 2 2 3 2 yy , , f x y x x y = + and ( ) ( )2 2 3 2 xy , , f x y xy x y − = + it follows that fxx (0, 0) is undefined and the Second-Derivative Test fails. Note that f (x, y) = x2 + y2 is nonnegative for all real numbers. So, (0, 0, 0) is a relative minimum. 27. f (x, y) = x3 + y3 The first partial derivatives of f, fx (x, y) = 3x2 and fy (x, y) = 3y2 , are zero at the critical point (0, 0). Because fxx (x, y) = 6x, fyy (x, y) = 6y, fxy (x, y) = 0, and ( ) ( ) ( ) 2 fxx 0, 0 fyy 0, 0 − ⎡⎣ fxy 0, 0 ⎤⎦ = 0, the Second-Partials Test fails. By testing “nearby” points, you can conclude that (0, 0, 0) is a saddle point. 28. f (x, y) = x3 + y3 − 3x2 + 6y2 + 3x + 12y + 7 The first partial derivatives of f, fx (x, y) = 3x2 − 6x + 3 = 3(x2 − 2x + 1) = 3(x − 1)2 fy (x, y) = 3y2 + 12y + 12 ( ) ( ) 3 2 4 4 3 2 y y y = + + = + are zero at the critical point (1, −2). Because fxx (x, y) = 6x − 6, fxx (1, −2) = 0, fyy (x, y) = 6y + 12, fyy (1, −2) = 0, fxy (x, y) = 0, the Second-Partials Test fails. Note that ( ) ( )3 ( )3 f x, y = x − 1 + y + 2 . By testing “nearby” points, you can conclude that (1, −2, 0) is a saddle point. 29. f (x, y) = x2 3 + y2 3 The first partial derivatives of f, ( ) 3 , 2 3 fx x y x = and ( ) 3 , 2 , 3 fy x y y = are undefined at the point (0, 0). Because ( ) 4 3 , 2 , 9 fxx x y x = − ( ) 4 3 , 2 , 9 fyy x y y = − fxy (x, y) = 0 and fxx (0, 0) is undefined, the Second-Derivative Test fails. Because f (x, y) ≥ 0 for all points in the xy-coordinate plane, (0, 0, 0) is a relative minimum. 30. ( ) ( )f x, y = x2 + y2 2 3 The first partials of f , ( ) ( )2 2 1 3 , 4 3 x f x y x x y = + and ( ) ( )2 2 1 3 , 4 3 y f x y y x y = + are zero at the critical point (0, 0). Because ( ) ( ) ( ) 2 2 2 2 4 3 4 3 , , 9 xx x y f x y x y + = + ( ) ( ) 2 2 2 2 4 3 4 3 , 9 yy y x f x y + = + and ( )2 2 4 3 8 9 xy f xy x y = − + are zero at the point (0, 0), the Second-Partials Test fails. Note that ( ) ( )f x, y = x2 + y2 2 3 is positive for all points (x, y) ≠ (0, 0). So (0, 0, 0) is a relative minimum. 31. f (x, y, z) = (x − 1)2 + (y + 3)2 + z2 Critical point: (x, y, z) = (1, −3, 0) Relative minimum 32. ( ) ( )( ) 2 f x, y, z = 6 − ⎡⎣x y + 2 z − 1 ⎤⎦ Critical points: Any points of the form (0, y, z), (x, −2, z), or (x, y, 1). They all correspond to relative maxima because f (x, y, z) ≤ 6. 33. The sum is x + y + z = 45, or z = 45 − x − y, and the product is P = xyz, or P = xy(45 − x − y) = 45xy − x2 y − xy2. The first partial derivatives of P are Px (x, y) = 45y − 2xy − y2 = y(45 − 2x − y) Py (x, y) = 45x − x2 − 2xy = x(45 − x − 2y). Setting these equal to zero produces the system 2 45 2 45. x y x y + = + = Solving the system, you have x = 15, y = 15, and z = 45 − 15 − 15 = 15. Section 7.5 Extrema of Functions of Two Variables 449 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 34. The sum is 32 32 x y z z x y + + = = − − and the product is P = xy2z = 32xy2 − x2 y2 − xy3. The first partial derivatives of P are Px = 32y2 − 2xy2 − y3 = y2 (32 − 2x − y) Py = 64xy − 2x2 y − 3xy2 = xy(64 − 2x − 3y). Setting these equal to zero produces the system 2 32 2 3 64. x y x y + = + = Solving this system, you have x = 8, y = 16, and z = 8. 35. The sum is x + y + z = 60, or z = 60 − x − y, and the sum of the squares is S = x2 + y2 + z2 = x2 + y2 + (60 − x − y)2. The first partial derivatives of S are Sx (x, y) = 2x + 2(60 − x − y)(−1) = 4x + 2y − 120 and Sy (x, y) = 2y + 2(60 − x − y)(−1) = 2x + 4y − 120. Setting these equal to zero produces the system 4 2 120 2 4 120. x y x y + = + = Solving this system, you have x = 20, y = 20, and z = 60 − 20 − 20 = 20. 36. The sum is x + y + z = 2, or z = 2 − x − y, and the sum of the squares is S = x2 + y2 + z2 = x2 + y2 + (2 − x − y)2. The first partial derivatives of S are Sx (x, y) = 2x + 2(2 − x − y)(−1) = 4x + 2y − 4 and Sy (x, y) = 2y + 2(2 − x − y)(−1) = 2x + 4y − 4. Setting these equal to zero produces the system 4 2 4 2 4 4. x y x y + = + = Solving this system, you have 2 2 3 3 x = , y = , and 2 2 2 3 3 3 z = 2 − − = . 37. The revenue function is ( ) 2 2 R x1, x2 = −5x1 − 8x2 − 2x1x2 + 42x1 + 102x2 and the first partial derivatives of R are 1 1 2 Rx = −10x − 2x + 42 and 2 2 1 Rx = −16x − 2x + 102. Setting these equal to zero produces the system 1 2 1 2 5 21 8 51. x x x x + = + = Solving the system, you have x1 = 3 and x2 = 6. Because 1 1 Rx x = −10, 1 2 Rx x = −2, and 2 2 Rx x = −16, it follows that 1 1 Rx x < 0 and ( )2 1 1 2 2 1 2 Rx x Rx x − Rx x > 0. So, the revenue is maximized when x1 = 3 and x2 = 6. 38. The revenue function is 2 2 R = 515 p1 + 805p2 + 1.5p1p2 − 1.5p1 − p2 and the first partial derivatives of R are 2 1 1 Rp = 515 + 1.5 p − 3p and 2 1 2 Rp = 805 + 1.5 p − 2 p . Setting these equal to zero produces the system 1 2 1 2 3 1.5 515 1.5 2 805. p p p p − = − + = Solving the system, you have 1 23 p = 596 and p2 = 850. So, the revenue is maximized when 1 23 p = 596 and p2 = 850. 39. The revenue function is 1 1 2 2 2 2 1000 1 1500 2 3 1 2 2 1 1.5 2 R xp xp p p pp p p = + = + + − − and the first partial derivatives of R are 2 1 1 Rp = 1000 + 3p − 4 p and 2 1 2 Rp = 1500 + 3p − 3p . Setting these equal to zero produces the system 1 2 1 2 4 3 1000 3 3 1500. p p p p + = − + = Solving this system, you have p1 = 2500 and p2 = 3000, and by the Second-Partials Test you can conclude that the revenue is maximized when p1 = 2500 and p2 = 3000. 450 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 40. The revenue function is 2 2 R = x1p1 + x2 p2 = 1000 p1 − 4 p1 + 6 p1p2 + 900 p2 − 3p2 and the first partial derivatives of R are 1 2 1 Rp = 1000 − 8p + 6 p and 1 2 2 Rp = 6 p + 900 − 6 p . Setting these equal to zero produces the system 1 2 1 2 8 6 1000 6 6 900. p p p p − = − + = Solving this system, you have p1 = $950.00 and p2 = $1100.00, and by the Second-Derivative Test you can conclude that the revenue is maximized when p1 = $950.00 and p2 = $1100.00. 41. The profit is ( )( ) ( ) ( ) 1 2 2 2 1 2 1 2 1 1 2 2 2 2 1 2 1 2 1 2 225 0.4 0.05 15 5400 0.03 15 6100 0.45 0.43 0.8 210 210 11,500 P R C C x x x x x x x x x x xx x x = − − = ⎡⎣ − + ⎤⎦ + − + + − + + = − − − + + − and the first partial derivatives of P are 1 2 1 Px = −0.9x − 0.8x + 210 and 2 2 1 Px = −0.86x − 0.8x + 210. Setting these equal to zero produces the system 1 2 1 2 0.9 0.8 210 0.8 0.86 210. x x x x + = + = Solving this system, you have x1 ≈ 94 and x2 ≈ 157, and by the Second-Partials Test you can conclude that the profit is maximized when x1 ≈ 94 and x2 ≈ 157. 42. The profit function is ( ) ( ) ( ) ( ) 1 2 1 2 1 2 2 2 1 2 1 1 2 2 2 2 1 2 1 2 , 15 15 15 0.02 4 500 0.05 4 275 0.02 0.05 11 11 775 P x x x x c c x x x x x x x x x x = + − − = + − + + − + + = − − + + − and the first partial derivatives of P are 1 1 Px = −0.04x + 11 and 2 2 Px = −0.10x + 11. Setting these equal to zero produces the system 1 2 0.04 11 0 0.10 11 0. x x − + = − + = Solving this system, you have x1 = 275 and x2 = 110. Because 1 1 Px x = −0.04, 1 2 Px x = 0, and 2 2 Px x = −0.10, it follows that 1 1 Px x < 0 and ( )2 1 1 2 2 1 2 Px x Px x − Px x > 0. So, the profit is maximized when x1 = 275 and x2 = 110. 43. Let x = length, y = width, and z = height. The sum of length and girth is (2 2 ) 96 96 2 2 . x y z x y z + + = = − − The volume is V = xyz = 96yz − 2zy2 − 2yz2 and the first partial derivatives are Vy = 96z − 4zy − 2z2 = 2z(48 − 2y − z) and Vz = 96y − 2y2 − 4yz = 2y(48 − y − 2z). Setting these equal to zero produces the system 2 48 2 48. y z y z + = + = So, x = 32. Solving this system, you have y = 16 and z = 16. The volume is a maximum when the length is 32 inches and the width and height are each 16 inches. Section 7.5 Extrema of Functions of Two Variables 451 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 44. Let x = length, y = width, and z = height. The sum of length and girth is (2 2 ) 144 144 2 2 . x y z x y z + + = = − − The volume is V = xyz = 144yz − 2zy2 − 2yz2 and the first partial derivatives are Vy = 144z − 4zy − 2z2 = 2z(72 − 2y − z) and Vz = 144y − 2y2 − 4yz = 2y(72 − y − 2z). Setting these equal to zero produces the system 2 72 2 72. y z y z + = + = Solving this system, you have y = 24 and z = 24. So, x = 48. The volume is a maximum when the length is 48 inches and the width and height are each 24 inches. 45. Let x = length y = width, h = height and C = cost. The volume is xyz 18 or z 18. xy = = The cost is 0.2 0.15(2) 0.15(2) 0.2 0.3 0.3 0.2 0.3 18 0.3 18 0.2 5.4 5.4. C xy xz yz xy xz yz xy x y xy xy xy y x = + + = + + ⎛ ⎞ ⎛ ⎞ = + ⎜ ⎟ + ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = + + The first partial derivatives of C are ( ) 2 , 0.2 5.4 x C x y y x = = − and ( ) 2 , 0.2 5.4. y C x y x y = = − Setting these equal to zero produces the system 2 2 5.4 0.2 0 0.2 5.4 0. y x x y − + = − = Solving this system, you have x = 3, y = 3, and ( )( ) 18 2. 3 3 z = = The cost is a minimum when x = 3 feet, y = 3 feet, and z = 2 feet. The minimum cost is 0.2(3)(3) 5.4 5.4 $5.40. 3 3 C = + + = 46. Let x = length, y = width, h = height, and C = cost. The volume is xyz 1584 or z 1584. xy = = The paint cost (per coat) is 0.11 0.06(2) 0.06(2) 0.11 0.12 0.12 0.11 0.12 1584 0.12 1584 0.11 190.08 190.08. C xy xz yz xy xz yz xy x y xy xy xy y x = + + = + + ⎛ ⎞ ⎛ ⎞ = + ⎜ ⎟ + ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = + + The first partial derivatives of C are ( ) 2 , 0.11 190.08 Cx x y y x = = − and ( ) 2 , 0.11 190.08. Cy x y x y = = − Setting these equal to zero produces the system 2 2 190.08 0.11 0 0.11 190.08 0. y x x y − + = − = Solving this system, you have x = 12, y = 12, and ( )( ) 1584 11. 12 12 z = = The cost is a minimum when x = 12 feet, y = 12 feet, and z = 11 feet. The minimum cost (per coat) is 0.11(12)(12) 190.08 190.08 $47.52. 12 12 C = + + = 452 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 47. The total cost function is C(x, y) = 2x2 + 3y2 − 15x − 20y + 4xy + 39 and the first partial derivatives are Cx = 4x − 15 + 4y and Cy = 6y − 20 + 4x. Setting these equal to zero produces the system 4x + 4y = 15 4x + 6y = 20. Solving this system, you have x = 1.25 and y = 2.5. So, the minimum total cost is ( )2 ( )2 ( ) ( ) ( )( ) 2 1.25 + 3 2.5 − 15 1.25 − 20 2.5 + 4 1.25 2.5 + 39 = $4.625million. 48. The duration function is D(x, y) = x2 + 2y2 − 18x − 24y + 2xy + 120 and the first partial derivatives are Dx = 2x − 18 + 2y and Dy = 4y − 24 + 2x. Setting these equal to zero produces the system 2x + 2y = 18 2x + 4y = 24. Solving this system, you have x = 6 and y = 3. So, to minimize the duration of the infection you should take 600 milligrams of the first drug and 300 milligrams of the second drug. 49. The total weight function is ( ) ( ) 1 2 2 2 3 0.002 0.001 4.5 0.004 0.005 3 0.002 4.5 0.005 0.005 T xW yW x x y y x y x x y y xy = + = − − + − − = − + − − and the first partial derivatives are Tx = 3 − 0.004x − 0.005y and Ty = 4.5 − 0.010y − 0.005x. Setting these equal to zero produces the system 0.004 0.005 3 0.005 0.01 4.5. x y x y + = + = Solving this system, you have x = 500 and y = 200. The lake should be stocked with approximately 500 smallmouth bass and 200 largemouth bass. 50. Points A and B are relative extrema. Points C and D are saddle points. 51. The population function is P( p, q, r) = 2 pq + 2 pr + 2qr. Because p + q + r = 1, r = 1 − p − q. So, ( ) ( ) 2 2 2 2 1 2 1 2 2 2 2 2 P pq p p q q p q p p q q pq = + − − + − − = − + − + − and the first partial derivatives are Pp = −4 p + 2 − 2q Pq = −4q + 2 − 2 p. Setting these equal to zero produces the system 4 p + 2q = 2 2 p + 4q = 2. Solving this system, you have 1 3 p = and 1 3q = .So, 1 3r = . The proportion is a maximum when 1 3p = , 1 3q = , and 1 3r = . The maximum proportion is (1)(1) (1)(1) (1)(1) 6 2 3 3 3 3 3 3 9 3 P = 2 + 2 + 2 = = . Chapter 7 Quiz Yourself 453 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 52. Because x + y + z = 1, you have z = 1 − x − y. Then H = − x ln x − y ln y − (1 − x − y) ln(1 − x − y). The first partial derivatives of H are ( ) ( ) ( ) ( )( ) ( ) ( ) , 1 1 ln 1 1 ln 1 1 1 1 ln 1 ln1 ln ln 1 Hx x y x x x y x y x xy x x y x x y ⎛ ⎞ ⎡ ⎛ − ⎞ ⎤ = − ⎜ ⎟ + − − ⎢ − − ⎜ ⎟ + − − − ⎥ ⎝ ⎠ ⎣ ⎝ − − ⎠ ⎦ = − − − ⎡⎣− − − − ⎤⎦ = − + − − and ( ) ( ) ( ) ( )( ) ( ) ( ) , 1 1 ln 1 1 ln 1 1 1 1 ln 1 ln1 ln ln 1 . Hy x y y y x y x y y xy y x y y x y ⎛ ⎞ ⎡ ⎛ − ⎞ ⎤ = − ⎜ ⎟ + − − ⎢ − − ⎜ ⎟ + − − − ⎥ ⎝ ⎠ ⎣ ⎝ − − ⎠ ⎦ = − − − ⎡⎣− − − − ⎦⎤ = − + − − Setting these equal to zero produces the system ( ) ( ) ln ln 1 0 ln ln 1 0. x x y y x y − + − − = − + − − = Solving this system, you have ln x = ln y ⇒ x = y. So, using Hx , ( ) ( ) ln ln 1 2 0 ln ln 1 2 1 2 3 1 1. 3 x x x x x x x x − + − = = − = − = = So, 1, 1, 3 3 x = y = and 1 1 1 1. 3 3 3 z = − − = The maximum value of H is 1 ln 1 1 ln 1 1 ln 1 ln 1 ln 3 1.0986. 3 3 3 3 3 3 3 H = − ⎛⎜ ⎞⎟ − ⎛⎜ ⎞⎟ − ⎛⎜ ⎞⎟ = − ⎛⎜ ⎞⎟ = ≈ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 53. True 54. False. Relative maxima sometimes occur at points where one or more of the partial derivatives do not exist. Chapter 7 Quiz Yourself 1. (a) (b) ( )2 ( )2 ( )2 d = −1 − 1 + 2 − 3 + 0 − 2 = 3 (c) 1 ( 1) 3 2 2 0 Midpoint , , 2 2 2 0, 5, 1 2 ⎛ + − + + ⎞ =⎜ ⎟ ⎝ ⎠ = ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 2. (a) (b) ( ( )) ( ) ( ) 2 2 2 5 1 1 3 6 4 140 2 35 d = − − + − + − − = = (c) ( ) ( ) 5 1 1 3 6 4 Midpoint , , 2 2 2 2, 2, 1 ⎛ + − + − + ⎞ =⎜ ⎟ ⎝ ⎠ = − (1, 3, 2) (−1, 2, 0) 2 −2 −1 1 2 1 2 3 y x z x y z (−1, 3, 4) (5, 1, −6) −2 −4 −6 −2 −2 2 4 2 4 2 4 454 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 3. (a) (b) ( )2 ( )2 ( )2 d = 3 − 0 + 0 + 3 + −3 − 3 = 54 = 3 6 (c) Midpoint 0 3, 3 0, 3 3 3, 3, 0 2 2 2 2 2 ⎛ + − + + − ⎞ ⎛ ⎞ = ⎜ ⎟ = ⎜ − ⎟ ⎝ ⎠ ⎝ ⎠ 4. ( )2 ( )2 ( )2 x − 2 + y + 1 + z − 3 = 16 5. Center: 0 2, 3 5, 1 ( 5) (1, 4, 2) 2 2 2 ⎛ + + + − ⎞ ⎜ ⎟ = − ⎝ ⎠ ( )2 ( )2 ( )2 Radius = 1 − 0 + 4 − 3 + −2 − 1 = 11 ( )2 ( )2 ( )2 Standard form: x − 1 + y − 4 + z + 2 = 11 6. x2 + y2 + z2 − 8x − 2y − 6z − 23 = 0 (x2 − 8x + 16) + (y2 − 2y + 1) + (z2 − 6z + 9) = 23 + 16 + 1 + 9 ( )2 ( )2 ( )2 x − 4 + y − 1 + z − 3 = 49 Center: (4, 1, 3) Radius: 49 = 7 7. 2x + 3y + z = 6 To find the x-intercept, let y = 0 and z = 0. 2x = 6 ⇒ x = 3 To find the y-intercept, let x = 0 and z = 0. 3y = 6 ⇒ y = 2 To find the z-intercept, let x = 0 and y = 0. z = 6 8. x − 2z = 4 To find the x-intercept, let z = 0. x = 4 Because the y-coefficient is 0, there is no y-intercept. The plane is parallel to the y-axis. To find the z-intercept, let x = 0. −2z = 4 ⇒ z = −2 9. The only intercept is y = 3. The plane is parallel to the xz-plane. 10. The graph of 2 2 2 1 4 9 16 x + y + z = is an ellipsoid. 11. The graph of z2 − x2 − y2 = 25 or 2 2 2 1 25 25 25 z − x − y = is a hyperboloid of two sheets. 12. The graph of 81z − 9x2 − y2 = 0 or 2 2 9 81 z = x + y is an elliptic paraboloid. 13. f (x, y) = x − 9y2 ( ) ( )2 f 1, 0 = 1 − 9 0 = 1 ( ) ( )2 f 4, −1 = 4 − 9 −1 = −5 x y z (3, 0, 0) (0, 2, 0) (0, 0, 6) 1 1 2 3 4 6 3 4 3 4 5 1 −1 x y z (4, 0, 0) (0, 0, −2) 1 1 −3 −4 −5 4 2 5 1 x y z (0, 3, 0) 2 6 2 4 6 2 4 6 x y z (0, −3, 3) (3, 0, −3) 1 1 −1 −2 −1 −1 1 −2 2 3 −3 −2 −3 −4 2 2 3 4 Chapter 7 Quiz Yourself 455 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 14. f (x, y) = 4x2 + y ( ) ( )2 f 1, 0 = 4 1 + 0 = 2 ( ) ( )2 ( ) f 4, −1 = 4 4 + −1 = 63 = 3 7 15. f (x, y) = ln(x − 2y) f (1, 0) = ln(1) = 0 f (4, −1) = ln(4 − 2(−1)) = ln 6 ≈ 1.79 16. (a) The temperatures in the Great Lakes region range from about 30° to about 50°. (b) The temperatures in the United States range from 40° to 80°. (c) The temperatures in Mexico range from about 70° to almost 90°. 17. f (x, y) = x2 + 2y2 − 3x − y + 1 fx (x, y) = 2x − 3 fx (−2, 3) = 2(−2) − 3 = −7 fy (x, y) = 4y − 1 fy (−2, 3) = 4(3) − 1 = 11 18. ( ) 3 2 f x, y x y x y − = + ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 3 3 3 3 x , , x y x y y y y y f x y x y x y x y + − − + + = = = + + + ( ) ( ) ( ) 2 2 3 33 2, 3 18 2 3 fx + − = = − + ( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 3 2 3 y , , x y y x y y xy x f x y x y x y + − − − − − − = = + + ( ) ( )( ) ( ) ( ) 2 2 3 2 2 3 3 2 2, 3 9 2 3 fy − − − − − − = = − + 19. f (x, y) = x3e2 y ( , ) 3 2 2 y , fx x y = x e ( ) ( ) 2, 3 3 2 2 2(3) 12 6 4841.15 fx − = − e = e ≈ ( , ) 2 3 2 y , fy x y = x e ( ) ( ) 2, 3 2 2 3 2(3) 16 6 6454.86 fy − = − e = − e ≈ − 20. f (x, y) = ln(2x + 7 y) ( , ) 2 , 2 7 fx x y x y = + ( ) ( ) ( ) 2, 3 2 2 0.118 2 2 73 17 fx − = = ≈ − + ( , ) 7 , 2 7 fy x y x y = + ( ) ( ) ( ) 2, 3 7 7 0.412 2 2 73 17 fy − = = ≈ − + 21. f (x, y) = 3x2 + y2 − 2xy − 6x + 2y The first partial derivatives of f, fx (x, y) = 6x − 2y − 6 and fy (x, y) = 2y − 2x + 2, are zero at the point (1, 0).Moreover, because fxx (x, y) = 6, fyy (x, y) = 2, and fxy (x, y) = −2, it follows that fxx (1, 0) > 0 and ( ) ( ) ( ) 2 fxx 1, 0 fyy 1, 0 − ⎡⎣ fxy 1, 0 ⎤⎦ = 8 > 0. So, (1, 0, −3) is a relative minimum. 22. f (x, y) = −x3 + 4xy − 2y2 + 1 The first partial derivatives of f, fx (x, y) = −3x2 + 4y and fy (x, y) = 4x − 4y, are zero at the points (0, 0) and (4 4) 3 3 , .Moreover, because fxx (x, y) = −6x, fyy (x, y) = −4, and fxy = 4, it follows that fxx (0, 0) = 0 and ( ) ( ) ( ) 2 fxx 0, 0 fyy 0, 0 − ⎡⎣ fxy 0, 0 ⎤⎦ = −16 < 0, (4 4) 3 3fxx , = −8 and ( ) ( ) ( ) 2 4 4 4 4 4 4 3 3 3 3 3 3 fxx , fyy , − ⎣⎡ fxy , ⎦⎤ = 16 > 0. So, (0, 0, 1) is a saddle point and (4 4 59 ) 3 3 27 , , is a relative maximum. 23. The cost function is ( ) 1 2 2 16 C x, y = x + y − 10x − 40y + 820 and the first partial derivatives are 1 8 Cx = x − 10 and Cy = 2y − 40. Setting these equal to zero and solving for x and y gives you x = 80 and y = 20. The minimum combined cost is 1 ( )2 2 ( ) ( ) 16 80 + 20 − 10 80 − 40 20 + 820 = 20, or $20,000. 456 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Section 7.6 Lagrange Multipliers Skills Warm Up 1. 4 6 3 2 3 2 x y x y − = ⎧⎨ ⎩ + = Equation 1 Equation 2 Multiply Equation 2 by 2: Add to Equation 1: Simplify: 78 4 6 4 8 7 x y x x + = = = 7 (7 ) 8 8 1 12 Substitute for in Equation 2: 2 3 2 Solve for : x y y y + = = The solution is (7 1 ) 8 12 , . 2. 6 6 5 3 1 x y x y − = ⎧⎨ ⎩− − = Equation 1 Equation 2 Multiply Equation 2 by 2: Add to Equation 1: Simplify: 78 6 2 2 8 7 x y y y − − = − = = − Substitute 78 − for x in Equation 2: ( ) 78 −3x − − = 1 Solve for x: 1 24 x = − The solution is ( 1 7 ) 24 8 − , − . 3. 5 25 5 15 x y x y − = ⎧⎨ ⎩ − = Equation 1 Equation 2 Multiply Equation 2 by 5: Add to Equation 1: Simplify: − 25 12 5 25 75 24 50 x y y y − + = − = − = − Substitute 25 12 − for y in Equation 1: ( 25) 12 5x − − = 25 Solve for x: 55 12 x = The solution is (55 25) 12 12 , − . 4. 4 9 5 8 2 x y x y − = ⎧⎨ ⎩− + = − Equation 1 Equation 2 Multiply Equation 2 by 4: Add to Equation 1: Simplify: 3 23 4 32 8 23 3 x y y y − + = − = − = − Substitute 3 23 − for y in Equation 2: ( 3 ) 23 −x + 8 − = −2 Solve for x: 22 23 x = The solution is (22 3 ) 23 23 , − . 5. 2 3 2 2 4 2 3 1 x y z x y z x y z − + = ⎧⎪ + + = ⎨⎪ ⎩− + + = − Equation 1 Equation 2 Equation 3 Multiply Equation 3 by 2: Add to Equation 2: Multiply Equation 3 by 2: Add to Equation 1: Subtract New Equation 2 from New Equation 1: Simplify: 1 3 2 4 6 2 6 7 2 2 4 6 2 3 7 1 3 1 x y z y z x y z y z y y − + + = − + = − + + = − + = = = New Equation 1 New Equation 2 Substitute 1 3 for y in New Equation 2: (1) 3 3 + 7z = 1 Solve for z: z = 0 Substitute 1 3 for y and 0 for z in Equation 3: (1) ( ) 3 −x + 2 + 3 0 = −1 Solve for x: 5 3 x = The solution is (5 1 ) 3 3 , , 0. Section 7.6 Lagrange Multipliers 457 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. F(x, y, λ ) = xy − λ (x + y − 10) Fx = y − λ = 0, y = λ Fy = x − λ = 0, x = λ Fλ = −(x + y − 10) = 0, 2λ = 10, λ = 5 So, λ = 5, x = 5, y = 5, and f (x, y) has a maximum at (5, 5). The maximum is f (5, 5) = 25. 2. F(x, y, λ ) = xy − λ (x + 3y − 6) Fx = y − λ = 0, y = λ Fy = x − 3λ = 0, x = 3λ Fλ = −(x + 3y − 6) = 0, −6λ = −6, λ = 1 So, λ = 1, x = 3, and y = 1, and f (x, y) has a maximum at (3, 1).The maximum is f (3, 1) = 3. Skills Warm Up —continued— 6. 4 6 2 3 3 4 3 3 0 x y z x y z x y z − − + = − ⎧⎪ − − = ⎨⎪ ⎩ + + = Equation 1 Equation 2 Equation 3 Add Equation 2 and Equation 3: Multiply Equation 2 by 2: Add to Equation 1: Multiply New Equation 1 by 5: Subtract New Equation 2 from New Equation 3: Simplify: 14 19 4 2 4 2 6 6 8 10 6 20 10 20 19 14 x y x y z x y x y x x − = − − = − = − = = = New Equation 1 New Equation 2 New Equation 3 Substitute 14 19 for x in New Equation 2: 14 19 − 10y = 6 Solve for y: 10 19 y = − Substitute 14 19 for x and 10 19 − for y in Equation 3: (14) ( 10) 19 19 3 + − + 3z = 0 Solve for z: 32 57 z = − The solution is (14 10 32) 19 19 57 , − , − . 7. ( ) ( ) ( ) 2 2 2 2 , , 2 , 2 x y f x y x y xy f x y xy y f x y x xy = + = + = + 8. ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )( ) 2 2 2 2 2 , 25 , 50 50 , 50 2 50 2 x y f x y xy y f x y xy y y y x y f x y xy y x y y x y x y = + = + = + = + + = + + 9. ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) 2 2 2 2 2 2 , , 2 , , 2 2 2 1 3 4 , , 2 2 0 2 , , 2 0 x y z f x y z x x xy yz f x y z x x y x xy yz x xy yz f x y z x x z x xy yz x xz f x y z x y x xy yz xy = − + = − + − + = − + = − + + − + = − + = + − + = 10. f (x, y, z) = z(xy + xz + yz) ( ) ( ) ( )( ) 2 fx x, y, z z y z xy xz yz 0 z yz = + + + + = + ( ) ( ) ( )( ) 2 fy x, y, z z x z xy xz yz 0 z xz = + + + + = + ( , , ) ( ) ( )(1) 2 2 fz x y z z x y xy xz yz xy xz yz = + + + + = + + 458 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 3. F(x, y, λ ) = x2 + y2 − λ (x + y − 8) Fx = 2x − λ = 0, 1 2 x = λ Fy = 2y − λ = 0, 1 2 y = λ Fλ = −(x + y − 8) = 0, −λ = −8, λ = 8 So, λ = 8, x = 4, and y = 4, and f (x, y) has a minimum at (4, 4). The minimum is f (4, 4) = 32. 4. F(x, y, λ ) = x2 + y2 − λ (−2x − 4y + 5) Fx = 2x + 2λ = 0, x = −λ Fy = 2y + 4λ = 0, y = −2λ Fλ = −(−2x − 4y + 5) = 0, −10λ = 5, 1 2 λ = − So, 1 2λ = − , 1 2x = , and y = 1, and f (x, y) has a minimum at (1 ) 2, 1 . The minimum is (1 ) 5 2 4 f , 1 = . 5. F(x, y, λ ) = x2 − y2 − λ (2y − x2 ) Fx = 2x + 2λ x = 0, 2x(1 + λ ) = 0, λ = −1 Fy = −2y − 2λ = 0, y = −λ Fλ = −(2y − x2 ) = 0, x = 2y So, λ = −1, x = 2, y = 1, and f (x, y) has a maximum at ( 2, 1).The maximum is f ( 2, 1) = 1. 6. F(x, y, λ ) = x2 − y2 − λ (x − 2y + 6) Fx = 2x − λ = 0, 1 2 x = λ Fy = −2y + 2λ = 0, y = λ ( ) 2 6 0, F x y λ = − − + = 32 λ = 6, λ = 4 So, λ = 4, x = 2, and y = 4, and f (x, y) has a minimum at (2, 4). The minimum is f (2, 4) = −12. 7. F(x, y, λ ) = 2x + 2xy + y − λ (2x + y − 100) Fx = 2 + 2y − 2λ = 0, y = λ − 1 Fy = 2x + 1 − λ = 0, 1 2 x λ − = Fλ = −(2x + y − 100) = 0, 2 1 ( 1) 100, 51 2 λ λ λ ⎛ − ⎞ − ⎜ ⎟ − − = − = ⎝ ⎠ So, λ = 51, x = 25, y = 50, and f (x, y) has a maximum at (25, 50).The maximum is f (25, 50) = 2600. 8. F(x, y, λ ) = 3x + y + 10 − λ (x2 y − 6) Fx = 3 − 2xyλ = 0 Fy = 1 − x2λ = 0, 2 1 x λ = Fλ = −(x2 y − 6) = 0, 2 y 6 x = For Fx , you can write 2 2 3 3 3 2 6 1 3 12 4. x x x x x = ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ = = So, x = 3 4, 3 3 4 , 2 y = and f (x, y) has a minimum at 3 3 4, 3 4 . 2 ⎛ ⎞ ⎜⎜ ⎟⎟ ⎝ ⎠ The minimum is 3 3 3 4, 3 4 9 4 20 17.143. 2 2 f ⎛ ⎞ + ⎜⎜ ⎟⎟ = ≈ ⎝ ⎠ 9. Note: f (x, y) has a maximum value when g(x, y) = 6 − x2 − y2 is maximum. F(x, y, λ ) = 6 − x2 − y2 − λ (x + y − 2) 2 0, 2 0, x y F x F y λ λ = − − = = − − = 2 2 x x y y λ λ − = ⎫ ⎬ = − = ⎭ Fλ = −(x + y − 2) = 0, 2x = 2, x = 1 So, x = 1, y = 1, and f (x, y) has a maximum at (1, 1). The maximum is f (1, 1) = 2. 10. Note: f (x, y) has a minimum when g(x, y) = x2 + y2 is minimum. F(x, y, λ ) = x2 + y2 − λ (2x + 4y − 15) Fx = 2x − 2λ = 0, x = λ Fy = 2y − 4λ = 0, y = 2λ Fλ = −(2x + 4y − 15) = 0, −10λ = −15, 3 2 λ = So, 3, 2 λ = 3, 2 x = y = 3, and f (x, y) has a minimum at 3, 3 . 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ The minimum is 3, 3 3 5. 2 2 f ⎛⎜ ⎞⎟ = ⎝ ⎠ Section 7.6 Lagrange Multipliers 459 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 11. F(x, y, λ ) = exy − λ (x2 + y2 − 8) 2 0, 2 2 0, 2 xy xy x xy xy y F ye x e x y F xe y e y x λ λ λ λ = − = = ⎫⎪⎪⎬⎪ = − = = ⎪⎭ x = y F (x2 y2 8) 0, λ = − + − = 2x2 = 8, x = 2 So, x = 2, y = 2, and f (x, y) has a maximum at (2, 2). The maximum is f (2, 2) = e4. 12. F(x, y, λ ) = 2x + y − λ (xy − 32) 2 0, 1 0, x y F y F x λ λ = − = = − = 2 2 1 y y x x λ λ = ⎫ ⎬ = = ⎭ Fλ = −(xy − 32) = 0, −2x2 = −32, x = 4 So, x = 4, y = 8, and f (x, y) has a minimum at (4, 8).The minimum is f (4, 8) = 16. 13. F(x, y, z, λ ) = 2x2 + 3y2 + 2z2 − λ (x + y + z − 24) Fx = 4x − λ = 0, λ = 4x Fy = 6y − λ = 0, λ = 6y Fz = 4z − λ = 0, λ = 4z Fλ = −(x + y + z − 24) = 0 24 4 6 4 8 288 36 λ λ λ λ λ + + = = = So, λ = 36, x = 9, y = 6, z = 9, and f (x, y, z) has a minimum at (9, 6, 9). The minimum is f (9, 6, 9) = 432. 14. F(x, y, z, λ ) = xyz − λ (x + y + z − 6) 0 0 0 x y z F yz F xz x y z F xy λ λ λ ⎫ = − =⎪ = − = = = ⎬⎪ = − =⎭ Fλ = −(x + y + z − 6) = 0, −3x = −6, x = 2 So, x = 2, y = 2, z = 2, and f (x, y, z) has a maximum at (2, 2, 2). The maximum is f (2, 2, 2) = 8. 15. F(x, y, z, λ ) = x2 + y2 + z2 − λ (x + y + z − 1) 2 0 2 0 2 0 x y z F x F y x y z F z λ λ λ ⎫ = − =⎪ = − = = = ⎬⎪ = − =⎭ ( ) 1 3 Fλ = − x + y + z − 1 = 0, 3x = 1, x = So, 1 3x = , 1 3y = , 1 3z = , and f (x, y, z) has a minimum at (1 1 1) 3 3 3 , , . The minimum is (1 1 1) 1 3 3 3 3 f , , = . 16. F(x, y, λ ) = x2 − 8x + y2 − 12y + 48 − λ (x + y − 8) 2 8 0 2 12 0 x y F x F y λ λ = − − = ⎫⎪⎬ = − − =⎪⎭ 2 8 2 12 2 x y y x − = − = + Fλ = −(x + y − 8) = 0, x + (x + 2) = 8, x = 3 So, x = 3, y = 5, and f (x, y) has a minimum at (3, 5). The minimum is f (3, 5) = −2. 17. F(x, y, z, λ ) = x + y + z − λ (x2 + y2 + z2 − 1) 1 2 0 1 2 0 1 2 0 x y z F x F y x y z F z λ λ λ = − = ⎫ ⎪ = − = = = ⎬⎪ = − = ⎭ Fλ = −(x2 + y2 + z2 − 1) = 0, 3 2 1, 3 3 x = x = So, 3, 3 x = 3, 3 y = 3, 3 z = and f (x, y, z) has a maximum at 3, 3, 3 . 3 3 3 ⎛ ⎞ ⎜⎜ ⎟⎟ ⎝ ⎠ The maximum is 3, 3, 3 3. 3 3 3 f ⎛ ⎞ ⎜⎜ ⎟⎟ = ⎝ ⎠ 18. F(x, y z, λ ) = x2 y2z2 − λ (x2 + y2 + z2 − 1) 2 2 2 2 2 2 2 2 0, 2 2 0, 2 2 0, x y z F xyz x F xyz y F xyz z λ λ λ = − = = − = = − = 2 2 2 2 2 2 y z xz x y z x y λ λ λ = ⎫ ⎪⎪ = = = ⎬⎪ = ⎪⎭ ( 2 2 2 1) 0, 3 2 1, 3 3 Fλ = − x + y + z − = x = x = So, 3, 3 x = 3, 3 y = 3, 3 z = and f (x, y, z) has a maximum at 3, 3, 3 . 3 3 3 ⎛ ⎞ ⎜⎜ ⎟⎟ ⎝ ⎠ The maximum is 3, 3, 3 1 . 3 3 3 27 f ⎛ ⎞ ⎜⎜ ⎟⎟ = ⎝ ⎠ 460 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 19. Maximize f (x, y, z) = xyz subject to the constraint x + y + z = 60. F(x, y z, λ ) = xyz − λ (x + y + z − 60) 0, 0, 0, x y z F yz yz F xz xz yz xz xy x y z F xy xy λ λ λ λ λ λ = − = = ⎫⎪ = − = = = = ⇒ = = ⎬⎪ = − = =⎭ Fλ = −(x + y + z − 60) = 0, −3x = −60, x = 20 So, x = 20, y = 20, and z = 20. 20. Maximize f (x, y, z) = x2 yz subject to the constraint x + y + z = 80. F(x, y z, λ ) = x2 yz − λ (x + y + z − 80) Fx = 2xyz − λ = 0, λ = 2xyz 2 2 2 0, 2 0, x y F xyz xyz y z F xz xz λ λ λ λ = − = = ⎫⎪ ⎬ = = − = = ⎪⎭ Fλ = −(x + y + z − 80) = 0 Using Fx and Fz , 2xy2 = x2 y ⇒ 2y = x and using Fx , −(2y + y + y − 80) = 0 ⇒ y = 20. So, x = 40, y = 20, and z = 20. 21. Minimize f (x, y, z) = x2 + y2 + z2 subject to the constraint x + y + z = 120. F(x, y, z, λ ) = x2 + y2 + z2 − λ (x + y + z − 120) 2 0 2 0 2 0 x y z F x F y x y z F z λ λ λ ⎫ = − =⎪ = − = = = ⎬⎪ = − =⎭ Fλ = −(x + y + z − 120) = 0, 3x = 120, x = 40 So, x = 40, y = 40, and z = 40. 22. Minimize f (x, y, z) = x3 + y3 + z3 subject to the constraint x + y + z = 36. F(x, y z, λ ) = x3 + y3 + z3 − λ (x + y + z − 36) 2 2 2 2 2 2 3 0, 3 3 0, 3 3 0, 3 x y z F x x F y y x y z F z z λ λ λ λ λ λ = − = = ⎫ ⎪ = − = = = = ⎬⎪ = − = = ⎭ Fλ = −(x + y + z − 36) = 0 Using Fλ , −(x + x + x − 36) = 0, x = 12. So, x = 12, y = 12, and z = 12. Section 7.6 Lagrange Multipliers 461 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 23. F(x, y, λ ) = x2 + y2 − λ (x + y − 6) Fx = 2x − λ = 0, 2 x λ = Fy = 2y − λ = 0, 2 y x λ = = Fλ = −(x + y − 6) = 0, x + x = 6 ⇒ x = 3 So, x = 3, y = 3, and d = x2 + y2 = 9 + 9 = 18 = 3 2. 24. F(x, y, λ ) = x2 + (y − 10)2 − λ ⎡⎣(x − 4)2 + y2 − 4⎤⎦ ( ) ( ) 2 2 4 0, 4 2 10 2 0, 10 x y F x x x x F y y y y λ λ λ λ = − − = = ⎫⎪ − ⎪⎬ − ⎪ = − − = = ⎪⎭ 5x + 2y = 20 Fλ (x 4)2 y2 4 0, = −⎡ − + − ⎤ = ⎣ ⎦ ( ) 2 4 2 20 5 4 2 x x ⎛ − ⎞ − + ⎜ ⎟ = ⎝ ⎠ 116 4 29 29 x ± = (x ≈ 3.2572, 4.7428) So, 116 4 29 29 x − = (the other value of x results in a negative y-value), 10 29 , 29 y = and 2 2 116 4 29 10 29 10 8.770. 29 29 d ⎛ − ⎞ ⎛ ⎞ = ⎜⎜ ⎟⎟ + ⎜⎜ − ⎟⎟ ≈ ⎝ ⎠ ⎝ ⎠ 25. ( ) ( )2 ( )2 ( )2 ( ) F x, y, z, λ = x − 2 + y − 1 + z − 1 − λ x + y + z − 1 ( ) ( ) ( ) 2 2 0 2 1 0 2 1 0 x y z F x F y F z λ λ λ ⎫ = − − =⎪⎪ = − − =⎬⎪ = − − =⎪⎭ 2 1 1 1 x y z x y z − = − = − − = = Fλ = −(x + y + z − 1) = 0 So, y = 0, z = 0, x = 1, and ( )2 ( )2 ( )2 d = 1 − 2 + 0 − 1 + 0 − 1 = 3. 26. F(x, y, z, λ ) = (x − 4)2 + y2 + z2 − λ (x2 + y2 − z2 ) Fx = 2(x − 4) − 2xλ = 0, 2x(1 − λ ) = 8 2 2 0 2 2 0 y z F y y F z z λ λ = − = ⎫⎪⎬ = + = ⎪⎭ ( ) ( ) 2 1 0 2 1 0 y z λ λ − = + = Fλ = −(x2 + y2 − z2 ) = 0, z = x2 + y2 From Fy , you have y = 0 or λ = 1.From Fx , you know that λ ≠ 1 (since 0 ≠ 8), so, y = 0. From Fλ and Fz , you now have x = z and λ = −1. So, x = 2, y = 0, z = 2, and ( )2 ( )2 ( )2 d = 2 − 4 + 0 + 2 = 2 2. 462 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 27. Maximize f (x, y, z) = xyz subject to the constraint 2x + 3y + 5z − 90 = 0. F(x, y z, λ ) = xyz − λ (2x + 3y + 5z − 90) 1 2 1 1 2 1 2 3 3 3 1 1 3 1 3 5 5 5 2 0, 3 0, 5 0, x y z F yz yz y x y x F xz xz z y z y F xy xy λ λ λ λ λ λ = − = = ⎫ ⎪⎪ = ⇒ = = − = = ⎬ ⎪ = ⇒ = = − = = ⎪⎭ Fλ = −(2x + 3y + 5z − 90) = 0 Using Fλ , ( (2 ) 3(2 ) ) 3 5 3 2 3 5 90 0 2 2 2 90 0 15. x x x x x x x − + + ⎡ ⎤ − = ⎣ ⎦ − − − + = ⇒ = So, x = 15, y = 10, and z = 6. The rectangular box has dimensions 15 units by 10 units by 6 units. 28. F(x, y, z, λ ) = xyz − λ (x + 2y + 2z − 108) 0 2 2 0 x y F yz x y F xz λ λ = − = ⎫⎪ ⎬ = = − =⎪ ⎭ Fz = xy − 2λ = 0, y = z Fλ = −(x + 2y + 2z − 108) = 0, 6y = 108, y = 18 So, x = 36, y = 18, and z = 18. The volume is maximized when the dimensions are 36 × 18 × 18 inches. 29. F(x, y, z, λ ) = xyz − λ (3xy + 2xz + 2yz − 1296) ( ) ( ) 3 2 0 or 3 3 2 0 x y F yz y z x y z F xz x z λ λ λ λ λ = − + = ⎫⎪ ⎬ = = = − + =⎪ ⎭ Fz = xy − (2λ x + 2λ y) = 0, x = 4λ (if x = y) Fλ = −(3xy + 2xz + 2yz − 1296) = 0 From , y F 32 z = 6λ = x and Fλ gives x = 12. So, 12, x = 12, y = and ( ) 32 z = 12 = 18. (The case z = 3λ results in λ = 0 or z = 0, which are impossible.) The volume is maximized when the dimensions are 12 × 12 × 18 feet. 30. Minimize C(x, y, z) = 5xy + 3(xy + 2xz + 2yz) = 8xy + 6xz + 6yz subject to the constraint xyz = 480. F(x, y, z, λ ) = 8xy + 6xz + 6yz − λ (xyz − 480) 8 6 0 8 6 0 6 6 0 x y z F y z yz F x z xz F x y xy λ λ λ ⎫ = + − =⎪ = + − =⎬⎪ = + − =⎭ 43 4 3, x y y z z y = = = ( ) ( )( ) 43 Fλ = − xyz − 480 = 0, y y y = 480, y = 2 3 45 So, x = 2 3 45, y = 2 3 45, and 8 3 3 z = 45. The dimensions for minimizing cost are 3 3 8 3 3 2 45 × 2 45 × 45 feet. y x z Section 7.6 Lagrange Multipliers 463 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 31. ( ) 2 2 ( ) F x1, x2 , λ = 0.25x1 + 25x1 + 0.05x2 + 12x2 − λ x1 + x2 − 1000 1 1 Fx = 0.5x + 25 − λ = 0, x1 = 2λ − 50 2 2 Fx = 0.1x + 12 − λ = 0, x2 = 10λ − 120 Fλ = −(x1 + x2 − 1000) = 0 (2 50) (10 120) 1000 12 1170 97.5 λ λ λ λ − + − = = = So, x1 = 145 and x2 = 855. To minimize cost, let x1 = 145 units and x2 = 855 units. 32. ( ) 2 2 ( ) F x1, x2 , λ = 0.25x1 + 10x1 + 0.15x2 + 12x2 − λ x1 + x2 − 2000 1 1 Fx = 0.5x + 10 − λ = 0, x1 = 2λ − 20 2 2 Fx = 0.3x + 12 − λ = 0, 2 10 3 x = λ − 40 Fλ = −(x1 + x2 − 2000) ( ) (10 ) 3 16 3 2 20 40 2000 2060 386.25 λ λ λ λ − + − = = = So, x1 = 752.5 and x2 = 1247.5. To minimize cost, let x1 = 753 units and x2 = 1247 units. 33. (a) Maximize f (x, y) = 100x0.25 y0.75 subject to the constraint 48x + 36y = 100,000. F(x, y, λ ) = 100x0.25 y0.75 − λ (48x + 36y − 100,000) ( ) 0.75 0.75 0.25 0.25 25 48 0 75 36 0 48 36 100,000 0 x y F x y F x y Fλ x y λ λ − − = − = = − = = − + − = Using Fx , 25 0.75 0.75 48 λ x y − = and Fy , 75 0.25 0.25 , 36 λ x y − = so 0.75 0.75 0.25 0.25 0.75 0.25 0.75 0.25 25 75 48 36 25 75 48 36 3 3 4 4 . x y x y y x x y y x y x − − = = = = Then using Fλ , (48 36(4 ) 100,000) 0 3125. 6 − x + x − = ⇒ x = So, 3125 6 x = and 6250, 3 y = and f (x, y) has a maximum at 3125, 6250 . 6 3 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ The maximum production level is 3125, 6250 147,314 units. 6 3 f ⎛⎜ ⎞⎟ ≈ ⎝ ⎠ 464 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (b) Using Fx , 25x−0.75 y0.75 − 48λ = 0, so 25 0.75 0.75 25 3125 0.75 6250 0.75 1.4731. 48 48 6 3 λ x y − ⎛ ⎞− ⎛ ⎞ = = ⎜ ⎟ ⎜ ⎟ ≈ ⎝ ⎠ ⎝ ⎠ (c) 147,314 + (125,000 − 100,000)λ ≈ 147,314 + 25,000(1.4731) ≈ 181,142 units (d) 147,314 + (350,000 − 100,000)λ ≈ 147,314 + 250,000(1.4731) ≈ 515,589 units 34. (a) Maximize f (x, y) = 100x0.6 y0.4 subject to the constraint 48x + 36y − 100,000 = 0. F(x, y, λ ) = 100x0.6 y0.4 − λ (48x + 36y − 100,000) ( ) 0.4 0.4 0.6 0.6 60 48 0 40 36 0 48 36 100,000 0 x y F x y F x y Fλ x y λ λ − − = − = = − = = − + − = Using Fx , 60 0.4 0.4 48 λ x y − = and Fy , 40 0.6 0.6 , 36 λ x y − = so 0.4 0.4 0.6 0.6 0.4 0.6 0.4 0.6 5 10 4 9 5 10 4 9 45 40 8 . 9 x y x y y x x y y x y x − − = = = = Then using Fy , 48 36 8 100,000 0 1250. 9 x x x ⎛ ⎛ ⎞ ⎞ −⎜ + ⎜ ⎟ − ⎟ = ⇒ = ⎝ ⎝ ⎠ ⎠ So, x = 1250 and 10,000, 9 y = and f (x, y) has a maximum at 1250, 10,000 . 9 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ The maximum production level is 1250, 10,000 119,247 units. 9 f ⎛⎜ ⎞⎟ ≈ ⎝ ⎠ (b) Using Fx , 0.4 0.4 60y 48 0, x − λ = so 0.4 0.4 0.4 10,000 60 60 9 1.192. 48 48 1250 y x λ ⎛ ⎞ ⎜ ⎟ = = ⎜ ⎟ ≈ ⎜⎜ ⎟⎟ ⎝ ⎠ (c) 119,247 + (125,000 − 100,000)(1.192) ≈ 149,047 units (d) 119,247 + (350,000 − 100,000)(1.192) ≈ 417,247 units Section 7.6 Lagrange Multipliers 465 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 35. (a) Minimize f (x, y) = 50x + 100y subject to 100x0.7 y0.3 − 20,000 = 0. F(x, y, λ ) = 50x + 100y − λ (100x0.7 y0.3 − 20,000) ( ) 0.3 0.3 0.7 0.7 0.7 0.3 50 70 0 100 30 0 100 20,000 0 x y F x y F x y Fλ x y λ λ − − = − = = − = = − − = Using Fx , 0.3 50 70 y x λ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ and using Fy , 0.7 30 . 100 y x ⎛ ⎞ λ ⎜ ⎟ = ⎝ ⎠ So, 0.3 0.7 50 30 70 100 3 3 . 14 14 y y y x x x y y x x λ λ ⎛ ⎞ ⎛ ⎞ = = ⎛ ⎞⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ = ⇒ = Since 0.7 0.3 0.3 0.7 0.3 0.3 100 20,000, 100 3 20,000 14 3 200 14 200 317 units. 3 14 x y x x x x = ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = ≈ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ So, 317 units and 3 (317) 68 units. 14 x ≈ y = ≈ (b) ( ) ( ) ( ) 0.7 0.3 0.3 0.3 0.7 0.7 , 100 , 70 , 30 x y f x y x y f x y x y f x y x y − − = = = ( ) ( ) 0.3 0.3 0.7 0.7 , 70 50 , 30 100 70 50 30 100 3 14 x y f x y x y f x y x y y x y x − − = = = = So, 3 14 y = x and the conditions are met. 36. (a) Minimize f (x, y) = 50x + 100y subject to 100x0.4 y0.6 − 20,000 = 0 F(x, y, λ ) = 50x + 100y − λ (100x0.4 y0.6 − 20,000) ( ) 0.6 0.6 0.4 0.4 0.4 0.6 50 40 0 100 60 0 100 20,000 0 x y F x y F x y Fλ x y λ λ − − = − = = − = = − − = Using Fx , 0.6 50 40 y x λ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ and using Fy , 0.4 60 . 100 y x ⎛ ⎞ λ ⎜ ⎟ = ⎝ ⎠ So, 0.4 0.6 50 60 40 100 3 3. 4 4 y y y x x x y y x x λ λ ⎛ ⎞ ⎛ ⎞ = = ⎛ ⎞⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ = ⇒ = Since 0.4 0.6 0.6 0.4 0.6 0.6 100 20,000 100 3 20,000 4 3 200 4 200 238 units. 3 4 x y x x x x = ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = ≈ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ So, 238 units and 3(238) 179 units. 4 x = y = ≈ (b) ( ) ( ) ( ) 0.4 0.6 0.6 0.6 0.4 0.4 , 100 , 40 , 60 x y f x y x y f x y x y f x y x y − − = = = ( ) ( ) 0.6 0.6 0.4 0.4 , 40 50 , 60 100 40 50 60 100 3 4 x y f x y x y f x y x y y x y x − − = = = = So, 3 4 y = x and the conditions are met. 466 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 37. (a) Minimize f (x, y) = 10(2x + 2y) + 4x = 24x + 20y subject to the constraint g(x, y) = 2xy − 6000 = 0. F(x, y, λ ) = 24x + 20y − λ (2xy − 6000) 24 2 0 20 2 0 x y F y F x λ λ = − = ⎫⎪⎬ = − =⎪ ⎭ 12 10 y x λ λ = = Fλ = −2xy + 6000 = 0 2 2 6000 2 10 12 6000 1 25 1 5 xy λ λ λ λ = ⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ = = So, x = 50 and y = 60. To minimize the cost of the fencing, make the fence 50 feet by 120 feet. (b) f (50, 60) = $2400 The minimum cost is $2400. 38. (a) Minimize C(x, y) = 600x + 240y subject to the constraint 2x − 2y = 360. F(x, y, λ ) = 600x + 240y − λ (4xy − 360) 600 4 0 240 4 0 x y F y F x λ λ = − = ⎫⎪⎬ = − =⎪⎭ 150 60 y x λ λ = = Fλ = −(4xy − 360) = 0 2 360 4 60 150 100 10 λ λ λ λ = ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ = = So, x = 6 and y = 15. To minimize the cost, make the partitions 6 feet by 15 feet. (b) The minimum cost is $7200. C(6, 15) = 7200 39. Minimize C(x, y, z) = x + 2y + 3z subject to the constraint 12xyz = 0.13. F(x, y, z, λ ) = x + 2y + 3z − λ (12xyz − 0.13) 1 12 0, 2 12 0, 3 12 0, x y z F yz F xz F xy λ λ λ = − = = − = = − = 12 1 12 2 12 3 yz xz xy λ λ λ = ⎫⎪ = ⎬⎪ = ⎭ 2 3 x y x z = = Fλ = −(12xyz − 0.13) = 0, 12 0.13 2 3 x⎛ x ⎞⎛ x ⎞ = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ 3 3 2 0.13 0.065 0.402 x x = = ≈ So, x = 3 0.065 ≈ 0.402, 1 3 0.065 0.201, 2 y = ≈ 1 3 0.065 0.134, 3 z = ≈ and f (x, y, z) is a minimum at about (0.402, 0.201, 0.134). To minimize the cost, use 0.402 liter of solution x, 0.201 liter of solution y, and 0.134 liter of solution z. 40. F(x, y, z, λ ) = x + 2y + 3z − λ (0.01x2 y2z2 − 0.13) 2 2 2 2 1 0.02 0, 2 0.02 0, x y F xy z F xyz λ λ = − = = − = 2 2 2 2 100 2 2 200 2 xy z x y x yz λ λ = ⎪⎫ ⎬ = = ⎪⎭ Fz = 3 − 0.02λ x2 y2z = 0, 300 = 2λ x2 y2z, x = 3z Fλ = −(0.01x2 y2z2 − 0.13) = 0, x2 y2z2 = 13 2 2 2 13 2 3 x ⎛ x ⎞ ⎛ x ⎞ = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 6 6 468 468 2.786 x x = = ≈ So, x = 6 468 ≈ 2.786, 1 6 468 1.393, 2 y = ≈ 1 6 468 0.929, 3 z = ≈ and f (x, y, z) is a minimum at about (2.786, 1.393, 0.929). To minimize the cost, use 2.786 liters of solution x, 1.393 liters of solution y, and 0.929 liter of solution z. Barn x x x y y Section 7.7 Least Squares Regression Analysis 467 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 41. Maximize G(x, y, z) = 0.05x2 + 0.16xy + 0.25z2 subject to the constraint 9x + 4y + 4z = 400. F(x, y, z, λ ) = 0.05x2 + 0.16xy + 0.25z2 − λ (9x + 4y + 4z − 400) Fx = 0.1x + 0.16y − 9λ = 0 0.16 4 0 0.5 4 0 y z F x F z λ λ = − = ⎫⎪⎬ = − = ⎪⎭ 25 8 x z λ λ = = Fλ = −(9x + 4y + 4z − 400) = 0 From Fλ , you have 9(25λ ) + 4y + 4(8λ ) = 400. So, y = 100 − 64.25λ. From Fx , you have 0.1(25λ ) + 0.16(100 − 64.25λ ) − 9λ = 0. So, 800 839 λ = ≈ 0.953516. So, x ≈ 23.8, y ≈ 38.7, z ≈ 7.6, and f (x, y, z) has a maximum at about (23.8, 38.7, 7.6). To maximize the amount of ice cream you can eat, have f (23.8, 38.7, 7.6) ≈ 190.1 grams. 42. (a) z(1, 2) = 2 (b) z(2, 2) = 8 43. (a) Maximize A = 0.0001t2 pr1.5 subject to the constraint 30t + 12 p + 15r = 2700. ( ) ( ) ( ) 2 1.5 1.5 2 1.5 2 1.5 2 0.5 0.0001 30 12 15 2700 0.0002 30 0 0.0001 12 0, 0.0001 12 0.00015 15 0 30 12 15 2700 0 t p r F A t pr t p r F tpr F t r t r F tpr Fλ t p r λ λ λ λ λ = − + + − = − = = − = = = − = = − + + − = From Ft , you obtain t = 0.8 p. From Fr , you obtain r = 1.2 p. From Fλ , you obtain p = 50. So, t = 40 and r = 60.To maximize the number of responses, spend 30(40) = $1200 on a cable television ad, 12(50) = $600 on a newspaper ad, and $15(60) = $900 on a radio ad. (b) ( )2 ( )( )1.5 A = 0.0001 40 50 60 ≈ 3718 The maximum number of responses is about 3718. Section 7.7 Least Squares Regression Analysis Skills Warm Up 1. ( )2 ( )2 ( )2 ( )2 ( )2 ( )2 2.5 − 1 + 3.25 − 2 + 4.1 − 3 = 1.5 + 1.25 + 1.1 = 2.25 + 1.5625 + 1.21 = 5.0225 2. ( )2 ( )2 ( )2 ( )2 ( )2 ( )2 1.1 − 1 + 2.08 − 2 + 2.95 − 3 = 0.1 + 0.08 + −0.05 = 0.01 + 0.0064 + 0.0025 = 0.0189 3. 2 6 2 4 8 4 6 2 4 4 12 8 4 S a b a b ab S a b a S b a b = + − − − + ∂ = − − ∂ ∂ = − − ∂ 4. 4 2 9 2 6 4 2 8 8 6 2 18 4 2 S a b a b ab S a b a S b a b = + − − − + ∂ = − − ∂ ∂ = − − ∂ 468 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. Linear model sum of the squared errors: ( )2 ( )2 ( )2 ( )2 ( )2 S = 1.2 − 2 + 2.8 − 2 + 4.4 − 4 + 6 − 6 + 7.6 − 8 = 1.6 Quadratic model sum of the squared errors: ( )2 ( )2 ( )2 ( )2 ( )2 S = 2.01 − 2 + 2.76 − 2 + 4.09 − 4 + 6 − 6 + 8.49 − 8 = 0.8259 2. Linear model sum of the squared errors: ( )2 ( )2 ( )2 ( )2 S = 4.1 − 4 + 2.7 − 2 + 1.3 − 1 + −0.1 − 0 = 0.6 Quadratic model sum of the squared errors: ( )2 ( )2 ( )2 ( )2 S = 3.64 − 4 + 1.76 − 2 + 0.36 − 1 + −0.56 − 0 = 0.9104 Skills Warm Up —continued— 5. 5 1 1 2 3 4 5 15 i i = Σ = + + + + = 6. ( ) ( ) ( ) ( ) ( ) ( ) 6 1 2 21 22 23 24 25 26 42 i i = = + + + + + = Σ 7. 4 1 1 1 1 1 1 25 i= i 1 2 3 4 12 Σ = + + + = 8. 3 2 2 2 2 1 1 2 3 14 i i = Σ = + + = 9. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 6 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 1 2 2 2 3 2 4 2 5 2 6 1 0 1 2 3 4 31 i i = − = − + − + − + − + − + − = + + − + − + − + − = Σ 10. ( ) ( ) ( ( )) ( ( )) ( ( )) ( ( )) 5 2 2 2 2 2 1 30 30 1 30 2 30 3 30 4 30 5 29 26 21 14 5 95 i i = − = − + − + − + − + − = + + + + = Σ x-values −3 −2 −1 0 1 Actual y-values 2 2 4 6 8 Linear model, f (x) = 1.6x + 6 1.2 2.8 4.4 6 7.6 Quadratic model, g(x) = 0.29x2 + 2.2x + 6 2.01 2.76 4.09 6 8.49 x-values −3 −1 1 3 Actual y-values 4 2 1 0 Linear model, f (x) = −0.7x + 2 4.1 2.7 1.3 −0.1 Quadratic model, g(x) = 0.06x2 − 0.7x + 1 3.64 1.76 0.36 −0.56 Section 7.7 Least Squares Regression Analysis 469 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 3. Linear model sum of the squared errors: ( )2 ( )2 ( )2 ( )2 S = 11 − 10 + 7.7 − 9 + 4.4 − 6 + 1.1 − 0 = 6.46 Quadratic model sum of the squared errors: ( )2 ( )2 ( )2 ( )2 S = 10 − 10 + 9.25 − 9 + 6 − 6 + 0.25 − 0 = 0.125 4. Linear model sum of the squared errors: ( ( )) ( ( )) ( ) ( ) ( ) 2 2 2 2 2 S = −5 − −4 + −1 − −3 + 1 − 0 + 5 − 5 + 9 − 9 = 6 Quadratic model sum of the squared errors: ( ( )) ( ( )) ( ) ( ) ( ) 2 2 2 2 2 S = −4.16 − −4 + −1.56 − −3 + 0.16 − 0 + 4.44 − 5 + 9.84 − 9 = 3.144 5. The sum of the squared errors is as follows. ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) 2 2 2 2 1 0 2 3 2 2 1 2 2 2 3 2 16 16 2 2 1 2 2 2 3 6 4 S a b a b a b S a b a b a a S a b b a b b b = − + + + + + + − ∂ = − + + − + + − = − ∂ ∂ = − + + + + + − = − ∂ Setting these partial derivatives equal to zero produces a = 1 and 2. 3 b = So, 2. 3 y = x + 6. The sum of the squared errors is as follows. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 3 1 1 3 2 6 3 2 1 2 1 63 2 40 12 2 3 2 1 2 1 23 2 8 8 S a b a b a b a b S a b a b a b a b a a S a b a b a b a b b b = − + + − + − + + − + + − ∂ = − − + − − + − + + − + + − = − ∂ ∂ = − + + − + − + + − + + − = − ∂ Setting these partial derivatives equal to zero produces 3 10 a = and b = 1. So, 3 1. 10 y = x + x-values 0 1 2 3 Actual y-values 10 9 6 0 Linear model, f (x) = −3.3x + 11 11 7.7 4.4 1.1 Quadratic model, g(x) = −1.25x2 + 0.5x + 10 10 9.25 6 0.25 x-values −1 1 2 4 6 Actual y-values −4 −3 0 5 9 Linear model, f (x) = 2.0x − 3 −5 −1 1 5 9 Quadratic model, g(x) = 0.14x2 + 1.3x − 3 −4.16 −1.56 0.16 4.44 9.84 −4 −3 −2 1 2 3 4 −1 −2 −3 1 2 3 4 5 x y (2, 3) (0, 0) (−2, −1) −3 −2 −1 1 2 3 −1 −2 2 3 4 x y (1, 1) (3, 2) (−3, 0) (−1, 1) 470 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 7. The sum of the squared errors is as follows. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 4 1 1 3 4 2 4 2 1 2 3 12 4 24 2 2 4 2 1 2 1 2 3 4 8 2 S a b a b b a b S a b a b a b a b a S a b a b b a b a b b = − + − + − + − + + + + + ∂ = − − + − − − + − + + + = − + ∂ ∂ = − + − + − + − + + + + + = − + − ∂ Setting these partial derivatives equal to zero produces: 12 4 24 4 8 2 a b a b − = − − + = So, a = −2.3 and b = −0.9, and y = −2.3x − 0.9. 8. The sum of the squared errors is as follows. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 5 3 4 2 2 1 1 10 5 3 8 4 2 4 2 1 2 1 92 24 48 2 5 3 2 4 2 2 2 1 2 1 24 8 10 S a b a b a b a b S a b a b a b a b a a b S a b a b a b a b b a b = − + + + − + + + − + + + − + − ∂ = − − + + − − + + − − + + − − + − ∂ = − − ∂ = − + + + − + + + − + + + − + − ∂ = − + + Setting these partial derivatives equal to zero produces: 92 24 48 24 8 10 a b a b − = − + = − So, a = 0.9 and b = 1.45, and y = 0.9x + 1.45. 9. y = 0.8x + 2 10. y = 0.4118x + 3 11. y = −1.1824x + 6.3851 12. y = −0.4519x + 5.6 13. (a) y = 4.13t + 11.6, t = 2 is 2002. (b) 2014: Let t = 14. y = 4.13(14) + 11.6 = 69.42 or $69.42 billion (c) Let y = 85 and find t. 85 4.13 11.6 73.4 4.13 17.77 or 2018 t t t = + = ≈ 14. (a) y = −1.78x + 127.6 (b) Let x = 32.95. y = −1.78(32.95) + 127.6 = 70 (c) Let y = 83 and find x. 83 1.78 127.6 44.6 1.78 $25 x x x = − + − = − ≈ 15. (a) y = 0.138x + 22.1 (b) Let x = 160. y = 0.138(160) + 22.1 = 44.18 bushels acre 16. (a) y = 0.22x − 7.5matches (iv). The positive slope, y-intercept, and/or one of the data points can be used. (b) y = −0.35x + 11.5 matches (i). The negative slope, y-intercept, and/or one of the data points can be used. (c) y = 0.09x + 19.8 matches (iii). The positive slope, y-intercept, and/or one of the data points can be used. (d) y = −1.29x + 89.8 matches (ii). The negative slope, y-intercept, and/or one of the data points can be used. 17. Positive correlation r ≈ 0.9981 −5 −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 1 2 3 4 5 x y (−1, 1) (0, −1) (1, −3) (−2, 4) −6 −5 −4 −3 −1 −1 −2 −3 −4 2 x y (−4, −2) (−2, −1) (−5, −3) (−1, 1) 14 16 12 10 8 6 4 2 1 2 3 4 5 6 x y Section 7.8 Double Integrals and Area in the Plane 471 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 18. Negative correlation r ≈ −0.9652 19. No correlation r = 0 20. Positive correlation r ≈ 0.9276 21. No linear correlation r ≈ 0.0750 22. Negative correlation r ≈ −0.9907 23. False, the slope is positive, which means there is a positive correlation. 24. True 25. True 26. True 27. True 28. False, there is a strong negative correlation, so the regression line will fit the data well. 29. Answers will vary. Section 7.8 Double Integrals and Area in the Plane Skills Warm Up 1. ] 1 1 0 0 ∫ dx = x = 1 2. ] 2 2 0 0 ∫ 3 dy = 3y = 6 3. ( ) ( ) 4 2 3 4 3 3 1 1 2 2 2 3 3 3 ∫ 2x dx = x ⎤⎦ = ⎡⎣ 4 − 1 ⎤⎦ = 42 4. 1 3 4 1 0 0 1 1 2 2 ∫ 2x dx = x ⎤⎦ = 5. ( ) ( ) ( ) 2 3 4 2 2 1 1 1 4 1 4 19 4 2 4 4 4 4 8 1 4 x − x + dx = ⎡⎣ x − x + x⎤⎦ = − + − − + = ∫ 6. ( ) 2 2 3 2 0 0 1 16 3 3 ∫ 4 − y dy = ⎡⎣4y − y ⎤⎦ = 7. 2 2 1 2 1 2 2 2 2 1 7 7 14 7 7 dx x x = − ⎤ = − + = ⎥⎦ ∫ 8. 4 4 1 1 2 dx 4 x 8 4 4 x = ⎤ = − = ∫ ⎦ 9. ( ) 2 2 2 0 2 0 2 ln 1 1 ln 5 ln 1 ln 5 1.609 x dx x x = + ⎤+ ⎦ = − = ≈ ∫ x 4 6 8 2 2 4 6 8 y 12 10 8 6 4 2 1 2 3 4 5 6 x y x 2 3 4 1 1 2 3 4 y 36 30 24 18 12 6 1 2 3 4 5 6 x y 1 2 3 4 5 2 4 6 8 10 x y 472 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. ( ) 2 2 0 0 2 2 3 2 2 x x x y dy xy y x ⎡ ⎤ − = ⎢ − ⎥ = ⎣ ⎦ ∫ 2. ( ) ( ) ( ) 2 0 0 2 2 2 5 2 5 21 2 2 5 8 8 8 0 y y x y dx x xy y y y + = ⎡⎣ + ⎤⎦ = + − = ∫ 3. ( ) 2 2 2 3 2 1 2 2 2 2 x x x x y dy y x x x x x x ⎤ = = − = − ⎥⎦ ∫ 4. 2 2 1 1 ln ln 2 y y dx y x y y y x ∫ = ⎤⎦ = 5. ( ) ( ) ( ) 2 2 3 2 2 2 4 3 2 4 3 2 6 2 2 16 2 2 2 16 y y x y y dx x y xy y y y y y y y y + = ⎡⎣ + ⎤⎦ = + − + = + − − ∫ 6. ( ) ( ) ( ) 3 4 2 4 4 5 2 5 2 1 4 1 4 1 4 4 2 2 64 32 2 64 32 x x xy y dy xy y x x x x x x + = ⎡⎣ + ⎤⎦ = + − + = + − − ∫ 7. ( ) ( ) ( ) 2 2 2 33 3 5 2 3 2 5 9 9 5 52 32 3 x x x x x y dy xy y x x x x x x x x + = ⎡⎣ + ⎤⎦ = + − + = − − + + ∫ 8. ( ) ( ) ( ) ( ) 1 2 1 2 2 2 3 2 1 2 1 2 2 3 2 2 2 2 2 2 2 2 1 3 23 23 23 1 2 1 1 1 3 1 1 2 y y y y x y dx x xy y y y y y y y y − − − − − − + = ⎡⎣ + ⎤⎦ = − + − = − ⎡ − + ⎤ ⎣ ⎦ = − + ∫ Skills Warm Up —continued— 10. ( ) ( ) ( ) ( ) 2 2 1 ln 1 1 ln 1 ln 2 1 ln 1 0.541 e e dy y y e e = − ⎤⎦ − = − − − = − ≈ ∫ 11. 2 2 2 2 1 1 0 0 5 1 2 1 1 2 2 72.847 xex dx ex e e + = + ⎤⎥⎦ = − ≈ ∫ 12. 1 2 2 1 0 0 2 1 2 1 1 2 2 0.432 e y dy e y e − − − = − ⎤⎦ = − + ≈ ∫ 13. y = x, y = 0, x = 3 14. y = x, y = 3, x = 0 15. y = 4 − x2 , y = 0, x = 0 16. y = x2 , y = 4x 1 2 3 4 1 2 3 4 x y 1 2 3 4 1 2 3 4 x y 1 2 3 4 5 4 8 12 16 20 x y 1 2 3 4 1 2 3 4 x y Section 7.8 Double Integrals and Area in the Plane 473 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 9. ( ) ( ) ( ) 2 1 1 2 2 3 ln ln 2 ln ln 1 2 2 2 ey ey y y x y x dx x y e y y ⎤ = ⎥ ⎥⎦ = − = ∫ 10. ( ) 3 3 2 2 2 2 1 1 10 1 10 1 y y xy dx y x x y y y y y = + ⎤ + ⎦ = − + = − + ∫ 11. ( ) ( ) 2 2 1 2 1 0 0 0 0 1 0 2 1 0 2 2 2 2 3 x y dy dx xy y dx x dx x x ⎡ ⎤ + = ⎢ + ⎥ ⎣ ⎦ = + = ⎡⎣ + ⎤⎦ = ∫ ∫ ∫ ∫ 12. ( ) ( ) ( ) 2 2 2 2 2 2 0 0 0 0 2 2 0 3 2 0 1 3 6 6 2 6 2 6 56 3 x dy dx x y dx x dx x x − = − ⎤⎦ = − = ⎡⎣ − ⎤⎦ = ∫ ∫ ∫ ∫ 13. 3 4 3 2 4 0 0 0 0 3 0 2 3 0 1 2 8 4 36 xy dx dy x y dy y dy y = ⎡⎣ ⎤⎦ = = ⎡⎣ ⎤⎦ = ∫ ∫ ∫ ∫ 14. ( ) 3 2 1 2 2 2 1 2 1 2 1 2 1 2 1 3 1 1 3 4 16 3 4 16 3 3 8 x y dy dx x y y dx x dx x x − − − − − − ⎡ ⎤ − = ⎢ − ⎥ ⎣ ⎦ = ⎛ − ⎞ ⎜ ⎟ ⎝ ⎠ ⎡ ⎤ = ⎢ − ⎥ ⎣ ⎦ = − ∫ ∫ ∫ ∫ 15. 2 6 2 2 6 3 2 3 0 0 0 0 2 5 0 6 2 0 6 64 x x x dy dx x y dx x dx x = ⎤⎦ = = ⎤⎦ = ∫ ∫ ∫ ∫ 16. ] ( ) 2 2 2 2 2 2 0 3 2 6 0 2 3 6 2 2 3 0 3 4 2 0 3 3 24 12 8 3 16 y y y y y y y y y dx dy xy dy y y dy y y − − − − = = − = ⎡⎣ − ⎤⎦ = ∫ ∫ ∫ ∫ 17. ( ) 1 1 2 0 0 0 0 1 2 0 3 1 0 2 3 2 2 1 2 y y x y dx dy x xy dy y dy y ⎡ ⎤ + = ⎢ + ⎥ ⎣ ⎦ = ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ = ∫ ∫ ∫ ∫ 18. ( ) ( ) 1 2 2 1 2 2 2 0 0 0 0 2 3 0 2 4 2 0 5 5 2 5 2 5 2 2 4 52 4 5 2 y y xy dx dy x y dy y y dy y y − − ⎤ − = − ⎥⎦ − = − ⎡ ⎤ = − ⎢ − ⎥ ⎣ ⎦ = − − = ∫ ∫ ∫ ∫ 19. ( ) ( ) ( ) 1 3 2 2 1 2 3 3 0 0 0 0 1 3 3 0 1 3 0 4 2 1 0 3 21 2 2 3 3 1 3 9 27 3 36 3 9 x x x y dydx xy y y dx x x xdx x xdx x x + + = ⎡⎣ + + ⎤⎦ = + + = + = ⎡⎣ + ⎤⎦ = ∫ ∫ ∫ ∫ ∫ 474 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 20. ( ) 3 2 2 4 1 1 2 2 2 1 2 1 3 0 0 0 0 1 2 2 2 2 20 5 13 3 3 2 3 6 y y y y x y dx dy x x xy dy y y dy y y ⎡ ⎤ ⎛ ⎞ ⎡ ⎤ + + = ⎢ + + ⎥ = ⎜ + ⎟ = ⎢ + ⎥ = ⎣ ⎦ ⎝ ⎠ ⎣ ⎦ ∫ ∫ ∫ ∫ 21. ( ) ( ) 1 2 1 2 0 0 0 0 1 2 0 3 2 1 2 0 1 2 1 3 1 3 1 1 1 2 1 x x x dy dx x y dx x x dx x − = − ⎤⎦ = − − − = − − ⎤⎥⎦ = ∫ ∫ ∫ ∫ 22. 4 4 0 0 2 0 2 0 4 0 2 2 4 0 2 2 1 1 2 1 ln 1 ln 17 2.833 x x dy dx y dx x x x dx x x = ⎤+ + ⎥⎦ = + = + ⎤⎦ = ≈ ∫ ∫ ∫ ∫ 23. Because (for a fixed x) ( ) 2 2 0 lim 2 2 , x y b x b e− + e− →∞ − ⎤ = ⎦ you have ( ) 2 2 0 0 0 2 0 2 lim 4 4. x y x x b b e dydx e dx e ∞ ∞ − + ∞ − − →∞ = = − ⎤⎦ = ∫ ∫ ∫ 24. Because (for a fixed y) ( 2 2) 2 0 1 1 2 2 lim , b x y y b ye ye − + − →∞ ⎤ − = ⎥⎦ you have ( 2 2) 2 0 0 0 2 0 1 2 1 4 1 4 lim . x y y b y b xye dx dy ye dy e ∞ ∞ − + ∞ − − →∞ = = − ⎤⎥⎦ = ∫ ∫ ∫ 25. [ ] [ ] 8 3 0 0 8 3 0 0 8 0 8 0 3 3 24 A dy dx y dx dx x = = = = = ∫ ∫ ∫ ∫ 26. ] 3 3 3 3 1 1 1 1 A = ∫ ∫ dy dx = ∫ 2 dx = 2x = 4 27. [ ] 4 0 0 4 0 0 4 0 2 4 0 1 2 8 x x A dy dx y dx x dx x = = = = ⎡⎣ ⎤⎦ = ∫ ∫ ∫ ∫ 28. [ ] 6 2 0 0 6 2 0 0 6 0 6 2 0 2 1 4 9 x x A dy dx y dx x dx x = = = = ⎡ ⎤ ⎢⎣ ⎥⎦ = ∫ ∫ ∫ ∫ 29. ( ) 2 4 2 0 0 2 2 0 3 2 0 4 4 3 16 3 x A dy dx x dx x x − = = − ⎡ ⎤ = ⎢ − ⎥ ⎣ ⎦ = ∫ ∫ ∫ 30. [ ] 4 0 0 4 0 0 4 0 3 2 4 0 23 16 3 x x A dy dx y dx x dx x = = = = ⎡⎣ ⎤⎦ = ∫ ∫ ∫ ∫ 31. ( ) 3 9 2 3 0 3 2 3 3 3 3 9 9 3 36 x A dy dx x dx x x − − − − = = − ⎡ ⎤ = ⎢ − ⎥ ⎣ ⎦ = ∫ ∫ ∫ −2 −1 1 2 3 2 4 6 8 10 x y y = 9 − x 2 Section 7.8 Double Integrals and Area in the Plane 475 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 32. ( ) 1 0 3 2 1 3 2 0 2 1 5 2 0 2 2 5 1 10 x x A dy dx x x dx x x = = − ⎡ ⎤ = ⎢ − ⎥ ⎣ ⎦ = ∫ ∫ ∫ 33. ( ) ( ) 2 5 0 32 2 0 2 0 2 2 0 32 5 2 5 5 5 5 4 5 y y A dx dy y ydy y dy y y − + = = − + − = − + ⎡ ⎤ = ⎢− + ⎥ ⎣ ⎦ = ∫ ∫ ∫ ∫ 34. The point of intersection of the two graphs is found by equating y = 9 x and y = x, which yields x = y = 3. ( ) 3 9 9 0 0 3 0 3 9 0 3 2 3 9 3 0 9 9 ln 2 9 9 ln 9 ln 3 2 9 9 ln 3 14.388 2 x x A dy dx dy dx x dx dx x x x = + = + ⎤ = ⎥ + ⎤⎦ ⎦ = + − = + ≈ ∫ ∫ ∫ ∫ ∫ ∫ 35. The point of intersection of the two graphs is found by equating y = 2x and y = x, which yields x = y = 0. ( ) 2 2 0 2 0 2 0 2 2 0 2 2 2 x x A dy dx x x dx x dx x = = − = ⎤ = = ⎥⎦ ∫ ∫ ∫ ∫ 36. The points of intersection of the two graphs are found by equating y = 4 − x2 and y = x + 2,which yields (−2, 0) and (1, 3). [ ] ( ) 1 4 2 2 2 1 4 2 2 2 1 2 2 3 2 1 2 1 1 3 2 9 2 2 2 x x x x A dy dx y dx x x dx x x x − − + − − + − − = = = − − + = ⎡⎣− − + ⎤⎦ = ∫ ∫ ∫ ∫ 37. 1 2 1 0 0 0 2 1 2 0 0 0 2 2 2 dy dx dx dx dy dy = = = = ∫ ∫ ∫ ∫ ∫ ∫ 38. ] ] ] ] 2 4 2 4 2 2 1 2 1 2 1 1 4 2 4 2 4 4 2 1 2 1 2 2 2 2 2 2 dx dy x dy dy y dy dx y dx dx x = = = = = = = = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ 39. ( ) 1 2 1 2 1 0 2 0 0 2 2 2 1 y ∫ ∫ dx dy = ∫ − y dy = ⎡⎣ y − y ⎤⎦ = 2 2 2 2 2 0 0 0 0 1 2 4 x dy dx x dx x ⎤ = = = ⎥⎦ ∫ ∫ ∫ x 1 1 y = x y = x3/2 y x y 1 2 3 4 5 1 2 3 4 5 (3, 2) x = 3 y 2 x = − y + 5 x 4 6 2 4 6 8 2 −2 (3, 3) (9, 1) y = x y = 9x y x y − 1 1 2 3 4 5 − 1 1 2 3 4 5 y = x y = 2x 2 2 x y x 2 3 4 1 2 3 4 1 y 1 2 1 2 x (2, 1) x 2 y = y −3 −1 1 3 −1 1 2 3 5 x y (−2, 0) (1, 3) y = x + 2 y = 4 − x2 476 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 40. ( ) 4 4 4 3 2 0 0 0 0 3 2 2 4 2 2 0 2 0 0 2 16 3 3 4 4 16 3 3 x y dy dx x dx x dx dy y dy y y = = ⎤ = ⎥⎦ ⎡ ⎤ = − = ⎢ − ⎥ = ⎣ ⎦ ∫ ∫ ∫ ∫ ∫ ∫ 41. 2 2 2 1 2 0 2 0 0 1 2 1 2 1 0 0 0 0 1 1 2 4 2 1 x y dy dx x dx x x dx dy y dy y ⎛ ⎞ ⎡ ⎤ = ⎜ − ⎟ = ⎢ − ⎥ = ⎝ ⎠ ⎣ ⎦ = = ⎤⎦ = ∫ ∫ ∫ ∫ ∫ ∫ 42. ( ) 4 2 4 0 0 4 3 2 0 2 2 2 3 8 3 x dy dx x dx x x = − = ⎡ − ⎤ ⎢⎣ ⎥⎦ = ∫ ∫ ∫ 2 3 2 2 22 0 0 0 0 8 3 3 y dx dy y dy y ⎤ = = = ⎥⎦ ∫ ∫ ∫ 43. ( ) 1 3 1 3 2 0 2 0 3 1 4 3 0 3 4 3 5 12 y y dx dy y y dy y y = − ⎡ ⎤ = ⎢ − ⎥ ⎣ ⎦ = ∫ ∫ ∫ ( ) 1 1 3 0 3 0 4 1 3 2 0 2 3 4 5 12 x x dy dx x x dx x x = − ⎡ ⎤ = ⎢ − ⎥ ⎣ ⎦ = ∫ ∫ ∫ 44. ( ) 2 4 2 2 2 2 0 2 3 2 2 4 4 32 3 3 y dx dy y dy y y − − − − = − ⎡ ⎤ = ⎢ − ⎥ = ⎣ ⎦ ∫ ∫ ∫ ( ) 4 4 4 0 4 0 4 3 2 0 2 4 4 4 32 3 3 x x dy dx x dx x − − − = − = − − ⎤ = ⎥⎦ ∫ ∫ ∫ x 2 3 4 1 2 3 4 1 y = x (4, 2) y 2 2 x (2, 1) y 1 1 x 2 y = y x 2 3 1 2 3 4 1 − 1 y = x (4, 2) 2 2 x x = y2 (1, 1) y 1 1 x = 3 y x 2 1 2 3 1 − 1 − 2 x = 4 − y2 y Section 7.8 Double Integrals and Area in the Plane 477 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 45. (a) 3 3 2 0 x y ∫ ∫ e dx dy cannot be evaluated in the order as given since no antiderivative for ex2 can be found. The region bounded by y ≤ x ≤ 3 and 0 ≤ y ≤ 3 is rewritten as 0 ≤ y ≤ x and 0 ≤ x ≤ 3 so that the integral can be changed to 3 2 0 0 . ∫ ∫ x ex dy dx ( ) 3 2 3 2 0 0 0 0 3 2 0 2 3 0 9 1 2 1 2 1 4051.042 x x x x x x e dy dx e y dx xe dx e e = ⎡ ⎤ ⎢⎣ ⎥⎦ = = ⎡ ⎤ ⎢⎣ ⎥⎦ = − ≈ ∫ ∫ ∫ ∫ (b) 2 2 2 0 y x e− dy dx ∫ ∫ cannot be evaluated in the order as given since no antiderivative for e− y2 can be found. The region bounded by x ≤ y ≤ 2 and 0 ≤ x ≤ 2 is rewritten as 0 ≤ x ≤ y and 0 ≤ y ≤ 2 so that the integral can be changed to 2 2 0 0 . y e− y dx dy ∫ ∫ ( ) 2 2 2 2 0 0 0 0 2 2 0 2 2 0 4 1 2 1 2 1 0.491 y y y y y y e dxdy xe y dy ye dy e e − − − − − = ⎡ ⎤ ⎢⎣ ⎥⎦ = = ⎡− ⎤ ⎢⎣ ⎥⎦ = − − ≈ ∫ ∫ ∫ ∫ 46. (a) 2 2 0 2 Area y y = ∫ ∫ dx dy (b) 4 0 2 Area x x = ∫ ∫ dy dx 47. 1 2 2 2 0 0 ∫ ∫ e−x − y dx dy ≈ 0.6588 48. ( ) 2 2 3 2 0 2 3 15.8476 x x ∫ ∫ x + y dy dx ≈ 49. 2 1 0 8.1747 ∫ ∫ x exy dy dx ≈ 50. ( ) 2 2 1 ln 2.0006 y y ∫ ∫ x + y dx dy ≈ 51. 1 1 2 0 1 0.4521 x ∫ ∫ − x dy dx ≈ 52. 3 2 0 0 1 24.3082 x ∫ ∫ x + x dy dx ≈ 53. 2 4 2 4 0 4 2 2 2 1.1190 1 x x xy dy dx x y − − ≈ + + ∫ ∫ 54. ( )( ) 4 0 0 2 2.5903 1 1 y dx dy x y ≈ + + ∫ ∫ 1 2 4 1 2 4 x y y = 3 y = 0 x = y x = 3 1 3 1 3 x y y = 2 y = x x = 2 x = 0 478 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 55. True 2 2 1 2 1 1 1 2 1 1 2 0 0 2 y dy dx y dx dx − − − − − ⎤ = = = ⎥⎦ ∫ ∫ ∫ ∫ ] 1 2 1 2 1 2 1 1 2 1 2 1 1 − − y dx dy − xy − dy − 4y dy 2y − 0 ∫ ∫ = ∫ = ∫ = ⎤⎦ = 56. True ] 2 5 5 6 5 6 5 2 1 2 1 2 2 5 5 105 2 2 x dy dx yx dx x dx x ⎤ = = = = ⎥⎦ ∫ ∫ ∫ ∫ 2 5 6 6 5 6 6 1 2 1 1 2 1 21 21 105 2 2 2 2 x dx dy x dy dy y ⎤ ⎤ = = = = ⎥ ⎥⎦ ⎦ ∫ ∫ ∫ ∫ Section 7.9 Applications of Double Integrals Skills Warm Up 1. 2. 3. 4. 5. ] ] 1 2 1 2 1 1 0 1 0 1 0 0 ∫ ∫ dy dx = ∫ y dx = ∫ dx = x = 1 6. ] ] 3 3 3 3 3 3 0 1 0 1 0 0 ∫ ∫ dx dy = ∫ x dy = ∫ 2 dy = 2y = 6 7. ] 1 1 0 0 0 0 1 2 3 1 0 0 1 3 1 3 x x x dy dx xy dx x dx x = = = ⎤⎦ = ∫ ∫ ∫ ∫ 8. ( ) 4 4 4 2 0 1 0 1 0 3 2 4 0 1 1 3 2 40 3 y y y dx dy xy dy y y dy y y = ⎤ = − ⎥⎦ = ⎡⎣ − ⎤⎦ = ∫ ∫ ∫ ∫ 9. ] ( ) 3 2 3 2 1 1 3 2 1 3 2 3 1 23 1 3 28 3 2 2 2 2 9 x x x x dy dx y dx x x dx x x = = − = ⎡⎣ − ⎤⎦ = + = ∫ ∫ ∫ ∫ 10. ] ( ) 1 2 2 1 2 2 0 0 1 2 0 3 2 1 0 1 1 3 2 7 6 2 2 x x x x dy dx y dx x x dx x x x − + − + = = − + − = ⎡⎣− + − ⎤⎦ = ∫ ∫ ∫ ∫ 2 1 1 2 x y 4 3 2 1 1 2 3 4 x y 10 8 6 4 2 1 2 3 4 5 x y 4 3 2 1 1 2 3 4 x y Section 7.9 Applications of Double Integrals 479 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. ( ) ( ) 2 1 2 2 1 0 0 0 0 2 0 2 2 0 32 3 4 3 2 3 2 2 10 x y dy dx xy y dx x dx x x + = ⎡⎣ + ⎤⎦ = + = ⎡⎣ + ⎤⎦ = ∫ ∫ ∫ ∫ 2. ( ) ( ) 3 1 3 2 1 0 0 0 0 3 0 2 3 0 2 6 2 3 2 3 3 18 x y dy dx xy y dx x dx x x + = ⎡⎣ + ⎤⎦ = + = ⎡⎣ + ⎤⎦ = ∫ ∫ ∫ ∫ 3. ( ) ( ) 1 1 2 1 1 2 2 22 1 0 1 0 1 2 2 1 1 2 4 1 3 5 1 1 1 2 1 2 1 2 1 1 1 2 3 5 2 15 1 x x x y dy dx x y dx x x dx x x dx x x − − − − − − − = ⎡⎣ ⎤⎦ = − = − = ⎡⎣ − ⎤⎦ = ∫ ∫ ∫ ∫ ∫ 4. ( ) ( ) 4 2 2 4 2 2 3 2 0 0 0 0 2 2 3 0 2 2 4 0 3 4 3 1 4 24 32 3 x x xy dy dx xy dx x x dx x − − ⎤ = ⎥⎦ = − = − − ⎤⎥⎦ = ∫ ∫ ∫ ∫ 5. ( ) ( ) 1 2 2 1 3 2 0 2 0 2 1 3 6 4 0 4 7 5 1 0 1 3 4 1 3 3 1 1 1 3 21 5 3 35 y y y y x y dx dy x xy dy y y y dy y y y + = ⎡⎣ + ⎤⎦ = − − = ⎡⎣ − − ⎤⎦ = ∫ ∫ ∫ ∫ 6. ( ) 2 3 6 3 6 0 2 0 2 6 2 0 2 3 6 0 2 9 3 5 2 8 9 3 5 2 2 24 36 y y x y dx dy x xy dy y y dy y y y ⎡ ⎤ + = ⎢ + ⎥ ⎣ ⎦ ⎛ ⎞ = ⎜ + − ⎟ ⎝ ⎠ ⎡ ⎤ = ⎢ + − ⎥ ⎣ ⎦ = ∫ ∫ ∫ ∫ 2 1 1 2 x y 4 3 2 1 1 2 3 4 x y x −1 1 1 2 2 3 4 3 y = 4 − x2 y 1 1 x y x = y2 x = y x 2 4 6 6 4 2 (3, 6) y y = 2x x = 3 −1 1 −1 x y y = 1 − x2 480 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 7. ] 2 2 2 2 2 2 2 2 2 2 2 2 (area of circle) a a x a a x a a x a a x a a dy dx y dx a x dx π a − − − − − − − − − = = − = ∫ ∫ ∫ ∫ 8. 2 2 2 2 0 0 0 2 1 area of circle 4 4 a a x a dy dx a x dx π a − = − = ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ∫ ∫ ∫ 9. 3 5 5 3 0 0 0 0 2 5 3 5 3 0 0 0 0 3 0 3 2 0 2 25 2 25 4 225 4 xy dy dx xy dx dy xy dy dx xy dx x dx x = ⎤ = ⎥⎦ = = ⎤⎥⎦ = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ 10. ( ) 5 25 2 5 25 2 0 0 0 0 25 2 5 25 2 5 2 0 0 0 0 5 2 0 3 5 0 2 2 2 2 2 25 25 3 250 3 x y y y x dy dx x dx dy x dx dy x dy y dy y y − − − − = ⎤ = ⎥⎦ = − ⎡ ⎤ = ⎢ − ⎥ ⎣ ⎦ = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ 11. 2 2 2 4 2 0 2 2 0 2 2 2 2 2 2 2 x y x y y y dy dx y dx dy y dx dy x y x y x y = + + + + ∫ ∫ ∫ ∫ ∫ ∫ ( ) ( ) ( ) 2 2 2 2 2 2 0 2 2 0 2 2 2 2 2 0 0 0 1 ln 2 1 ln 5 ln 2 1 ln 5 1 ln 5 ln 5 0.916 2 2 2 2 2 2 x x x x y dy dx x y dx x y x x dx dx x = + ⎤+ ⎥⎦ ⎡ ⎤ ⎛ ⎞ ⎤ = ⎣ − ⎦ = = ⎜ ⎟ ⎥ = ≈ ⎝ ⎠ ⎦ ∫ ∫ ∫ ∫ ∫ 12. ( ) ( ) ( ) 4 24 0 0 2 0 2 2 4 4 2 0 0 2 0 2 0 4 0 2 2 4 0 1 1 1 2 1 2 1 1 ln 1 4 1 ln 17 4 0.708 x y x x y dy dx y dx dy x x y dy dx y dx x x x dx x x = + + ⎤⎥= + + ⎥⎦ = + = + ⎤⎦ = ≈ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ 13. 2 4 0 0 4 2 0 0 2 0 2 2 0 2 2 2 4 V y dx dy xy dy y dy y = = ⎤⎥⎦ = = ⎤⎦ = ∫ ∫ ∫ ∫ x −a a −a a y y = a2 − x2 a a x y y = a2 − x2 x 2 3 4 1 2 3 4 1 y Section 7.9 Applications of Double Integrals 481 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 14. ( ) [ ] ( ) 2 4 0 0 2 4 0 0 2 0 2 2 0 6 2 6 2 24 8 24 4 32 V y dx dy x xy dy y dy y y = − = − = − = ⎡⎣ − ⎤⎦ = ∫ ∫ ∫ ∫ 15. ( ) ( ) 6 23 4 0 0 6 2 3 4 2 0 0 6 2 0 6 3 2 0 12 2 3 4 1 12 2 3 4 2 1 2 6 6 1 6 18 12 x x V x y dy dx y xy y dx x x dx x x x − + − + ⎛ − − ⎞ = ⎜ ⎟ ⎝ ⎠ = ⎡ − − ⎤ ⎢⎣ ⎥⎦ = ⎛ − + ⎞ ⎜ ⎟ ⎝ ⎠ = ⎡ − + ⎤ ⎢⎣ ⎥⎦ = ∫ ∫ ∫ ∫ 16. ( ) ( ) ( ) 2 2 0 0 2 2 2 0 0 2 2 0 2 3 0 2 2 2 1 2 2 1 2 4 6 3 x x V x y dy dx y xy y dx x dx x − − = − − ⎡ ⎤ = ⎢ − − ⎥ ⎣ ⎦ = − = − − ⎤ = ⎥⎦ ∫ ∫ ∫ ∫ 17. ( ) 2 0 0 2 2 0 0 2 2 0 3 2 2 0 4 4 2 4 3 2 2 4 2 y y V x y dx dy x x xy dy y y dy y y = − − ⎡ ⎤ = ⎢ − − ⎥ ⎣ ⎦ ⎛ ⎞ = ⎜ − ⎟ ⎝ ⎠ ⎡ ⎤ = ⎢ − ⎥ = ⎣ ⎦ ∫ ∫ ∫ ∫ 18. ( ) 1 0 0 1 2 0 0 1 3 0 2 4 1 0 1 2 2 2 8 3 8 y y V xy dx dy x x y dy y y dy y y = − ⎡ ⎤ = ⎢ − ⎥ ⎣ ⎦ ⎛ ⎞ = ⎜ − ⎟ ⎝ ⎠ ⎡ ⎤ = ⎢ − ⎥ ⎣ ⎦ = ∫ ∫ ∫ ∫ 19. ( ) 1 1 2 2 0 0 3 1 1 2 0 0 1 2 0 1 2 0 3 1 0 4 4 4 4 3 4 4 1 3 4 11 3 4 11 40 3 3 3 V x y dy dx y x y y dx x dx x dx x x = − − ⎡ ⎤ = ⎢ − − ⎥ ⎣ ⎦ = ⎡ − − ⎤ ⎢⎣ ⎥⎦ = ⎛ − ⎞ ⎜ ⎟ ⎝ ⎠ ⎡ ⎤ = ⎢ − ⎥ = ⎣ ⎦ ∫ ∫ ∫ ∫ ∫ 20. ( ) 1 2 0 0 1 2 0 3 2 1 2 0 1 1 3 3 1 1 1 x V x dy dx x x dx x = − = − = − − ⎤ = ⎥⎦ ∫ ∫ ∫ 21. ( ) 3 2 0 0 3 2 2 0 0 3 3 0 4 3 0 1 2 1 2 1 2 81 2 4 x x V xy dy dx xy dx x dx x = = ⎡⎣ ⎤⎦ = = ⎡⎣ ⎤⎦ = ∫ ∫ ∫ ∫ 22. ] 4 0 0 4 0 0 4 2 0 3 4 0 64 3 3 x x V x dy dx xy dx x dx x = = = ⎤ = = ⎥⎦ ∫ ∫ ∫ ∫ x 2 3 4 1 2 3 4 1 y x 1 1 − 1 2 2 3 3 4 4 5 5 6 2 3 y = − x + 4 y 1 1 2 2 y = 2 − x y x 1 1 2 2 y = x y x x 1 1 y = x y −2 2 −2 2 x y x 1 1 y 1 2 3 4 1 2 3 4 x y 1 2 3 4 5 6 1 2 3 4 5 6 x y y = 2x 482 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 23. ( ) ( ) 2 2 2 0 0 2 2 2 0 0 2 3 2 0 4 3 2 2 0 1 2 9 4 3 2 134 3 9 9 2 9 18 18 x x V x dy dx y x y dx x x x dx x x x x + + = − = ⎡⎣ − ⎤⎦ = − − + + = ⎡⎣− − + + ⎤⎦ = ∫ ∫ ∫ ∫ 24. ( ) ( ) ( ) 2 4 2 0 0 2 4 2 2 0 0 2 2 2 0 3 2 2 2 3 0 1 2 1 2 1 1 16 3 6 3 4 2 4 2 x x V x y dy dx xy y dx x x x dx x x x − − = + = ⎡⎣ + ⎤⎦ = − + − = ⎡− − + − ⎤ = ⎢⎣ ⎥⎦ ∫ ∫ ∫ ∫ 25. P ( ) ( ) ( ) ( ) 2 2 0 0 3 2 2 2 0 0 2 0 2 2 2 0 120,000 2 60,000 2 60,000 1 1 4 2 60,000 1 1 4 2 10,000 people dy dx x y x y dx dx x x x x − = + + = − + + ⎤⎦ ⎛ ⎞ = − ⎜ − ⎟ ⎜ + + ⎟ ⎝ ⎠ = ⎡ − ⎤ ⎢⎣ + + ⎥⎦ = ∫ ∫ ∫ ∫ 26. P ( ) ( ) 0 4 2 0 2 0 4 2 0 2 0 2 4 2 0 0 2 0 2 2 5000 1 2 1250 4 1 2 1250 ln 1 2 1250 ln 33 1250 ln 33 1250 ln 33 1 3779 people y y y y y xe dx dy x e x dx dy x e x dy e dy e e − − − − − − = + = + = ⎡⎣ + ⎤⎦ = = ⎡⎣ ⎤⎦ = − ≈ ∫ ∫ ∫ ∫ ∫ ∫ 27. [ ] 5 3 0 0 5 3 2 0 0 5 0 5 0 1 15 1 1 15 2 1 30 1 30 32 Average 9 9 y dy dx y dx dx x = = ⎡⎣ ⎤⎦ = = = ∫ ∫ ∫ ∫ 28. 4 2 0 0 2 2 4 0 0 4 0 2 4 0 Average 1 8 1 8 2 1 2 8 2 8 xy dy dx xy dx x dx x = ⎤ = ⎥⎦ = ⎤ = = ⎥⎦ ∫ ∫ ∫ ∫ 29. ( ) 2 2 2 2 0 0 3 2 2 2 0 0 2 2 0 2 3 0 Average 1 4 1 4 3 1 8 2 4 3 1 8 2 4 3 3 8 3 x y dx dy x xy dy y dy y y = + ⎡ ⎤ = ⎢ + ⎥ ⎣ ⎦ = ⎛ + ⎞ ⎜ ⎟ ⎝ ⎠ = ⎡ + ⎤ ⎢⎣ ⎥⎦ = ∫ ∫ ∫ ∫ 30. ( ) ( ) 1 1 1 2 0 0 0 0 0 2 1 2 1 2 0 0 2 1 2 Average 2 2 2 2 2 2 1 1 2.952 y x y x y y y y y y y y e dx dy e dy e e dy e e e e e e e = + = + ⎤ = − ⎦ = ⎡⎣ − ⎤⎦ = ⎡⎣ − ⎤⎦ = − + = − ≈ ∫ ∫ ∫ ∫ x 2 1 2 1 y = 4 − x 2 y 1 2 3 4 1 2 3 4 x y y = x + 2 Section 7.9 Applications of Double Integrals 483 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 31. ( ) 50 50 2 2 45 40 1 2 1 2 1 2 1 2 3 50 50 2 1 2 2 45 1 1 2 1 2 1 2 1 2 40 50 2 45 2 2 2 3 50 2 2 2 45 Average 1 192 576 5 2 5000 50 1 96 576 5 5000 50 3 1 48,200 4860 50 50 3 1 48,200 2430 50 $13,40 50 3 3 x x x x xx dx dx x x x x x x x x x dx x x dx x x x = + − − − − ⎡ ⎤ = ⎢ + − − − − ⎥ ⎣ ⎦ = ⎛ + − ⎞ ⎜ ⎟ ⎝ ⎠ ⎡ ⎤ = ⎢ + − ⎥ = ⎣ ⎦ ∫ ∫ ∫ ∫ 0 32. ( ) 60 65 2 2 50 55 1 2 1 2 1 2 1 2 3 65 60 2 1 2 2 50 1 1 2 1 2 1 2 1 2 55 60 2 50 2 2 60 2 3 2 2 2 50 Average 1 200 580 5 2 7500 100 1 100 580 5 7500 100 3 1 26,750 4600 50 100 3 1 26,750 2300 50 100 3 3 $ x x x x xx dx dx x x x x x x x x x dx x x dx x x x = + − − − − ⎡ ⎤ = ⎢ + − − − − ⎥ ⎣ ⎦ = ⎛ + − ⎞ ⎜ ⎟ ⎝ ⎠ = ⎡ + − ⎤ ⎢⎣ ⎥⎦ = ∫ ∫ ∫ ∫ 11,025 33. ( ) ( ) 150 75 100 50 1 1 2 2 1 2 150 75 2 2 100 50 1 1 2 2 1 2 150 3 2 2 75 100 1 1 1 2 1 2 50 2 15 2 100 2 2 2 Average 1 500 3 750 2.4 1250 1 3 500 2.4 750 1250 1 250 2.4 750 1250 1 484,375 60 18,750 1250 p p p p dp dp p p p p dp dp p p p p p p dp p p dp = ⎡⎣ − + − ⎤⎦ = ⎡⎣− + − + ⎤⎦ = ⎡⎣− + − + ⎤⎦ = ⎡⎣ − + ⎤⎦ ∫ ∫ ∫ ∫ ∫ 0 3 2 150 2 2 2 100 1 484,375 20 9375 1250 $75,125 = ⎡⎣ p − p + p ⎤⎦ = ∫ 34. (a) The value of ( , ) R ∫ ∫ f x y dAwould represent the total (volume) of annual snowfall for Erie County, New York. (b) The value of ( , ) R R f x y dA dA ∫ ∫ ∫ ∫ or 1 ( , ) R R f x y dA dA ∫ ∫ ∫ ∫ would represent the annual average total (volume) snowfall for Erie County, New York. 35. Average ( ) 1.6 250 325 250 0.6 0.4 325 0.4 300 200 300 200 1.4 325 325 0.4 300 300 1 100 1 100 1250 1250 1.6 128,844.1 103.0753 25,645.24 1250 1.4 x y dx dy y x dy y dy y ⎤ = = ⎥⎦ ⎡ ⎤ = = ⎢ ⎥ ≈ ⎣ ⎦ ∫ ∫ ∫ ∫ 36. 1.25 250 325 250 0.25 0.75 325 0.75 300 200 300 200 1.75 325 325 0.75 300 300 Average 1 1 1250 1250 1.25 193.5742 193.5742 287.74 1250 1250 1.75 x y dx dy x y dy y dy y ⎤ = = ⎥⎦ ⎡ ⎤ ≈ ≈ ⎢ ⎥ ≈ ⎣ ⎦ ∫ ∫ ∫ ∫ 484 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Review Exercises for Chapter 7 1. 2. 3. ( )2 ( )2 ( )2 3 1 5 0 8 2 4 25 36 65 d = − + − + − = + + = 4. ( ) ( ) ( ) 2 2 2 1 4 3 1 7 5 25 4 4 33 d = ⎡⎣ − − ⎤⎦ + − + − = + + = 5. 2 ( 4) 6 2 4 8 ( ) Midpoint , , 1, 4, 6 2 2 2 ⎛ + − + + ⎞ = ⎜ ⎟ = − ⎝ ⎠ 6. 5 ( 1) 0 ( 2) 7 9 ( ) Midpoint , , 2, 1, 8 2 2 2 ⎛ + − + − + ⎞ = ⎜ ⎟ = − ⎝ ⎠ 7. ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 0 1 0 5 1 25 x y z x y z − + − + − = + − + = 8. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 4 5 3 10 4 5 3 100 x y z x y z − + ⎡⎣ − − ⎤⎦ + − = − + + + − = 9. Center 3 1, 4 0, 1 ( 5) (2, 2, 3) 2 2 2 ⎛ + − + − + − ⎞ = ⎜ ⎟ = − − ⎝ ⎠ ( )2 ( )2 ( )2 Radius = 2 − 3 + −2 + 4 + −3 + 1 = 1 + 4 + 4 = 3 ( )2 ( )2 ( )2 Sphere: x − 2 + y + 2 + z + 3 = 9 10. Center 3 5, 4 8, 0 2 (4, 6, 1) 2 2 2 ⎛ + + + ⎞ = ⎜ ⎟ = ⎝ ⎠ ( )2 ( )2 ( )2 Radius = 4 − 3 + 6 − 4 + 1 − 0 = 1 + 4 + 1 = 6 ( )2 ( )2 ( )2 Sphere: x − 4 + y − 6 + z − 1 = 6 11. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 8 4 6 20 0 8 16 4 4 6 9 20 16 4 9 4 2 3 49 x y z x y z x x y y z z x y z + + − + − − = − + + + + + − + = + + + − + + + − = Center: (4, −2, 3) Radius: 7 12. ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 4 10 7 0 4 4 10 25 7 4 25 2 5 36 x y z y z x y y z z x y z + + + − − = + + + + − + = + + + + + − = Center: (0, −2, 5) Radius: 6 x y z (−1, 3, −3) (−2, −2, 1) (2, −1, 4) (3, 1, 2) 1 −2 −2 −3 1 3 4 −3 −2 2 3 4 2 x y z (−4, −3, 5) (1, −2, −3) (−2, 2, 2) −2 −4 −4 4 2 2 4 (4, 5 , 1) 2 Review Exercises for Chapter 7 485 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 13. Let z = 0. ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 1 0 3 25 2 1 16 x y x y + + − + − = + + − = Circle of radius 4 14. Let z = 0. ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 1 3 0 6 72 1 3 36 x y x y − + + + − = − + + = Circle of radius 6 15. x + 2y + 3z = 6 To find the x-intercept, let y = 0 and z = 0. x = 6 To find the y-intercept, let x = 0 and z = 0. 2y = 6 ⇒ y = 3 To find the z-intercept, let x = 0 and y = 0. 3z = 6 ⇒ z = 2 x-intercept: (6, 0, 0) y-intercept: (0, 3, 0) z-intercept: (0, 0, 2) 16. 2y + z = 4 Because the coefficient of x is zero, there is no x-intercept. To find the y-intercept, let z = 0. 2y = 4 ⇒ y = 2 To find the z-intercept, let y = 0. z = 4 y-intercept: (0, 2, 0) z-intercept: (0, 0, 4) The plane is parallel to the x-axis. 17. 3x − 6z = 12 To find the x-intercept, let z = 0. 3x = 12 ⇒ x = 4 Because the coefficient of y is zero, there is no y-intercept. To find the z-intercept, let x = 0. −6z = 12 ⇒ z = −2 x-intercept: (4, 0, 0) z-intercept: (0, 0, −2) The plane is parallel to the y-axis. 18. 4x − y + 2z = 8 To find the x-intercept, let y = 0 and z = 0. 4x = 8 ⇒ x = 2 To find the y-intercept, let x = 0 and z = 0. −y = 8 ⇒ y = −8 To find the z-intercept, let x = 0 and y = 0. 2z = 8 ⇒ z = 4 x-intercept: (2, 0, 0) y-intercept: (0, −8, 0) z-intercept: (0, 0, 4) 19. The graph of x2 + y2 + z2 − 2x + 4y − 6z + 5 = 0 is a sphere whose standard equation is ( )2 ( )2 ( )2 x − 1 + y + 2 + z − 3 = 9. 20. The graph of 16x2 + 16y2 − 9z2 = 0 is an elliptic cone whose standard equation is 2 2 2 0. 1 16 1 16 1 9 x + y − z = 21. The graph of 2 2 2 1 16 9 x + y + z = is an ellipsoid. 22. The graph of 2 2 2 1 16 9 x − y − z = is a hyperboloid of two sheets. 23. The graph of 2 2 9 z = x + y is an elliptic paraboloid. 4 6 −2 4 x y z x y z 4 −6 −6 8 −4 −10 8 x y (0, 0, 2) (6, 0, 0) (0, 3, 0) z x y (0, 0, 4) (0, 2, 0) 3 4 5 5 5 z x y z (4, 0, 0) (0, 0, −2) 1 1 −3 −4 −5 4 2 5 1 x y (0, 0, 4) (0, −8, 0) (2, 0, 0) 3 1 − 9 5 3 4 2 z 486 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 24. The graph of −4x2 + y2 + z2 = 4 is a hyperboloid of one sheet whose standard equation is 2 2 2 1. 4 4 y + z − x = 25. The graph of z = x2 + y2 is the top half of a circular cone whose standard equation is x2 + y2 − z2 = 0. 26. The graph of 2 2 4 z = x − y is a hyperbolic paraboloid. 27. f (x, y) = xy2 (a) ( ) ( )2 f 2, 3 = 2 3 = 18 (b) ( ) ( )2 f 0, 1 = 0 1 = 0 (c) ( ) ( )2 f −5, 7 = −5 7 = −245 (d) ( ) ( )2 f −2, −4 = −2 −4 = −32 28. ( ) 2 f x, y x y = (a) ( ) 62 6, 9 4 9 f = = (b) ( ) 82 8, 4 16 4 f = = (c) ( ) 2 , 2 2 f t = t (d) ( ) 2 f r, r r r, r = = r ≠ 0 29. The domain of f (x, y) = 1 − x2 − y2 is the set of all points inside or on the circle x2 + y2 = 1. The range is [0, 1]. 30. The domain of f (x, y) = x2 + y2 − 3 is the set of all points in the xy-plane. The range is [−3, ∞). 31. The domain of f (x, y) = exy is the set of all points in the xy-plane. The range is (0, ∞). 32. The domain of f (x, y) 1 x y = + is the set of all points except those on the line y = −x. The range is all real numbers except 0. 33. z = 10 − 2x − 5y c = 0: 10 − 2x − 5y = 0, 2x + 5y = 10 c = 2: 10 − 2x − 5y = 2, 2x + 5y = 8 c = 4: 10 − 2x − 5y = 4, 2x + 5y = 6 c = 5: 10 − 2x − 5y = 5, 2x + 5y = 5 c = 10: 10 − 2x − 5y = 10, 2x + 5y = 0 The level curves are lines of slope 2 5− . 34. z = 9 − x2 − y2 c = 0: 9 − x2 − y2 = 0, x2 + y2 = 9 c = 1: 9 − x2 − y2 = 1, x2 + y2 = 8 c = 2: 9 − x2 − y2 = 2, x2 + y2 = 5 c = 3: 9 − x2 − y2 = 3, x2 + y2 = 0 (point) The level curves are circles (except c = 3). 35. ( )2 z = xy ( )2 c = 1: xy = 1, y 1 x = ± ( )2 2 4: 4, c xy y x = = = ± ( )2 3 9: 9, c xy y x = = = ± ( )2 c = 12: xy = 12, y 2 3 x = ± ( )2 4 16: 16, c xy y x = = = ± The level curves are hyperbolas. 1 − 1 1 − 1 x c = 1 c = 0 c = 2 c = 3 y 3 1 −3 −2 −1 3 −1 x c = 10 c = 5 c = 4 c = 2 c = 0 y 1 −1 −1 1 x c = 1 c = 4 c = 9 c = 12 c = 16 y Review Exercises for Chapter 7 487 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 36. z = y − x2 c = 0: y − x2 = 0, y = x2 c = −1: y − x2 = −1, y = x2 − 1 c = 1: y − x2 = 1, y = x2 + 1 c = −2: y − x2 = −2, y = x2 − 2 c = 2: y − x2 = 2, y = x2 + 2 The level curves are parabolas. 37. (a) No; the precipitation increments are 7.99 inches, 9.99 inches, 9.99 inches, 9.99 inches, and 19.99 inches. (b) You could increase the number of level curves to correspond to smaller increments of precipitation. 38. Southwest to northeast 39. z = −4.51 + 0.046x + 0.060y (a) (100, 40) 4.51 0.046(100) 0.060(40) $2.49 z = − + + = (b) Because z 0.060 z 0.46, y x ∂ ∂ = > = ∂ ∂ y has the greater influence on earnings per share. 40. z = 1.54 + 0.116x + 0.122y (a) (300, 130) 1.54 0.116(300) 0.122(130) $52.20 z = + + = (b) Because z 0.122 z 0.116, y x ∂ ∂ = > = ∂ ∂ y has the greater influence on shareholders’ equity. 41. ( ) 2 2 , 3 2 5 2 3 2 3 5 x y f x y x y xy x y f xy y f x x = + + − = + + = + − 42. ( ) 2 2 2 2 , 4 3 4 6 4 2 3 x y f x y xy xy x y f y y xy f x xy x = + − = + − = + − 43. 2 2 2 2 3 2 2 z x y z x x y z x y y = ∂ = ∂ ∂ = − ∂ 44. ( ) ( )( ) ( )( ) 2 2 4 2 2 4 2 2 2 4 4 z xy x y z xy x y y x z xy x y x y = + + ∂ = + + + ∂ ∂ = + + + ∂ 45. ( ) ( ) ( ) ( ) , ln5 4 , 5 5 4 , 4 5 4 x y f x y x y f x y x y f x y x y = + = + = + 46. f (x, y) = ln 2x + 3y ( , ) 1 2 1 2 2 3 2 3 fx x y x y x y ⎛ ⎞ = ⎜ ⎟ = ⎝ + ⎠ + ( ) ( ) , 1 3 3 2 2 3 2 2 3 fy x y x y x y ⎛ ⎞ = ⎜ ⎟ = ⎝ + ⎠ + 47. ( , ) y x x y x y x y f x y xe ye f ye e f xe e = + = + = + 48. ( ) ( ) ( ) 2 2 2 2 2 , , 2 , 2 y y x y y f x y x e f x y xe f x y x e − − − = = = − 49. w = xyz2 2 2 2 w yz x w xz y w xyz z ∂ = ∂ ∂ = ∂ ∂ = ∂ 50. w = 3xy − 5xz + 2yz 3 5 3 2 5 2 w y z x w x z y w x y z ∂ = − ∂ ∂ = + ∂ ∂ = − + ∂ 51. z = 3xy (a) z 3y; z ( 2, 3, 18) 9 x x ∂ ∂ = − − = − ∂ ∂ (b) z 3x; z ( 2, 3, 18) 6 y y ∂ ∂ = − − = − ∂ ∂ x c = 2 c = 1 c = 0 c = −1 c = −2 y −4 −3 −2 2 3 4 −2 −3 3 4 5 488 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 52. z = y2 − x2 (a) z 2x; z (1, 2, 3) 2 x x ∂ ∂ = − = − ∂ ∂ (b) z 2y; z (1, 2, 3) 4 y y ∂ ∂ = = ∂ ∂ 53. z = 8 − x2 − y2 (a) z 2x; z (1, 1, 6) 2 x x ∂ ∂ = − = − ∂ ∂ (b) z 2y; z (1, 1, 6) 2 y y ∂ ∂ = − = − ∂ ∂ 54. z = 100 − x2 − y2 (a) ( ) 2 2 ; 0, 6, 8 0 100 z x z x x y x ∂ ∂ = − = ∂ − − ∂ (b) ( ) 2 2 ; 0, 6, 8 3 100 4 z y z y x y y ∂ ∂ = − = − ∂ − − ∂ 55. f (x, y) = 3x2 − xy + 2y3 fx = 6x − y fy = −x + 6y2 fxx = 6 fyy = 12y fxy = fyx = −1 56. f (x, y) y x y = + x ( )2 f y x y = − + y ( )2 f x x y = + ( )3 2 xx f y x y = + ( )3 2 yy f x x y = − + xy yx ( )3 f f y x x y − = = + 57. f (x, y) = 1 + x + y 1 2 1 fx x y = + + 1 2 1 fy x y = + + ( )3 2 1 4 1 fxx fyy fxy fyx x y = = = = + + 58. ( ) f x, y = x2e− y2 2 2 2 2 2 4 y x y xx y xy yx f xe f e f f xye − − − = = = = − ( ) 2 2 2 2 2 2 2 2 1 y y y yy f xye f x y e − − = − = − 59. f (x, y, z) = xy + 5x2 yz3 − 3y3z 3 3 3 2 10 10 1 10 30 x xx xy xz f y xyz f yz f xz f xyz = + = = + = 2 3 2 3 2 2 2 5 9 1 10 18 15 9 y yx yy yz f x xz yz f xz f yz f xz y = + − = + = − = − 2 2 3 2 2 2 2 2 15 3 30 15 9 30 z zx zy zz f x yz y f xyz f x z y f xyz = − = = − = 60. f (x, y, z) 3yz x z = + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 2 3 2 2 3 2 3 3 3 3 6 3 3 3 0 3 3 3 6 x y z xx yx zx xy yy zy xz yz zz f yz f z f xy x z x z x z yz z y x z f f f x z x z x z f z f f x x z x z y x z x xy f f f x z x z x z = − = = + + + − = =− = − + + + = − = = + + − = − = = − + + + 61. ( )1 3 C = 15 xy + 99x + 139y + 2293 (a) C 5x 2 3y1 3 99; x ∂ − = + ∂ At (500, 250), C $99.50. x ∂ = ∂ C 5x1 3y 2 3 139; y ∂ − = + ∂ At (500, 250), C $140.00. y ∂ = ∂ (b) Downhill skis; this is determined by comparing the marginal costs for the two models of skis at the production level (500, 250). 62. 2 2 1 2 1 2 1 2 15 16 1 1 1 10 10 100 R = x + x − x − x − x x (a) 1 2 1 15 1 1 5 100 R x x x ∂ = − − ∂ ( ) 1 R 50, 40 $4.60 x ∂ = ∂ (b) 2 1 2 16 1 1 5 100 R x x x ∂ = − − ∂ ( ) 2 R 50, 40 $7.50 x ∂ = ∂ Review Exercises for Chapter 7 489 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 63. f (x, y) = x2 + 2y2 The first partial derivatives of f, fx (x, y) = 2x and fy (x, y) = 4y, are zero at the critical point (0, 0). Because fxx (xy) = 2, fyy (xy) = 4, and fxy (x, y) = 0, it follows that fxx (0, 0) > 0 and ( ) ( ) ( ) 2 fxx 0, 0 fyy 0, 0 − ⎡⎣ fxy x, y ⎤⎦ = 8 > 0. So, (0, 0, 0) is a relative minimum. 64. f (x, y) = x3 − 3xy + y2 The first partial derivatives of f, fx (x, y) = 3x2 − 3y and fy (x, y) = −3x + 2y, are zero at the critical points (0, 0) and (3 9 ) 2 4 , . Because fxx (x, y) = 6x, fyy (x, y) = 2, and fxy (x, y) = −3, it follows that fxx (0, 0) = 0, ( ) ( ) ( ) 2 fxx 0, 0 fyy 0, 0 − ⎡⎣ fxy 0, 0 ⎤⎦ = −9 < 0, (3 9 ) 2 4fxx , = 9 > 0, and ( ) ( ) ( ) 2 3 9 3 9 3 9 2 4 2 4 2 4 fxx , fyy , − ⎡⎣ fxy , ⎤⎦ = 9 > 0. So, (0, 0, 0) is a saddle point and (3 9 27 ) 2 4 16 , , − is a relative minimum. 65. ( ) ( )2 ( )2 f x, y = 1 − x + 2 + y − 3 The first partial derivatives of f, fx (x, y) = −2(x + 2) and fy (x, y) = 2(y − 3), are zero at the critical point (−2, 3).Because fxx (x, y) = −2, fyy (x, y) = 2, and fxy (x, y) = 0, it follows that fxx (−2, 3) > 0 and ( ) ( ) ( ) 2 fxx −2, 3 fyy −2, 3 − ⎡⎣ fxy −2, 3 ⎤⎦ = −4 < 0. So, (−2, 3, 1) is a saddle point. 66. f (x, y) = ex − x + y2 The first partial derivatives of f, ( , ) x 1 fx x y = e − and fy (x, y) = 2y, are zero at the critical point (0, 0). Because ( , ) x , fxx x y = e fyy (x, y) = 2, and fxy (x, y) = 0, it follows that fxx (0, 0) = 1 > 0 and ( ) ( ) ( ) 2 fxx 0, 0 fyy 0, 0 − ⎡⎣ fxy 0, 0 ⎤⎦ = 2 > 0. So, (0, 0, 1) is a relative minimum. 67. f (x, y) = x3 + y2 − xy The first partial derivatives of f, fx (x, y) = 3x2 − y and fy (x, y) = 2y − x, are zero at the critical points (0, 0) and (1 1 ) 6 12 , . Because fxx (x, y) = 6x, fyy (x, y) = 2, and fxy (x, y) = −1, it follows that fxx (0, 0) = 0, ( ) ( ) ( ) 2 fxx 0, 0 fyy 0, 0 − ⎣⎡ fxy 0, 0 ⎦⎤ = −1 < 0, (1 1 ) 6 12 fxx , = 1 > 0, and ( ) ( ) ( ) 2 1 1 1 1 1 1 6 12 6 12 6 12 fxx , fyy , − ⎡⎣ fxy , ⎤⎦ = 1 > 0. So, (0, 0, 0) is a saddle point and (1 1 1 ) 6 12 432 , , − is a relative minimum. 68. f (x, y) = y2 + xy + 3y − 2x + 5 The first partial derivatives of f, fx (x, y) = y − 2 and fy (x, y) = 2y + x + 3, are zero at the critical point (−7, 2).Because fxx (x, y) = 0, fyy (x, y) = 2, and fxy (x, y) = 1, it follows that fxx (−7, 2) > 0 and ( ) ( ) ( ) 2 fxx −7, 2 fyy −7, 2 − ⎣⎡ fxy −7, 2 ⎦⎤ = −1 < 0. So, (−7, 2, 15) is a saddle point. 69. f (x, y) = x3 + y3 − 3x − 3y + 2 The first partial derivatives of f, fx (x, y) = 3x2 − 3 and fy (x, y) = 3y2 − 3, are zero at the critical points (1, 1), (−1, −1), (1, −1), and (−1, 1).Because fxx (x, y) = 6x, fyy (x, y) = 6y, and fxy (x, y) = 0, it follows that fxx (1, 1) = 6 > 0, ( ) ( ) ( ) 2 fxx 1, 1 fyy 1, 1 − ⎣⎡ fxy 1, 1 ⎦⎤ = 36 > 0, fxx (−1, −1) = −6 < 0, ( ) ( ) ( ) 2 fxx −1, −1 fyy −1, −1 − ⎡⎣ fxy −1, −1 ⎤⎦ = 36 > 0, fxx (1, −1) = 6 > 0, ( ) ( ) ( ) 2 fxx 1, −1 fyy 1, −1 − ⎡⎣ fxy 1, −1 ⎤⎦ = −36 < 0, fxx (−1, 1) = −6 < 0, and ( ) ( ) ( ) 2 fxx −1, 1 fyy −1, 1 − ⎡⎣ fxy −1, 1 ⎤⎦ = −36 < 0. So, (1, 1, −2) is a relative minimum, (−1, −1, 6) is a relative maximum, (1, −1, 2) is a saddle point, and (−1, 1, 2) is a saddle point. 70. f (x, y) = − x2 − y2 The first partial derivatives of f, fx (x, y) = −2x and fy (x, y) = −2y, are zero at the critical point (0, 0). Because fxx (x, y) = −2, fyy (x, y) = −2, and fxy (x, y) = 0, it follows that fxx (0, 0) = 1 < 0 and ( ) ( ) ( ) 2 fxx 0, 0 fyy 0, 0 − ⎡⎣ fxy 0, 0 ⎤⎦ = 4 > 0. So, (0, 0, 0) is a relative minimum. 490 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 71. ( ) ( ) 1 1 2 2 1 1 2 2 2 2 1 2 1 2 100 200 0.5 0.5 100 200 R xp xp x x x x x x x x = + = − + − = − − + + 1 1 1 2 2 2 2 100 0 50 200 0 200 x x R x x R x x = − + = ⇒ = = − + = ⇒ = 1 1 Rx x = −2 2 2 Rx x = −1 1 2 Rx x = 0 2 d Rx1x1Rx2 x2 Rx1x2 = − ⎡ ⎤ ⎣ ⎦ ( )( ) ( )2 d = −2 −1 − 0 = 2 > 0 Since 1 1 Rx x < 0, R is maximum at (50, 200), or when x1 = 50 and x2 = 200. The maximum revenue is $22,500. 72. ( ) ( ) ( ) ( ) 1 2 1 2 2 2 1 2 1 1 2 2 10 10 0.03 4 300 0.05 7 175 P x x C C x x x x x x = + − − = + − + + − + + 1 1 1 2 2 2 6 0.06 0 100 3 0.1 0 30 x x P x x P x x = − = ⇒ = = − = ⇒ = By the Second-Partials Test, profit is maximized at x1 = 100 and x2 = 30. 73. F(x, y, λ ) = 2xy − λ (2x + y − 12) ( ) ( ) , , 2 2 0 2 , , 2 0 x y F x y y y x F x y x λ λ λ λ = − = ⎫⎪ ⎬ = = − =⎪ ⎭ Fλ (x, y, λ ) = −(2x + y − 12) = 0 Using Fx , (2 2 12) 0 4 12 3. x x x x − + − = − = − = So, x = 3 and y = 6, and f (x, y) has a maximum at (3, 6). The relative maximum is f (3, 6) = 36. 74. F(x, y, λ ) = 2x + 3xy + y − λ (x + 2y − 29) ( ) ( ) , , 2 3 0 2 1 , , 3 1 2 0 x y F x y y x y F x y x λ λ λ λ = + − = ⎫⎪ ⎬ − = = + − =⎪⎭ Fλ (x, y, λ ) = −(x + 2y − 29) = 0 Using , and Fλ 2 29 2 1 2 30 15. x y x y x x + = − = = = So, x = 15 and y = 7, and f (x, y) has a maximum at (15, 7). The relative maximum is f (15, 7) = 352. 75. F(x, y, λ ) = x2 + y2 − λ (x + y − 4) ( ) ( ) , , 2 0 , , 2 0 x y F x y x F x y y λ λ λ λ = − = ⎫⎪⎬ = − =⎪ ⎭ x = y Fλ (x, y, λ ) = −(x + y − 4) = 0, x = y = 2 Minimum f (2, 2) = 8. So, x = 2, y = 2, and f (x, y) has a minimum at (2, 2). The relative minimum is f (2, 2) = 8. 76. F(x, y, λ ) = 3x2 − y2 − λ (2x − 2y + 5) ( ) ( ) , , 6 2 0 3 , , 2 2 0 x y F x y x y x F x y y λ λ λ λ = − = ⎫⎪ ⎬ = = − + = ⎪⎭ Fλ (x, y, λ ) = −(2x − 2y + 5) = 0 Using Fλ , ( ( ) ) 5 4 2 23 5 0 4 5 . x x x x − − + = = = So, 5 4 x = and 15 4 y = , and f (x, y) has a minimum at (5 15) 4 4 , . The relative minimum is (5 15) 75 4 4 8 f , = − . Review Exercises for Chapter 7 491 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 77. F(x, y, z, λ ) = xyz − λ (x + 2y + z − 4) 0 2 0 0 x y z F yz F xz F xy λ λ λ = − = ⎫ ⎪ = − =⎬⎪ = − = ⎭ xz = 2yz = 2xy ⇒ 2 2 x y z y = = Fλ = −(x + 2y + z − 4) = 0 2y + 2y + 2y − 4 = 0 So, 23 , y = 43 , x = 43 z = , and f (x, y, z) has a maximum at (4 2 4) 3 3 3 , , .The relative maximum is (4 2 4) 32 3 3 3 27 f , , = . 78. F(x, y, z, λ ) = x2z + yz − λ (2x + y + z − 5) ( ) 2 2 2 0 0, 0 2 5 0 x y z F xz F z z F x y F x y z λ λ λ λ λ = − = = − = = = + − = = − + + − = Substituting z = λ into Fx yields x = 1. From Fz , you have y = λ − 1.Substituting that into Fλ yields λ = 2. So, x = 1, y = 1, z = 2, and f (x, y, z) has a maximum at (1, 1, 2). The relative maximum is f (1, 1, 2) = 4. 79. Minimize 2 2 C = 0.25x1 + 10x1 + 0.15x2 + 12x2 subject to the constraint x1 + x2 = 1000. ( ) 2 2 ( ) F x1, x2 , λ = 0.25x1 + 10x1 + 0.15x2 + 12x2 − λ x1 + x2 − 1000 ( ) ( ) 1 1 2 1 2 1 2 2 , , 0.50 10 0 , , 0.30 12 0 x x F x x x F x x x λ λ λ λ ⎫ = + − =⎪⎬ = + − =⎪ ⎭ 5x1 − 3x2 = 20 Fλ (x1, x2 , λ ) = −(x1 + x2 − 1000) = 0, x2 = 1000 − x1 So, x1 = 377.5 and x2 = 622.5, and f (x1, x2 ) has a minimum at (377.5, 622.5). The cost is minimized when 378 units of x1 and 623 units of x2 are ordered. 80. Maximize f (x, y) = 4x + xy + 2y subject to the constraint 20x + 4y = 2000. F(x, y, λ ) = 4x + xy + 2y − λ (20x + 4y − 2000) ( ) ( ) , , 4 20 0 , , 2 4 0 x y F x y y F x y x λ λ λ λ = + − = ⎫⎪⎬ = + − = ⎪⎭ 4 5( 2) 5 6 y x y x + = + = + Fλ (x, y, λ ) = −(20x + 4y − 2000) = 0, y = 500 − 5x Adding y = 5x + 6 and y = 500 − 5x yields y = 253. So, x = 49.4. f (x, y) has a maximum at (49.4, 253). The maximum production level is f (49.4, 253) ≈ 13,202. 81. Σxi = 1 Σyi = 0 Σxi2 = 15 Σxi yi = 15 ( ) ( )( ) ( ) ( )2 4 15 1 0 60 4 15 1 59 a − = = − 1 0 60(1) 15 4 59 59 b = ⎛⎜ − ⎞⎟ = − ⎝ ⎠ 60 15 59 59 y = x − 82. Σxi = −2 Σyi = 0 Σxi2 = 18 Σxi yi = 8 ( ) ( )( ) ( ) ( )2 5 8 2 0 20 5 18 2 43 a − − = = − − 1 0 20( 2) 8 5 43 43 b = ⎛⎜ − − ⎞⎟ = ⎝ ⎠ 20 8 43 43 y = x + −3 −2 −1 1 2 3 4 −2 −3 −4 1 2 3 x y (−2, −3) (−1, −1) (1, 2) (3, 2) y = 6 0 x − 59 15 59 −3 −2 1 2 3 −1 −2 −3 1 2 3 x y (−3, −1) (−2, −1) (2, 1) (1, 1) (0, 0) y = 2 0 x + 43 8 43 492 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 83. (a) y = −2.6x + 347 (b) Let x = 85. y = −2.6(85) + 347 = 126 cameras (c) Let y = 200 and find x. 200 2.6 347 147 2.6 $56.54 x x x = − + − = − ≈ 84. (a) y = 0.90x − 2.2 (b) Let x = 80. y = 0.90(80) − 2.2 ≈ 69.8 million women 85. ( ) 1 1 0 0 4 2 x x y dy dx + ∫ ∫ − ( ) ( ) ( ) 1 2 1 0 0 1 2 0 1 2 0 3 2 1 0 4 4 1 1 3 2 1 1 x xy y dx x x x dx x x dx x x x + = ⎡⎣ − ⎤⎦ = ⎡ + − + ⎤ ⎣ ⎦ = + − = ⎡⎣ + − ⎤⎦ = ∫ ∫ ∫ 86. ( ) ( ) ( ) ( ) 2 4 3 4 2 3 2 3 0 3 0 3 2 3 3 3 3 2 8 4 8 4 3 24 36 24 36 24 x y dx dy x xy dy y dy y y − − − − ⎡ ⎤ − = ⎢ − ⎥ ⎣ ⎦ = − = ⎡ − ⎤ ⎢⎣ ⎥⎦ = − − − + = − ∫ ∫ ∫ ∫ 87. 2 2 2 2 2 1 1 2 1 2 1 2 2 1 2 2 2 2 1 2 1 2 4 1 2 2 2 1 2 2 1 2 4 1 2 1 4 2 7 4 y y x dx dy x dy y y y dy y y y dy y y − ⎤ = ⎥⎦ ⎛ ⎞ = ⎜ − ⎟ ⎝ ⎠ = ⎛ − ⎞ ⎜ ⎟ ⎝ ⎠ ⎡ ⎤ = ⎢ + ⎥ ⎣ ⎦ = ⎛ + ⎞ − ⎛ + ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = ∫ ∫ ∫ ∫ ∫ 88. ] ( ) 4 16 2 4 16 2 0 0 0 0 4 2 0 3 2 4 2 0 23 128 3 2 2 2 16 16 x x x dy dx xy dx x x dx x − − = = − = − ⎤⎥⎦ = ∫ ∫ ∫ ∫ 89. [ ] ( ) 2 9 2 2 5 2 9 2 2 5 2 2 2 3 2 2 1 3 32 3 4 4 x x A dy dx y dx x dx x x − − − − − − = = = − = ⎡⎣ − ⎤⎦ = ∫ ∫ ∫ ∫ 90. [ ] 4 4 1 0 4 4 1 0 4 1 4 1 4 4 ln 4 ln 4 x x A dy dx y dx dx x x = = = = ⎡⎣ ⎤⎦ = ∫ ∫ ∫ ∫ 91. ( ) [ ] ( ) ( ) ( ) 6 3 3 13 1 6 3 3 1 3 1 6 3 3 2 6 2 3 1 3 2 1 3 6 9 2 3 1 3 x x x x A dy dx y dx x x dx x x x + − + + − + − − = = = + − − = ⎡ + − − ⎤ ⎣ ⎦ = ∫ ∫ ∫ ∫ 92. [ ] ( ) 2 1 2 2 2 2 1 2 2 2 2 2 1 2 3 2 1 2 1 1 2 3 2 9 2 x x x x x x A dy dx y dx x x dx x x x − − − − − − − − − − = = = − + + = ⎡− + + ⎤ ⎢⎣ ⎥⎦ = ∫ ∫ ∫ ∫ Review Exercises for Chapter 7 493 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 93. ( ) [ ] 4 2 0 0 4 2 2 0 0 4 0 4 0 1 2 1 4 3 3 5 5 20 V y dy dx y y dx dx x = − = ⎡⎣ − ⎤⎦ = = = ∫ ∫ ∫ ∫ 94. ( ) ( ) ( ) ( ) 12 6 1 2 0 0 12 2 6 1 2 0 0 12 2 0 3 2 12 0 2 4 3 3 2 2 3 3 1 6 1 18 8 8 4 24 2 24 96 x x V x y dy dx y xy y dx x x dx x x x − − = − − = ⎡⎣ − − ⎤⎦ = − + = ⎡⎣ − + ⎤⎦ = ∫ ∫ ∫ ∫ 95. [ ] 2 2 2 2 2 0 0 0 0 0 0 4 4 4 2 8 x x V = ∫ ∫ dy dx = ∫ y dx = ∫ x dx = ⎡⎣ x ⎤⎦ = 96. ( ) ( ) 2 2 2 2 3 2 2 3 2 4 2 0 0 0 0 1 16 1 16 1 3 3 3 3 12 4 4 4 2 4 x x V = ∫ ∫ − y dy dx = ∫ ⎡⎣ y − y ⎤⎦ dx = ∫ − x + x dx = ⎡⎣ x − x + x ⎤⎦ = 97. ( ) 2 3 4 2 3 4 4 4 2 4 4 2 2 4 4 0 0 0 0 0 0 0 0 64 64 4096 3 3 9 9 V xy dy dx x y dy dx x y dx x dx x ⎤ ⎤ = = = ⎥ = = ⎥ = ⎦ ⎦ ∫ ∫ ∫ ∫ ∫ ∫ 98. ( ) 2 2 2 3 3 3 3 3 2 3 0 0 0 0 0 0 0 3 27 2 2 2 2 2 x V x x y dy dx xy y dx x y dx x dx x ⎡ ⎤ ⎤ = + = ⎢ + ⎥ = + = = ⎥ = ⎣ ⎦ ⎦ ∫ ∫ ∫ ∫ ∫ 99. 4 3 0 0 4 2 3 0 0 4 0 2 4 0 1 12 1 1 12 2 1 24 1 9 24 2 Average 9 3 xy dy dx xy dx x dx x = = ⎡⎣ ⎤⎦ = = ⎡⎣ ⎤⎦ = ∫ ∫ ∫ ∫ 100. ( ) ( ) 2 5 2 2 0 0 2 2 2 3 5 0 0 2 2 0 2 2 2 0 1 10 1 1 10 3 1 125 10 3 1 5 25 125 10 3 2 3 44 3 Average 2 5 25 x xy y dy dx x y xy y dx x x dx x x x = + + = ⎡⎣ + + ⎤⎦ = + + = ⎡⎣ + + ⎤⎦ = ∫ ∫ ∫ ∫ 101. ( ) 50 40 2 2 40 30 1 2 1 2 1 2 1 2 40 50 2 3 2 2 40 1 1 2 1 1 2 1 2 1 2 30 50 2 40 2 2 2 50 2 3 2 2 2 40 Average 1 150 400 5 2 3000 100 1 75 400 1 5 3000 100 3 1 30,500 3300 50 100 3 1 30,500 1650 50 100 3 3 x x x x xx dx dx x x x x x x x x x dx x x dx x x x = + − − − − = ⎡ + − − − − ⎤ ⎢⎣ ⎥⎦ = ⎛ + − ⎞ ⎜ ⎟ ⎝ ⎠ = ⎡ + − ⎤ ⎢⎣ ⎥⎦ = ∫ ∫ ∫ ∫ $5700 102. ( ) ( ) 375 375 2 2 437.5 525 1 1 2 2 2 1 375 2 2 3 375 437.5 1 2 1 2 2 2 525 1 375 2 437.5 1 1 1 2 3 1 1 1 437.5 1 2 1 1250 5 3 1 2 1 1250 5 9 1 1250 1 1250 Average 200 250 200 125 30,000 60 6,656,250 15,000 20 6,656,250 x x x x dxdx xx xx x x dx x x dx x x x = − + − = ⎡⎣ − + − ⎤⎦ = − + − = ⎡⎣− + − ⎤⎦ ∫ ∫ ∫ ∫ 375 = $446,093.75 494 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 103. ( ) 5280 3960 2 3 3 4 0 0 3960 5280 2 3 7 4 0 0 5280 2 3 0 5 3 5280 0 2 Average 1 0.003 20,908,800 1 0.012 20,908,800 7 1 3388.830 20,908,800 1 2033.298 20,908,800 $155.69 ft x y dydx x y dx x dx x = = ⎡ ⎤ ⎢⎣ ⎥⎦ = = ⎡⎣ ⎤⎦ ≈ ∫ ∫ ∫ ∫ Chapter 7 Test Yourself 1. (a) (b) ( )2 ( )2 ( )2 3 1 1 3 0 0 4 4 0 2 2 d = − + − + + − = + + = (c) Midpoint 1 3, 3 1, 0 0 (2, 2, 0) 2 2 2 ⎛ + − − + ⎞ = ⎜ ⎟ = − ⎝ ⎠ 2. (a) (b) ( )2 ( )2 ( )2 4 2 0 2 2 3 4 4 1 3 d = − + + − + − = + + = (c) Midpoint 2 4, 2 0, 3 2 3, 1, 5 2 2 2 2 ⎛ − − + + ⎞ ⎛ ⎞ = ⎜ ⎟ = ⎜− ⎟ ⎝ ⎠ ⎝ ⎠ 3. (a) (b) ( )2 ( )2 ( )2 d = 5 − 3 + 11 + 7 + −6 − 2 = 4 + 324 + 64 = 14 2 (c) Midpoint 3 5, 7 11, 2 6 (4, 2, 2) 2 2 2 ⎛ + − + − ⎞ = ⎜ ⎟ = − ⎝ ⎠ 4. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 20 10 10 125 0 20 100 10 25 10 25 125 100 25 25 10 5 5 25 x y z x y z x x y y z z x y z + + − + − + = − + + + + + − + = − + + + − + + + − = Center: (10, −5, 5) Radius: 5 x y z (1, −3, 0) (3, −1, 0) 1 1 −2 −1 −1 1 −2 2 3 −3 −2 −4 2 2 3 4 x y z (−4, 0, 2) (−2, 2, 3) 1 1 −1 −2 −1 −1 −3 1 2 3 −3 −2 −3 2 3 2 x y z (3, −7, 2) (5, 11, −6) 2 2 −2 −2 −2 2 −4 4 6 −6 −4 −6 4 6 8 10 12 4 6 8 Chapter 7 Test Yourself 495 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 5. The graph of 4x2 + 2y2 − z2 = 16 is a hyperboloid of one sheet whose standard equation is 2 2 2 1. 4 8 16 x + y − z = 6. The graph of 36x2 + 9y2 − 4z2 = 0 is an elliptic cone whose standard equation is 2 2 2 0. 4 9 x + y − z = 7. The graph of 4x2 − y2 − 16z = 0 is a hyperbolic paraboloid whose standard equation is 2 2 . 4 16 z = x − y 8. ( ) ( ) ( ) ( ) ( ) 2 2 2 , 1 3, 3 3 3 3 1 19 1, 4 1 1 4 1 6 f x y x xy f f = + + = + + = = + + = 9. ( ) ( ) ( ) ( ) ( ) ( ) ( ) , 2 3 3 2 3 9 3 3, 3 3 3 3 6 2 1 24 1, 4 9 3 1 4 f x y x y x y f f + = − + = = = − + = = − − 10. ( ) ( ) ( ) ( ) ( ) ( ) , ln 3, 3 3 3 ln 3 9 ln 1 0 3 1, 4 1 4 ln 1 4 ln 1 5.5 4 4 f x y xy x y f f ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = ⎛ ⎞ = = ⎜ ⎟ ⎝ ⎠ = ⎛ ⎞ = ⎛ ⎞ ≈ − ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 11. f (x, y) = 3x2 + 9xy2 − 2 fx (x, y) = 6x + 9y2 ( ) ( ) ( )2 fx 10, −1 = 6 10 + 9 −1 = 69 fy (x, y) = 18xy fy (10, −1) = 18(10)(−1) = −180 12. f (x, y) = x x + y ( , ) 1 ( ) 1 2 (1) (1) 2 2 fx x y x x y x y x x y x y = ⎛ ⎞ + − + + ⎜ ⎟ ⎝ ⎠ = + + + ( ) ( ) 10, 1 10 10 1 14 2 10 1 3 fx − = + − = + − ( , ) 1 ( ) 1 2 (1) (0) 2 2 fy x y x x y x y x x y = ⎛ ⎞ + − + + ⎜ ⎟ ⎝ ⎠ = + ( ) ( ) 10, 1 10 5 2 10 1 3 fy − = = + − 13. f (x, y) = 3x2 + 4y2 − 6x + 16y − 4 The first partial derivatives of f, fx (x, y) = 6x − 6 and fy (x, y) = 8y + 16, are zero at the point (1, −2). Moreover, because fxx (x, y) = 6, fyy (x, y) = 8, and fxy (x, y) = 0, it follows that fxx (1, −2) > 0 and ( ) ( ) ( ) 2 fxx 1, −2 fyy 1, −2 − ⎡⎣ fxy 1, −2 ⎤⎦ = 48 > 0. So, (1, −2, −23) is a relative minimum. 14. f (x, y) = 4xy − x4 − y4 The first partial derivatives of f , fx (x, y) = 4y − 4x3 and fy (x, y) = 4x − 4y3, are zero at the points (0, 0), (1, 1), and (−1, −1). Moreover, because fxx (x, y) = −12x2 , fyy (x, y) = −12y2 , and fxy (x, y) = 4, it follows that fxx (0, 0) = 0, ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 fxx 0, 0 fyy 0, 0 − ⎡⎢⎣ fxy 0, 0 ⎤⎦ = −16 < 0, fxx 1, 1 = −12 < 0, fxx 1, 1 fyy 1, 1 − ⎡⎣ fxy 1, 1 ⎤⎥⎦ = 128 > 0, fxx (−1, −1) = −12 < 0, and ( ) ( ) ( ) 2 fxx −1, −1 fyy −1, −1 − ⎡⎣ fxy −1, −1 ⎤⎦ = 128 > 0. So, (0, 0, 0) is a saddle point, (1, 1, 2) is a relative minimum, and (−1, −1, 2) is a relative minimum. 496 Chapter 7 Functions of Several Variables © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 15. Maximize f (x, y) = 60x0.7 y0.3 subject to the constraint 42x + 144y = 240,000. F(x, y, λ ) = 60x0.7 y0.3 − λ (42x + 144y − 240,000) Fx = 42x−0.3y0.3 − 42λ = 0, 0.3 0.3 42 , 0.3 0.3 42 λ x y λ x y − = = − Fy = 18x0.7 y−0.7 − 144λ = 0 Fλ = −42x − 144y + 240,000 = 0 Substituting λ = x−0.3y0.3 into Fy yields 18x0.7 y−0.7 − 144x−0.3y0.3 = 0 ⇒ x = 8y. From Fλ , you can obtain y = 500. So, x = 4000. To maximize production, the company should use 4000 units of labor and 500 units of capital. 16. 2 29 22 231 160 i i i i i x y x x y = = = = Σ Σ Σ Σ ( ) ( )( ) ( ) ( )2 5 160 29 22 0.52 5 231 29 a − = ≈ − 1(22 0.52(29)) 1.4 5 b = − ≈ y = 0.52x + 1.4 17. ( ) ( ) ( ( ) ) ( ) 1 1 2 1 2 2 1 0 0 1 2 2 2 0 1 4 2 0 1 5 3 0 30 1 15 15 1 15 15 15 1 3 5 3 2 2 x x x y dy dx x y y dx x x x x dx x x x dx x x x x − = ⎡⎣ − ⎤⎦ = ⎡ − − − ⎤ ⎢⎣ ⎥⎦ = − + + − = ⎡− + + − ⎤ = ⎢⎣ ⎥⎦ ∫ ∫ ∫ ∫ ∫ 18. 2 1 2 1 1 2 1 0 0 2 0 2 0 2 0 0 1 2 ln 1 ln 1 1 1 1 y e y e x e y e dx dy dy dy y e y y y − − ⎡ ⎤ − ⎡ ⎤ − = ⎢ ⎥ = = ⎣ + ⎦ = = + ⎣ + ⎦ + ∫ ∫ ∫ ∫ 19. ( ) ( ) 3 2 2 3 2 2 2 2 2 0 2 2 3 0 0 0 3 2 3 2 4 square units x x 3 3 dy dx x x dx x x dx x x − + ⎡ ⎤ ⎡ ⎤ = ⎣ − − + ⎦ = − + = ⎢− + ⎥ = ⎣ ⎦ ∫ ∫ ∫ ∫ 20. ( ) [ ] ( ) 4 3 4 3 4 2 4 0 0 0 0 0 0 V = ∫ ∫ 8 − 2x dy dx = ∫ 8y − 2xy dx = ∫ 24 − 6x dx = ⎡⎣24x − 3x ⎤⎦ = 48 21. ( ) 2 3 1 1 2 1 1 3 2 2 3 0 0 0 0 0 0 9 9 2 3 Average 2 2 11 area 3 3 3 6 x y y dx x y dy dx x dx x x ⎡ ⎤ ⎛ ⎞ ⎡ ⎤ + ⎢ + ⎥ ⎜ + ⎟ ⎢ + ⎥ = = ⎣ ⎦ = ⎝ ⎠ = ⎣ ⎦ = ∫ ∫ ∫ ∫ [Show More]

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