Solution Manual for Calculus and Its Applications, 14th Edition

Chapter 2 Applications of the Derivative 2.1 Describing Graphs of Functions 1. (a), (e), (f) 2. (c), (d) 3. (b), (c), (d) 4. (a), (e) 5. Increasing for x .5, concave down, y-intercept (0, 0), x-intercepts (0, 0) and (1, 0). 6. Increasing for x –.4, concave down for x 3, y-intercept (0, 5), x-intercept (–3.5, 0). The graph approaches the x-axis as a horizontal asymptote. 7. Decreasing for x < 0, relative minimum point at x = 0, relative minimum value = 2, increasing for 0 < x 2, concave up for x 1, y-intercept at (0, 2), x-intercept (3.6, 0). 8. Increasing for x < –1, relative maximum at x = –1, relative maximum value = 5, decreasing for –1 < x 2.9, concave down for x 1, y-intercept (0, 3.3), x-intercepts (–2.5, 0), (1.3, 0), and (4.4, 0). 12. Increasing for x < –1.5, relative maximum at x = –1.5, relative maximum value = 3.5, decreasing for –1.5 < x < 2, relative minimum at x = 2, relative minimum value = –1.6, increasing from 2 < x 5.5, concave down for x < 0, inflection point at (0, 1), concave up for 0 < x 4, y-intercept (0, 1), x-intercepts (–2.8, 0), (.6, 0), (3.5, 0), and (6.7, 0). 13. The slope decreases for all x. 14. Slope decreases for x 3. 15. Slope decreases for x 1. Minimum slope occurs at x = 1. 16. Slope decreases for x 3. 17. a. c. 18. a. c. C, F A, E E 19. 20. 21. 11. Decreasing for 1 ≤ x 3, maximum value = 6 (at x = 1), minimum value = .9 (at x = 3), concave up for 1 ≤ x 4; the line y = 4 is an asymptote. 74 A, B, F b. D C 9. Decreasing for x 2, concave up for all x, no inflection point, defined for x > 0, the line y = x is an asymptote, the y-axis is an asymptote. 10. Increasing for all x, concave down for x 3, y-intercept (0, 1), x-intercept (–.5, 0). b. Copyright © 2018 Pearson Education Inc. Section 2.1 Describing Graphs of Functions 22. 75 28. Oxygen content decreases until time a, at which time it reaches a minimum. After a, oxygen content steadily increases. The rate at which oxygen content grows increases until b, and then decreases. Time b is the time when oxygen content is increasing fastest. 29. 1960 23. 30. 1999; 1985 31. The parachutist’s speed levels off to 15 ft/sec. 32. Bacteria population stabilizes at 25,000,000. 33. 24. 34. 25. 35. 26. 36. 27. 37. a. Yes; there is a relative minimum point between the two relative maximum points. b. Yes; there is an inflection point between the two relative extreme points. 38. No Copyright © 2018 Pearson Education Inc. 76 Chapter 2 Applications of the Derivative 39. 8. [0, 4] by [−15, 15] Vertical asymptote: x = 2 9. 40. [0, 50] by [−1, 6] c=4 10. 41. [−6, 6] by [−6, 6] The line y = x is the asymptote of the first 1 function, y   x. x 2.2 11. The First and Second Derivative Rules 1. (e) 2. (b), (c), (f) 12. 3. (a), (b), (d), (e) 4. (f) 5. (d) 6. (c) 13. 7. 14. Copyright © 2018 Pearson Education Inc. Section 2.2 The First and Second Derivative Rules 77 b. The function f(x) is increasing for 1 ≤ x < 2 because the values of f  ( x) are positive. The function f(x) is decreasing for 2 < x ≤ 3 because the values of f  ( x) are negative. Therefore, f(x) has a relative maximum at x = 2. Since f(2) = 9, the coordinates of the relative maximum point are (2, 9). 15. 16. 17. 18. c. The function f(x) is decreasing for 9 ≤ x < 10 because the values of f  ( x) are negative. The function f(x) is increasing for 10 < x ≤ 11 because the values of f  ( x) are positive. Therefore, f(x) has a relative minimum at x = 10. d. f (2)  0, so the graph is concave down. e. f ( x)  0, so the inflection point is at x = 6. Since f(6) = 5, the coordinates of the inflection point are (6, 5). f. The x-coordinate where f  ( x)  6 is x = 15. 24. a. f(2) = 3 b. t = 4 or t = 6 c. f f f  A POS POS NEG B 0 NEG 0 d. f(t) attains its least value after 5 minutes, at t = 5. To confirm this, observe that f (t )  0 for 4 ≤ t < 5 and f (t )  0 for 5 < t ≤ 6. C NEG 0 POS e. Since f (7.5)  1 , the rate of change is 1 unit per minute. f. The solutions to f (t )  1 are t = 2.5 and t = 3.5, so f (t ) is decreasing at the rate of 1 unit per minute after 2.5 minutes and after 3.5 minutes. g. The greatest rate of decrease occurs when f (t ) is most negative, at t = 3 (after 3 minutes). 19. 20. a. f ( x)  0 at x = 2 or x = 4; however f  ( x)  0 at x = 4, so there is a relative extreme point at x = 2. b. f  ( x)  0 at x = 3 or x = 4, so there are inflection points at x = 3 and at x = 4. 21. t = 1 because the slope is more positive at t = 1. 22. t = 2 because the v(2) is more positive than v(1). 23. a. f(t) attains its greatest value after 1 minute, at t = 1. To confirm this, observe that f (t )  0 for 0 ≤ t < 1 and f (t )  0 for 1 0. 40. f ( x)  3x 2  18 x  24  3( x 2  6 x  8)  3( x  2)( x  4) Graph I cannot be the graph because it does not have horizontal tangents at x = 2 and x = 4. 41. 5 3/ 2 15 x ; f ( x)  x1/ 2 2 4 Graph I could be the graph of f(x) since f  ( x)  0 for x > 0. f  ( x)  42. a. (C) b. (D) c. (B) d. (A) e. (E) 43. a. b. Since f (65)  .03 , the rate of change was –0.03 million farms per year. The number of farms was declining at the rate of about 30,000 farms per year. c. 36. f(0) = 3, f  (0)  1  y  3  1( x  0)  y  x3 37. a. 44. a. h(100.5)  h(100)  h  (100)(.5) The change = h(100.5) – h(100) 1 1 1  h  (100)(.5)     inch. 3 2 6 T (10)  T (10.75)  T  (10)(0.75)  4   3 degrees b. (ii) because the temperature is falling (assuming cooler is better). The graph of f (t ) reaches its minimum at t ≈ 35. Confirm this by observing that the graph of y  f  (t ) crosses the t-axis at t ≈ 35. The number of farms was decreasing fastest in 1960. Since f (5)  0 , the amount is decreasing. b. Since f (5)  0 , the graph of f(t) is concave up. b. (ii) because the water level is falling. 38. a. The solution of f(t) = 6 is t ≈ 15, so there were 6 million farms in 1940. d. The solutions of f  (t )  .06 are t ≈ 20 and t ≈ 53, so the number of farms was declining at the rate of 60,000 farms per year in 1945 and in 1978. e. f (0.25)  f (0)  f  (0)(.25)  3  (1)(.25)  3.25 Since f(65) ≈ 2, there were about 2 million farms. c. 3 4 The graph of f (t ) reaches its minimum at t = 4. Confirm this by observing that the graph of f  (t ) crosses the t-axis at t = 4. The level is decreasing fastest at t = 4 (after 4 hours). Copyright © 2018 Pearson Education Inc. Section 2.3 The First and Second Derivative Tests and Curve Sketching d. Since f (t ) is positive for 0 ≤ t 2, the greatest level of drug in the bloodstream is reached at t = 2 (after 2 hours). Critical Points, Intervals The solutions of f (t )  3 are t ≈ 2.6 and t ≈ 7, so the drug level is decreasing at the rate of 3 units per hour after 2.6 hours and after 7 hours. e. 45. x < −3 −3 < x < 3 3<x x–3 − − + x+3 − + + f  ( x) + − + f ( x) Increasing on , 3 Decreasing on  3,3 Increasing on 3,   f  x   3 x5  20 x 3  120 x y  f  ( x) Relative maximum at (−3, 54), relative minimum at (3, −54). y = f(x) 2. Note that f ( x)  15 x 4  60 x 2  120 , or use the calculator’s ability to graph numerical derivatives. Relative maximum: x ≈ –2.34 Relative minimum: x ≈ 2.34 Inflection point: x ≈ ±1.41, x = 0 f  x  x4  x2 y  f  ( x) f ( x)  x 3  6 x 2  1 f ( x)  3x 2  12 x  3x( x  4) f ( x)  0 if x  0 or x  4 f (0)  1; f (4)  31 Critical points: (0, 1), (4, −31) [−4, 4] by [−325, 325] 46. 79 y = f(x) Critical Points, Intervals x<0 0<x<4 4<x 3x − + + x−4 − − + f  ( x) + − + f ( x) Increasing on , 0 Decreasing on 0, 4 Increasing on  4,   Relative maximum at (0, 1), relative minimum at (4, −31). 3. [−1.5, 1.5] by [−.75, 1] Note that f ( x)  4 x 3  2 x, or use the calculator’s ability to graph numerical derivatives. Relative maximum: x = 0 Relative (and absolute) minimum: x ≈ ±.71 Inflection points: x ≈ ±.41 2.3 1. The First and Second Derivative Tests and Curve Sketching f ( x)  x 3  27 x f ( x)  3x 2  27  3( x 2  9)  3( x  3)( x  3) f ( x)  0 if x  3 or x  3 f (3)  54, f (3)  54 Critical points: (−3, 54), (3, −54) f ( x)   x 3  6 x 2  9 x  1 f ( x)  3 x 2  12 x  9  3( x 2  4 x  3)  3( x  1)( x  3) f ( x)  0 if x  1 or x  3 f (1)  3, f (3)  1 Critical points: (1, −3), (3, 1) Critical Points, Intervals x<1 1<x<3 3<x −3(x – 1) + − − x–3 − − + f  ( x) − + − f ( x) Decreasing on  ,1 Increasing on 1, 3 Decreasing on 3,   Relative maximum at (3, 1), relative minimum at (1, −3). Copyright © 2018 Pearson Education Inc. 80 Chapter 2 Applications of the Derivative f ( x)  6 x 3  4. 3 2 x  3x  3 2 6. Critical Points, Intervals x 2x + 1 + − + + − + Increasing on 1  ,    2 Decreasing on  1 1  ,  2 3  Increasing 1  on  ,   3  Critical Points, Intervals x 2x  1 − + 3x  1 − f  ( x) f ( x) 1 2  1 2 x 1 1 3 3 x 1 1 x 2 2 1 x 2 − + + 2x − 1 − − + f  ( x) + − + f ( x) Increasing on 1   ,   2 Decreasing on  1 1  ,  2 2  Increasing on 1   ,   2  1 5 minimum  ,  .  2 3  1 33  minimum at   ,   .  2 8 7. 1 3 x  x2 1 3 f ( x)   x 3  12 x 2  2 f ( x)  3 x 2  24 x  3x( x  8) f ( x)  0 if x  8 or x  0 f  8  258, f 0  2 f ( x)  x 2  2 x  x( x  2) f ( x)  0 if x  0 or x  2 1 f (0)  1; f (2)   3 1  Critical points: (0, 1),  2,    3 Critical Points, Intervals 1 2  1 7 Relative maximum   ,  , relative  2 3  1 43  Relative maximum at  ,   , relative  3 18  f ( x)  4 3 x x2 3 f ( x)  4 x 2  1  (2 x  1)(2 x  1) 1 1 f ( x)  0 if x   or x  2 2  1 7 1 5 f    ; f     2 3 2 3  1 7 1 5 Critical points:   ,  ,  ,   2 3  2 3 f ( x)  18 x 2  3x  3  3(6 x 2  x  1)  3(2 x  1)(3 x  1) 1 1 f ( x)  0 if x   or x  2 3 33  1  43  1 f     , f      2 3 8 18  1 33   1 43  Critical points:   ,   ,  ,    2 8   3 18  5. f ( x)  Critical points: 8, 258 , 0, 2 Critical Points, Intervals x<0 0<x<2 2 0) the entire graph is concave up and it has a minimum value. Unlike a parabola, it is not symmetric. Also, this graph has a vertical asymptote (x = 0), while a parabola does not have an asymptote. 3. y  2 x 2  5 x  2 5  5 2  4(2)(2) 5  3 1    ,2 2(2) 4 2  1  The x-intercepts are   , 0  and (–2, 0).  2  x 4. y  4  2 x  x 2 50.  (2)  (2) 2  4(1)(4) 2  2 5  2(1) 2  1  5 x [0, 25] by [0, 50] The relative minimum occurs at (5, 5). To determine this algebraically, observe that 75 f ( x)  3  2 . Solving f ( x)  0 gives x The x-intercepts are (1  5, 0) and (1  5, 0) . 5. y  4 x  4 x 2  1 4  4 2  4(4)(–1) 4  0  2(4) –8 1  The x-intercept is  , 0  . 2  x 2 x  25 , or x = ±5. This confirms that the relative extreme value (for x > 0) occurs at x = 5.To show that there are no inflection 150 points, observe that f  ( x)  3 . Since x f  ( x) changes sign only at x = 0 (where f(x) is undefined), there are no inflection points. 2.4 6. y  3x 2  10 x  3 10  10 2  4(3)(3) 10  8  2(3) 6 1   The x-intercepts are   , 0  and (–3, 0).  3  x Curve Sketching (Conclusion) 1. y  x 2  3 x  1 x  (3)  (3) 2  4(1)(1) 3  5  2(1) 2 7. x 3 5  , 0   2  8. 2. y  x  5 x  5 x 5  5 2  4(1)(5) 5  5  2(1) 2 1 3 x  2 x 2  5 x; f  ( x)  x 2  4 x  5 3  (4)  (4) 2  4(1)(5) 4  4  2(1) 2 Since f  ( x) has no real zeros, f(x) has no relative extreme points. 3 5  , 0  and The x-intercepts are   2  2 f ( x)  f ( x)   x 3  2 x 2  6 x  3 f  ( x)  3 x 2  4 x  6 4  4 2  4(3)(6) 4  56  2(3) 6 Since f  ( x) has noreal zeros, f(x) has no x relative extreme points. Since, f  ( x) < 0 for all x, f(x) is always decreasing. Copyright © 2018 Pearson Education Inc. Section 2.4 Curve Sketching (Conclusion) 9. f  x   x 3  6 x 2  12 x  6 Critical Points, Intervals f   x   3 x  12 x  12 f   x   6 x  12 2 To find possible extrema, set f ( x)  0 and solve for x. 2 3 x  12 x  12  0  2  3 x  4x  4  0 f  2  2 3  6  2 2  12  2  6  2 Thus, (2, 2) is a critical point. (continued) x2 2 x x−2 − + f  ( x) + + f ( x) Increasing on , 2 Increasing on .  2,   . No relative maximum or relative minimum. Since f  ( x)  0 for all x, the graph is always increasing. To find possible inflection points, set f  ( x)  0 and solve for x. 6 x  12  0  x  2 Since f   x   0 for x 2 (meaning the graph is concave up), the point (2, 2) is an inflection point. f 0  6, so the y-intercept is (0, −6). 10. f  x   x 3 x + − f  ( x) − − f ( x) Decreasing on , 0 Decreasing on 0,   is concave up) and f   x   0 for x > 0 (meaning the graph is concave down), the point (0, 0) is an inflection point. f 0  0, so the y-intercept is (0, 0). 11. f  x   x 3  3x  1 f   x   3x 2  3 f   x   6 x To find possible extrema, set f ( x)  0 and solve for x. 3x 2  3  0  no real solution Thus, there are no extrema. Since f  ( x)  0 for all x, the graph is always increasing. To find possible inflection points, set f  ( x)  0 and solve for x. 6x  0  x  0 f 0   1 Since f   x   0 for x 0 To find possible extrema, set f ( x)  0 and solve for x. 3x 2  0  x  0 x>0 No relative maximum or relative minimum. Since f  ( x)  0 for all x, the graph is always decreasing. To find possible inflection points, set f  ( x)  0 and solve for x. 6 x  0  x  0 Since f   x   0 for x < 0 (meaning the graph  x  2  0  x  2 2 Critical Points, Intervals x<0 89 (meaning the graph is concave up), the point (0, 1) is an inflection point. This is also the y-intercept. f  0   0 3  0 (continued on next page) Thus, (0, 0) is a critical point. Copyright © 2018 Pearson Education Inc. 90 Chapter 2 Applications of the Derivative (continued) To find possible inflection points, set f  ( x)  0 and solve for x. 12  6 x  0  x  2 f  2  5 Since f   x   0 for x  2 (meaning the graph is concave up) and f   x   0 for 12. x  2 (meaning the graph is concave down), the point (2, –5) is an inflection point. f 0  5, so the y-intercept is (0, 5). f  x  x3  2 x 2  4 x f   x   3x 2  4 x  4 f   x   6 x  4 To find possible extrema, set f ( x)  0 and solve for x. 3x 2  4 x  4  0  no real solution Thus, there are no extrema. Since f  ( x)  0 for all x, the graph is always increasing. To find possible inflection points, set f  ( x)  0 and solve for x. 6x  4  0  x   2 3 f  x  2x 3  x  2 f   x  6 x 2  1 f   x   12 x To find possible extrema, set f ( x)  0 and solve for x. 56  2 f      3 27 Since f   x   0 for x   23 (meaning the graph is concave down) and f   x   0 for x   23 (meaning the graph is concave up),  14.  is an inflection point. the point  23 ,  56 27 f 0  0, so the y-intercept is (0, 0). 6 x 2  1  0  no real solution Thus, there are no extrema. Since f  ( x)  0 for all x, the graph is always increasing. To find possible inflection points, set f  ( x)  0 and solve for x. 12 x  0  x  0 f 0  2 Since f   x   0 for x 0 (meaning the graph is concave up), the point (0, 2) is an inflection point. This is also the y-intercept. 13. f  x   5  13 x  6 x 2  x 3 f   x   13  12 x  3x 2 f   x   12  6 x To find possible extrema, set f ( x)  0 and solve for x. 13  12 x  3 x 2  0  no real solution Thus, there are no extrema. Since f  ( x)  0 for all x, the graph is always decreasing. Copyright © 2018 Pearson Education Inc. Section 2.4 Curve Sketching (Conclusion) 15. 4 3 x  2x 2  x 3 f   x   4 x 2  4 x  4 x  x  1 f   x   8 x  4 16. f  x  To find possible extrema, set f ( x)  0 and solve for x. 9 x 2  12 x  9  0  no real solution Thus, there are no extrema. Since f  ( x)  0 for all x, the graph is always decreasing. To find possible inflection points, set f  ( x)  0 and solve for x. 4 x 2  4 x  0  x  0, 1 f 0  0  0, 0 is a critical point 1  1 f 1   1,  is a critical point  3 3 Critical Points, Intervals x0 0  x 1 1 x 4x – + +  x  1 − – + f  ( x) + − + f ( x) Decreasing on  , 0 Decreasing on 0, 1 Increasing on 1,   18 x  12  0  x   Since f   x   0 for x   23 (meaning the graph is concave up) and f   x   0 for x   23 (meaning the graph is concave down),    1 8x  4  0  x  2 1 1 f   2 6 f 0  6, so the y-intercept is (0, –6). 17. f  x   1  3x  3x 2  x 3   f   x   3  6 x  3 x 2  3 x 2  2 x  1 f   x   6  6 x (meaning the graph is concave down) and f   x   0 for x  12 (meaning the graph is concave up), the point  the point  23 ,  169 is an inflection point. points. However, neither is a local maximum, nor a local minimum. Therefore, they may be inflection points. However, f  0  0 and f   0, so neither is an inflection point. Since f  ( x)  0 for all x, the graph is always increasing. To find possible inflection points, set f  ( x)  0 and solve for x. 2 3 16  2 f      3 9 We have identified (0, 0) and 1, 13 as critical Since f   x   0 f  x   3x 3  6 x 2  9 x  6 f   x   9 x 2  12 x  9 f   x   18 x  12 To find possible extrema, set f ( x)  0 and solve for x. for x  12 91  12 , 16  is an inflection point. f 0  0  (0, 0) is the y-intercept. To find possible extrema, set f ( x)  0 and solve for x. 3  6 x  3x 2  0  x  1 f 1  0 Since f  ( x)  0 for all x, the graph is always decreasing, and thus, there are no extrema. Therefore, (1, 0) may be an inflection point. Set f  ( x)  0 and solve for x. 6  6x  0  x  1 Since f   x   0 for x  1 (meaning the graph is concave up) and f   x   0 for x  1 (meaning the graph is concave down), the point (1, 0) is an inflection point. f 0  1, so the y-intercept is (0, 1). (continued on next page) Copyright © 2018 Pearson Education Inc. 92 Chapter 2 Applications of the Derivative  (continued)  Thus, (0, 0),  3, 9 , and  3, 9 are critical points. f  0  12, so the graph is concave down at x = 0, and (0, 0) is a relative maximum.      12  24 so the graph is concave up at x   3 , and   3, 9 , is a f   3  12  3 18. 1 3 x  2 x2 3 f   x  x2  4x f   x   2 x  4 f   x   0 if x  0 or x  4 f 0  0  0, 0 is a critical pt. 32 32   f  4     4,   is a critical pt.  3 3 f  x  relative minimum.  3   12  3   12  24 so the graph is concave up at x  3 , and  3, 9 , is a relative minimum. The concavity of this function reverses twice, so there must be at least two inflection points. Set f   x   0 and solve for x:   x  1 x  1  0  x  1 4 2 f  1   1  6  1  5 2 f 1  14  6 1  5  is a relative minimum. f   x   0 when x  2. 16 16   f  2     2,   is an inflection pt.  3 3 The y-intercept is (0, 0). Thus, the inflection points are (−1, −5) and (1, −5). 20. 19.  12 x 2  12  0  12 x 2  1  0  at x = 0, and (0, 0) is a relative maximum. f   4  4  0, so the graph is concave up at  2 f  f  0  4  0, so the graph is concave down x = 4, and 4,  32 3 2 f  x   3x 4  6 x 2  3 f  x  x 4  6 x 2 f   x   12 x 3  12 x f   x   12 x 2  12 To find possible extrema, set f ( x)  0 and solve for x. f   x   36 x 2  12 f   x   4 x 3  12 x To find possible extrema, set f ( x)  0 and solve for x. 12 x 3  12 x  0   4 x 3  12 x  0 12 x x 2  1  0  x  0, x  1 4 x x 2  3  0  x  0, x   3, x  3 f 0   3  0 4  6  0 2  3  3   f 0   0 4  6  0 2  0 4      6   3   9  18  9 4 2 f  3    3   6   3   9  18  9 f  3   3 4 f  1  3   1  6   1  3  0 2 2 f 1  3  1  6  1  3  0 Thus, (0, 3), (−1, 0) and (1, 0) are critical points. f  0  12, so the graph is concave down at 4 2 x = 0, and (0, 3) is a relative maximum. (continued on next page) Copyright © 2018 Pearson Education Inc. Section 2.4 Curve Sketching (Conclusion) (continued) Critical Points, Intervals x3 x–3 − + f  1  36 1  12  24 so the graph is f  ( x) – + concave up at x = 1, and (1, 0) is a relative minimum. The concavity of this function reverses twice, so there must be at least two inflection points. Set f   x   0 and solve for x: f ( x) Decreasing on ,3 Increasing on 3,   f   1  36  1  12  24 so the graph is 2 concave up at x = −1, and (−1, 0) is a relative minimum. 2 Thus, (3, 0) is local minimum. Since f   x   0, when x = 3, (3, 0) is also an inflection point. The y-intercept is (0, 81). 2 36 x  12  0  x 93 0  0 2  4(36)(12) 3  2  36 3 4 2    3 3 3 4 f   3   6  3    3  3   3   3  4 2  3  3  3 4 f  3  6    3 3  3   3   3   3 4 ,  and Thus, the inflection points are    3 3  3 4  3 , 3 .   22. f  x    x  2  1 4 f   x   4  x  2 f   x   12  x  2 3 To find possible extrema, set f ( x)  0 and solve for x. 4  x  2  0  x  2 3 f  2  1 Thus, (–2, –1) is a critical point. f   2  0, so we must use the first derivative rule to determine if (–2, –1) is a local maximum or minimum. 21. f  x    x  3 Critical Points, Intervals 4 f   x   4  x  3 f   x   12  x  3 3 To find possible extrema, set f ( x)  0 and solve for x. 4  x  3  0  x  3 3 f 3  0 Thus, (3, 0) is a critical point. f  3  0, so we must use the first derivative x –2 x+2 − + f  ( x) – + f ( x) Decreasing on , 2 Increasing on 2,   Thus, (–2, –1) is local minimum. Since f   x   0, when x = –2, (–2, –1) is also an inflection point. The y-intercept is (0, 15). rule to determine if (3, 0) is a local maximum or minimum. Copyright © 2018 Pearson Education Inc. 94 Chapter 2 Applications of the Derivative 1 1  x, x  0 x 4 1 1 y   2  4 x 2 y   3 x To find possible extrema, set y   0 and solve for x: 1 1  2  0 x2 4 x Note that we need to consider the positive solution only because the function is defined only for x > 0. When x = 2, y = 1, and 1 y    0, so the graph is concave up, and 4 (2, 1) is a relative minimum. Since y can never be zero, there are no 23. y  1 tells us that the x y-axis is an asymptote. As x  , the graph inflection points. The term approaches y  14 x, so this is also an asymptote of the graph. 9  x  1, x  0 x 9 y   2  1 x 18 y   3 x To find possible extrema, set y   0 and solve for x: 9 9  2  1  0   2  1  9  x 2  x  3 x x Note that we need to consider the positive solution only because the function is defined 9 only for x > 0. When x = 3, y   3  1  7, 3 18 and y   3  0, so the graph is concave up, 3 and (3, 7) is a relative minimum. Since y can never be zero, there are no 25. y  9 tells us that the x y-axis is an asymptote. As x  , the graph approaches y = x + 1, so this is an asymptote of the graph. inflection points. The term 2 , x0 x 2 y   2 x 4 y   3 x To find possible extrema, set y   0 and solve for x: 2  2  0  no solution, so there are no x extrema. Since y   0 for all x, the graph is always 24. y  decreasing. Since y can never be zero, there 2 tells us x that the y-axis is an asymptote. As x  , the graph approaches y = 0, so this is also an asymptote of the graph. are no inflection points. The term 12  3x  1, x  0 x 12 y   2  3 x 24 y   3 x To find possible extrema, set y   0 and solve for x: 12 12  2  3  0   2  3  4  x 2  x  2 x x 26. y  (continued on next page) Copyright © 2018 Pearson Education Inc. Section 2.4 Curve Sketching (Conclusion) (continued) Note that we need to consider the positive solution only because the function is defined only for x > 0. When x = 2, y = 13 and 24 y   3  0, so the graph is concave up, and 2 (2, 13) is a relative minimum. Since y can never be zero, there are no 12 tells us that the x y-axis is an asymptote. As x  , the graph approaches y = 3x + 1, so this is also an asymptote of the graph. inflection points. The term 2 x   2, x  0 x 2 2 1 y   2  2 x 4 y   3 x To find possible extrema, set y   0 and solve for x: 2 1 2 1  2  0 2   x2 2 2 x x Note that we need to consider the positive solution only because the function is defined only for x > 0. When x = 2, y = 4 and 4 y   3  0, so the graph is concave up, and 2 (2, 4) is a relative minimum. Since y can never be zero, there are no 27. y  2 tells us that the x y-axis is an asymptote. As x  , the graph x approaches y   2, so this is also an 2 asymptote of the graph. inflection points. The term x 5  , x0 4 4 x 2 1 y   3  4 x 6 y   4 x To find possible extrema, set y   0 and solve for x: 2 1 2 1  3   0   3    8  x3  x  2 4 4 x x Note that we need to consider the positive solution only because the function is defined only for x > 0. When x = 2, 1 2 5 1 6 y  2     , and y   4  0, so 4 4 2 2 x 1  the graph is concave up, and  2,   is a  2 relative minimum. Since y can never be zero, there are no inflection points. The term 1 tells us that the y-axis is an asymptote. As x2 x 5 x  , the graph approaches y   , so 4 4 this is an asymptote of the graph. If x = 1, then 1 1 5 y = y  2    0, so (1, 0) is an 4 4 1 x-intercept. 28. y  1 95 2  29. y  6 x  x, x  0 3 1 y   3x 1 2  1  x 3 y    x  3 2 2 To find possible extrema, set y   0 and solve for x: 3 1  0  x  9 x Note that we need to consider the positive solution only because the function is defined only for x > 0. When x = 9, y = 9, and y   0, so the graph is concave down, and (9, 9) is a relative maximum. Since y can never be zero, there are no inflection points. (continued on next page) Copyright © 2018 Pearson Education Inc. 96 Chapter 2 Applications of the Derivative (continued) When x = 0, y = 0, so (0, 0) is the y-intercept. y  6 x  x  0  36 x  x 2  x  36, so (36, 0) is an x-intercept 32. g ( x)  f  ( x) . The zeros of g(x) correspond to the extreme points of f(x). But the zeros of f(x) also correspond to the extreme points of g(x). Observe that at points where f(x) is decreasing, g(x) 0. But at points where g(x) is increasing, f(x) 0. 33. 1 x  , x0 2 x 1 1 y   3 2  2 2x 3 y   5 2 4x To find possible extrema, set y   0 and solve for x: 1 1  3 2   0  x 1 2 2x 3 3 When x = 1, y  , and y    0, so the 2 4 Therefore, f ( x)  ax 2  c; f (0)  c  1; f (2)  0  4a  2b  c  0  1 4a  1  0  a   : 4 1 2 Thus, f ( x)   x  1. 4 30. y    graph is concave up, and 1, 32 is a relative minimum. Since y can never be zero, there 1 tells us x that the y-axis is an asymptote. As x  , the are no inflection points. The term graph approaches y  2x , so this is an asymptote of the graph. The graph has no intercepts. f ( x)  ax 2  bx  c; f ( x)  2ax  b f (0)  b  0 (There is a local maximum at x  0  f (0)  0). 34. f ( x)  ax 2  bx  c; f ( x)  2ax  b f (1)  2a  b  0 (There is a local maximum at x  1  f  (1)  0); b  2a Therefore, f ( x )  ax 2  2ax  c; f (0)  c  1; f (1)  a  2a  1  1  a  2, b  4 : f ( x)  2 x 2  4 x  1. 35. Since f (a )  0 and f  ( x) is increasing at x = a, f   0 for x a. According to the first derivative test, f has a local minimum at x = a. 36. Since f (a )  0 and f  ( x) is decreasing at x = a, f   0 for x a. According to the first derivative test, f has a local maximum at x = a. 37. a. [0, 20] by [−12, 50] 31. g ( x)  f  ( x) . The 3 zeros of g(x) correspond to the 3 extreme points of f(x). f ( x)  g  ( x) , the zeros of f(x) do not correspond with the extreme points of g(x). b. Since f(7) = 15.0036, the rat weighed about 15.0 grams. c. Using graphing calculator techniques, solve f(t) = 27 to obtain t ≈ 12.0380. The rat’s weight reached 27 grams after about 12.0 days. Copyright © 2018 Pearson Education Inc. Section 2.5 Optimization Problems d. e. The solutions of f  (t )  .02 are t ≈ 64.4040, t ≈164.0962, and t ≈ 216.9885. The canopy was growing at the rate of .02 meters per day after about 64.4 days, after about 164.1 days, and after 217.0 days. f. Since the solution to f (t )  0 is t ≈ 243.4488, the canopy has completely stopped growing at this time and we may say that the canopy was growing slowest after about 243.4 days (see the graph in part (d)). (The growth rate also has a relative minimum after about 103.8 days.) g. The graph shown in part (d) shows that f (t ) was greatest at t = 32, after 32 days. (The growth rate also has a relative maximum after about 193.2 days.) f  t  [0, 20] by [−2, 5] Note that f (t )  .48  .34t  .0144t 2 . Since f (4)  1.6096 , the rat was gaining weight at the rate of about 1.6 grams per day. e. f. Using graphing calculator techniques, solve f  (t )  2 to obtain t ≈ 5.990 or t ≈ 17.6207. The rat was gaining weight at the rate of 2 grams per day after about 6.0 days and after about 17.6 days. The maximum value of f (t ) appears to occur at t ≈ 11.8. To confirm, note that f  (t )  .34  .0288 x , so the solution of f  (t )  0 is t ≈ 11.8056. The rat was growing at the fastest rate after about 11.8 days. 38. a. 2.5 Optimization Problems 1. g ( x)  10  40 x  x 2  g  ( x)  40  2 x  g  ( x)  2 The maximum value of g(x) occurs at x = 20; g(20) = 410. [32, 250] by [−1.2, 4.5] b. Since f(100) = 1.63, the canopy was 1.63 meters tall. c. The solution of f(t) = 2 is t ≈ 143.9334. The canopy was 2 meters high after about 144 days. d. Note that 97 2. f ( x)  12 x  x 2  f  ( x)  12  2 x  f  ( x)  2 The maximum value of f(x) occurs at x = 6; f(6) = 36. f (t )  .142  .0032t  .0000237t 2 .0000000532t 3 (Alternately, use the calculator’s numerical differentiation capability.) The graph of y  f  (t ) is shown. Since f (80)  .0104 , the canopy was growing at the rate of about .0104 meters per day. [32, 250] by [−.01, .065] Copyright © 2018 Pearson Education Inc. 98 Chapter 2 Applications of the Derivative 3. f (t )  t 3  6t 2  40  f  (t )  3t 2  12t  f  (t )  6t  12 The minimum value for t ≥ 0 occurs at t = 4; f(4) = 8. 4 The maximum value of Q(x) occurs at x  . 3 4 2 Then y  2   . 3 3 7. x  y  6  y  6  x Q( x)  x 2  (6  x) 2  2 x 2  12 x  36 dQ d 2Q  4 x  12; 4 dx dx 2 dQ  0  4 x  12  0  x  3 dx The minimum of Q(x) occurs at x = 3. The 4. f (t )  t 2  24t  f  (t )  2t  24  f  (t )  2 The minimum value of f(t) occurs at t = 12; f(t) = –144. minimum is Q(3)  3 2  (6  3) 2  18 9. xy  36  y  S ( x)  x  S  ( x)  1  d 2Q  2 2 dx The maximum value of Q(x) occurs at x = 1, y = 1. Q(1)  2(1)  (1) 2  1. 3 Q( x)  x (2  x)  2 x  x . dQ  4 x  3x 2 dx dQ 4  0  4 x  3x 2  0  x  0 or x  dx 3 dx d 2Q dx 2  4, x0 x2  0  x  6 or  6 3 , S  (6)  10. x  y  1  y  1  x y  z  2  z  2  y  1 x Q( x)  x(1  x)(1  x)  x  x 3 Q  ( x)  1  3 x 2  3 Q  ( x)  6 x, Q     2 3  3  Substituting into Q  x 2 y yields  4  6 x, 72 36 Q  ( x)  0  1  3 x 2  0  x  6. Solving x + y = 2 for y gives y = 2 – x. 2 x2 4 3 x The positive value x = 6 minimizes S(x), and 36 y  6. S (6, 6)  6  6  12 6 dQ  2  2x dx dQ  0  2  2x  0  x  1 dx d 2Q 36 x 36 S  ( x)  1  Q( x)  x(2  x)  2 x  x 2 . 2 36 x S  ( x)  0  1  5. Solving x + y = 2 for y gives y = 2 – x. Substituting into Q = xy gives 2 d 2Q  4, so the function is dx 2 concave upward at all points. 8. No maximum. d 2Q dx 2 x  4 3  4 Q(x) is a maximum when x  3 3 3 , 3 3 3 3 3 3 3  , and z  1   . 3 3 3 3 The maximum value of Q(x) is y  1 3  3 3  3 2 3 Q    .  3  3  9  3  Copyright © 2018 Pearson Education Inc. Section 2.5 Optimization Problems 11. Let A = area. a. Objective equation: A = xy Constraint equation: 8x + 4y = 320 14. a. b. Solving constraint equation for y in terms of x gives y = 80 – 2x. Substituting into objective equation yields b. Let P = perimeter. Objective: P = 2x + 2y Constraint: 100 = xy A = x(80 – 2x) = 2 x 2  80 x . c. dA d 2A  4 x  80   4 dx dx 2 The maximum value of A occurs at x = 20. Substituting this value into the equation for y in part b gives y = 80 – 40 = 40. Answer: x = 20 ft, y = 40 ft c. Objective equation: S  x 2  4 xh Constraint: x 2 h  32 b. From constraint equation, h  32 x2 . Thus, 100 . So x 200  100  P  2x  2   2x   x  x From the constraint, y  dP 200 d 2 P  2 2 ;  400 x 3 dx x dx 2 The minimum value of P for x > 0 occurs at x = 10. Solving for y gives 100 y  10 . 10 Answer: x = 10 m, y = 10 m 12. Let S = surface area. a. 99 15. 128  32  . S  x 2  4x  2   x 2  x  x c. dS 128 d 2S 256  2x  2 ;  2 3 2 dx dx x x The minimum value of S for x > 0 occurs 32 at x = 4. Solving for h gives h  2  2 . 4 Answer: x = 4 ft, h = 2 ft Let C = cost of materials. Objective: C = 15x + 20y Constraint: xy = 75 Solving the constraint for y and substituting 1500  75  gives C  15 x  20    15 x  ;  x x 13. a. dC 1500 d 2C 3000  15  2 ;  3 dx x dx 2 x The minimum value for x > 0 occurs at x = 10. Answer: x = 10 ft, y = 7.5 ft 16. b. length + girth = h + 4x c. Objective equation: V  x 2 h Constraint equation: h + 4x = 84 or h = 84 – 4x d. Substituting h = 84 – 4x into the objective equation, we have V  x 2 (84  4 x)  4 x 3  84 x 2 . e. V   12 x 2  168 x V   24 x  168 The maximum value of V for x > 0 occurs at x = 14 in. Solving for h gives h = 84 – 4(14) = 28 in. Let C = cost of materials. Constraint: x 2 y  12 Objective: C  2 x 2  4 xy  x 2  3 x 2  4 xy Solving the constraint for y and substituting 48  12  ; gives C  2 x 2  4 x  2   3 x 2  x  x dC 48 d 2C 96  6x  2 ;  6 3 dx x dx 2 x The minimum value of C for x > 0 occurs at x = 2. Answer: x = 2 ft, y = 3 ft Copyright © 2018 Pearson Education Inc. 100 Chapter 2 Applications of the Derivative 17. Let x = length of base, h = height, M = surface area. 8000 Constraint: x 2 h  8000  h  2 x 2 18. 21. Constraint: x + y = 100 Objective: P  xy Solving the constraint for y and substituting gives P  x(100  x)   x 2  100 x Objective: M  2 x  4 xh Solving the constraint for y and substituting 32, 000  8000  gives M  2 x 2  4 x  2   2x 2   x  x dP d 2P  2 x  100;  2 dx dx 2 The maximum value of P occurs at x = 50. Answer: x = 50, y = 50 dM 32, 000 d 2 M 64, 000 ;  4x   4 2 2 dx x dx x2 The minimum value of M for x > 0 occurs at x = 20. Answer: 20 cm  20 cm  20 cm 22. Constraint: xy = 100 Objective: S  x  y Solving the constraint for y and substituting 100 gives S  x  x dS 100 d 2 S 200  1 2 ;  3 dx x dx 2 x The minimum value of S for x > 0 occurs at x = 10. Answer: x = 10, y = 10 23. y x x Let C = cost of materials. Constraint: x 2 y  250 Objective: C  2 x 2  2 xy Solving the constraint for y and substituting 500 gives C  2 x 2  x dC 500 d 2C 1000  4x  2 ;  4 3 2 dx x dx x The minimum value of C for x > 0 occurs at x = 5. Answer: x = 5 ft, y = 10 ft 19. Let x = length of side parallel to river, y = length of side perpendicular to river. Constraint: 6x + 15y = 1500 Objective: A = xy Solving the constraint for y and substituting 2  2  gives A  x   x  100   x 2  100 x 5  5  dA 4 d 2A 4   x  100; 2   dx 5 5 dx The minimum value of A for x > 0 occurs at x = 125. Answer: x = 125 ft, y = 50 ft Constraint: 2x + 2h + πx = 14 or (2 + π)x + 2h = 14 Objective: A  2 xh   x2 2 Solving the constraint for h and substituting gives 2   2  A  2x 7  x  x   2 2   2  14 x    2  x 2  dA d 2A  14  (4   ) x;  4 –  dx dx 2 The maximum value of A occurs at x  Answer: x  20. Let x = length, y = width of garden. Constraint: 2x + 2y = 300 Objective: A  xy Solving the constraint for y and substituting gives A  x(150  x)   x 2  150 x dA d 2A  2 x  150;  2 dx dx 2 The maximum value of A occurs at x = 75. Answer: 75 ft  75 ft Copyright © 2018 Pearson Education Inc. 14 ft 4 14 . 4 Section 2.5 Optimization Problems 24. 101 27. Let S = surface area. Constraint:  x 2 h  16 or x 2 h  16 Objective: S  2 x 2  2 xh Solving the constraint for h and substituting 16   16   gives S  2 x 2  2 x  2   2  x 2    x  x 2 dS 16  d S 32     2  2 x  2  ;  2  2  3  2    dx x dx x  The minimum value of S for x > 0 occurs at x = 2. Answer: x = 2 in., h = 4 in. 1 dA d 2A 25. A  20w  w 2 ;  20  w;  1 2 dw dw 2 The maximum value of A occurs at w = 20. 1 1 x  20  w  20  (20)  10 2 2 Answer: w = 20 ft, x = 10 ft 26. Let x miles per hour be the speed. d = s · t , so 500 time of the journey is hours. Cost per x hour is 5 x 2  2000 dollars. Cost of the journey is 1, 000, 000  500  C  (5 x 2  2000)    2500 x    x  x dC 1, 000, 000 dC  2500   0 , and we . Set dx dx x2 If x = distance from C to P, let y = be the distance from P to M. Then cost is the objective: C  6 x  10 y and the constraint y 2  (20  x) 2  24 2  976  40 x  x 2 . Solving the constraint for y and substituting  gives C  6 x  10 976  40 x  x 2  . 12 dC  6  5(976  40 x  x 2 ) 1 2 (40  2 x) . dx 5(40  2 x)  6 976  40 x  x 2 dC Solve  0: dx 5(40  2 x) 6 0 976  40 x  x 2 5(40  2 x)  6 976  40 x  x 2 200  10 x  6 976  40 x  x 2 40000  4000 x  100 x 2  36 x 2  1440 x  35136 64 x 2  2560 x  4864  0 x 2  40 x  76  0  x  2, x  38. But x  20  x  38 and d 2C dx 2 x  2  0. Therefore, the value of x that minimizes the cost of installing the cable is x = 2 meters and the minimum cost is C = $312. obtain x 2  400  x  20. The speed is 20 miles per hour. Copyright © 2018 Pearson Education Inc. 102 Chapter 2 Applications of the Derivative 28. x x 2  22 x  137  ( x  11) x 2  36  x x  22x  137   ( x  11) x  36  2 2 2 2 x 4  22 x 3  137 x 2  x 4  22 x 3  157 x 2  792 x  4356 20 x 2  792 x  4356  0 x Let P be the amount of paper used. The objective is P  ( x  2)( y  1) and the constraint is x  y  50 . Solving the constraint for y and substituting gives 100  50  and P  ( x  2)   1  52  x   x  x dP 100 dP  1  2 . Solve  0: dx dx x 100 1  2  0  x  10, x  10. But x > 0 and x d 2P dx 2 x 10  0. Therefore, x  10, y  5 and the dimensions of the page that minimize the amount of paper used: 6 in. × 12 in. 2 29. Distance = ( x  2)  y By the hint we minimize 2 792  7922  4 204356 2  20  33 or 6.6 Since 0 ≤ x ≤ 11, we have x = 6.6. The minimum total distance is D(6.6)  6.6 2  36  16  (11  6.6) 2  221  14.87 miles. 31. Distance  x 2  y 2  x 2  ( 2 x  5) 2  5 x 2  20 x  25 The distance has its smallest value when 5 x 2  20 x  25 does, so we minimize D( x)  5 x 2  20 x  25  D  ( x)  10 x  20 Now set D ( x)  0 and solve for x: 10 x  20  0  x  2 y  2(2)  5  1 The point is (2, 1). 32. Let A = area of rectangle. Objective: A = 2xy D  ( x  2) 2  y 2  ( x  2) 2  x , since Constraint: y  9  x 2 Substituting, the area of the rectangle is given y x. by A  2 x 9  x 2 . dD  2( x  2)  1 dx dD Set  0 to give: 2x = 3, or dx x 3 3 3 . So the point is  , . 2 2 2 3 ,y 2 30. Let D be the total distance. D( x)  d1  d 2  x 2  36  16  (11  x) 2 x x  11 D ( x)   2 2 x  36 x  22 x  137 Now set D ( x)  0 and solve for x: x 2 x  36  x  11 2 x  22 x  137 [0, 3] by [−2, 10] Using graphing calculator techniques, this function has its maximum at x ≈ 2.1213. To confirm this, use the calculator’s numerical differentiation capability to graph the derivative, and observe that the solution of dA  0 is x ≈ 2.1213. dx 0 x x 2  22 x  137  ( x  11) x 2  36  0 Copyright © 2018 Pearson Education Inc. (continued on next page) Section 2.6 Further Optimization Problems b. The order quantity multiplied by the number of orders per year gives the total number of packages ordered per year. The constraint function is then rx = 800. (continued) c. Solving the constraint function for r gives 800 r . Substituting into the cost x 12,800 equation yields C ( x)  2 x  . x 12,800 12,800 C ( x)  2   2 0 2 x x2 12,800 x2   6400  x  80, r  10 2 The minimum inventory cost is C (80)  $320. 4. a. The order cost is 160r, and the carrying x cost is 32   16 x. The inventory cost C 2 is C = 160r + 16x. [0, 3] by [−10, 10] The maximum area occurs when x ≈ 2.12. 2.6 Further Optimization Problems 1. a. At any given time during the orderreorder period, the inventory is between 180 pounds and 0 pounds. The average is 180  90 pounds. 2 b. The maximum is 180 pounds. c. The number of orders placed during the year can be found by counting the peaks in the figure. b. The order quantity times the number of orders per year gives the total number of sofas ordered. The constraint function is rx = 640. c. There were 6 orders placed during the year. d. There were 180 pounds of cherries sold in each order-reorder period, and there were 6-order-reorder periods in the year. So there were 6 · 180 = 1080 pounds sold in one year. 2. a. There are 6 orders in a year, so the ordering cost is 6 · 50 = $300. The average inventory is 90 pounds, so the carrying cost is 90 · 7 = $630. The inventory cost is $300 + $630 = $930. b. The maximum inventory is 180 pounds, so the carrying cost is 7 · 180 = $1260. The inventory cost is $300 + $1260 = $1560. 3. a. 103 The order cost is 16r, and the carrying x cost is 4   2 x. The inventory cost C is 2 C = 2x + 16r. Solving the constraint function for r gives 640 r . Substituting into the cost x equation yields 102, 400 C ( x)   16 x. x 102, 400 C ( x)  16  x2 102, 400 C ( x)  0  16   0  x  80 x2 The minimum inventory cost is C (80)  $2560. 5. Let x be the order quantity and r the number of orders placed in the year. Then the inventory cost is C = 80r + 5x. The constraint is 10, 000 rx = 10,000, so r  and we can write x 800, 000 C ( x)   5x . x a. C (500)  Copyright © 2018 Pearson Education Inc. 800, 000  5(500)  $4100 500 104 Chapter 2 Applications of the Derivative b. C ( x)    800, 000 x 800, 000 2 5 5 0 x2 800, 000 x2   160, 000  x  400 5 The minimum value of C(x) occurs at x = 400. 6. Let x be the number of tires produced in each production run, and let r be the number of runs in the year. Then the production cost is C = 15,000r + 2.5x. The constraint is 600, 000 600, 000 rx = 600,000, so x  and r  . r x 1, 500, 000 Then C (r )  15, 000r  and r 15, 000(600, 000) C ( x)   2.5 x . x a. C (10)  15, 000(10)  b. C ( x)  –9  10 x 2 9 –9  10 9 x 2 1, 500, 000  300, 000 10 x 9. The inventory cost is C  hr  s   where r 2 is the number of orders placed and x is the Q order size. The constraint is rx = Q, so r  x hQ sx and we can write C ( x)   . x 2  hQ s C  ( x)  2  . Setting C  ( x)  0 gives 2 x  hQ  2.5   2.5  0  x 2  8. Let x be the size of each order and let r be the number of orders placed in the year. Then the inventory cost is C = 40r + 2x and rx = 8000, 8000 1600 so x  , C (r )  40r  r r 16, 000 16, 000 C (r )  40   40  0 r2 r2 . 16, 000 r2   r  20 40 The minimum value for C occurs at r = 20 (for r > 0). x2 –9  10 9  2.5 9  10 9 3  x   10 4  x  60, 000 .25  10 .5 Each run should produce 60,000 tires. x2  7. Let x be the number of microscopes produced in each run and let r be the number of runs. The objective function is x C  2500r  15 x  20    2500r  25 x . 2 The constraint is xr = 1600, x  1600 , so r 40, 000 C (r )  2500r  . r 40, 000 C (r )  2500   r2 . 40, 000 40, 000 2 2500   0  r   r  4 2500 r2 C has a minimum at r = 4. There should be 4 production runs.  2hQ s 2hQ  0, x 2  , x . The s 2 s 2hQ gives the minimum s value for C(x) for x > 0. positive value 10. In this case, the inventory cost becomes for x  600 75r  4 x C  (75  ( x  600))r  4 x for x  600 Since r  1200 , x  90,000  4 x for x  600  x C ( x)   810,000  x  4 x  1200 for x  600 Now the function 810, 000 f ( x)   4 x  1200 has x 810, 000 f ( x)    4, f  (450)  0 and x2 f ( x)  0 for x > 450. Thus, C(x) is increasing for x > 600 so the optimal order quantity does not change. Copyright © 2018 Pearson Education Inc. Section 2.6 Further Optimization Problems 11. C ( x)  10  648 x2 18 5 x 18 The optimal dimensions are x  m, 5 C ( x)  0  10  The objective is A = (x + 100)w and the constraint is x + (x + 100) + 2w = 2x + 2w + 100 = 400; or x + w = 150, w = 150 – x. A( x)  ( x  100)(150  x)   x 2  50 x  15, 000 A ( x)  2 x  50, A (25)  0 The maximum value of A occurs at x = 25. Thus the optimal values are x = 25 ft, w = 150 – 25 = 125 ft. 15. a. C ( x)  10 x  54 and x 0 x (0,1000), (5,1500)  1500  1000  100 . 50 y  1500  100  x  5 ; y  100 x  1000  A( x )  100 x  1000 . b. Let x be the discount per pizza. Then, for 0  x  18, revenue  R ( x)  (100 x  1000)(18  x)  18000  800 x  100 x 2 R ‘( x)  800  200 x  800  200 x  0  x  4 Therefore, revenue is maximized when the discount is x  $4 . c. 13. The constraint is xw = 54, so w  2 m A  x    x  10050  x    x 2  50 x  5000 , A ( x)  2 x  50 A ( x)  0  2 x  50  0  x  25 . In this case, the maximum value of A occurs at x = –25, and A(x) is decreasing for x > –25. Thus, the best non-negative value for x is x = 0. The optimal dimensions are x = 0 ft, w = 50 ft. 14. Refer to the figure for exercise 13. The objective is C  2(5 x)  2(5w)  2 w  10 x  12 w. 648 w  3 5 m. 12. Refer to the figure for exercise 11. The objective remains A = (x + 100)w, but the constraint becomes 2x + 2w + 100 = 200; or x + w = 50, so A(x) = (x + 100)(50 – x) The objective is F = 2x + 3w, and the 54 constraint is xw = 54, or w  , so x 162 F ( x)  2 x  , x 162 162 F  ( x)  2  2  2  2  0  x x . 162 2 x   x9 2 The minimum value of F for x > 0 is x = 9. The optimal dimensions are thus x = 9 m, w = 6 m. 105 Let each pizza cost $9 and let x be the discount per pizza. Then A( x )  100 x  1000 and, for 0  x  9, revenue  R( x)  (100 x  1000)(9  x). R ( x )  9000  100 x  100 x 2 R  ( x)  100  200 x  100  200 x  0  x  .5 In this case, revenue is maximized when the discount is x = −$.50. Since 0  x  9, the revenue is maximized when x = 0. 16. The objective is S  2 x 2  3xy where x and y are the dimensions of the box. The constraint 36 is x 2 y  36 , so y  2 and x 108  36  S ( x)  2 x 2  3 x  2   2 x 2  . x  x 648 . x (continued on next page) Copyright © 2018 Pearson Education Inc. 106 Chapter 2 Applications of the Derivative (continued) S  ( x)  4 x  The objective is A = xh and the constraint is  x  2h  440  h  220  108 x2 S  ( x)  0  4 x  108 2 2 x.     A( x )  x  220  x   220 x  x 2 ,  2  2 0 x3. x x 2 y  36  9 y  36  y  4 The optimal dimensions are 3 in.  3 in.  4 in. A ( x)  220   x  220   x  0  x  The optimal dimensions are x  17. Let x be the length and width of the base and let y be the height of the shed. The objective is C  4 x 2  2 x 2  4  2.5 xy  6 x 2  10 xy . The constraint is x 2 y  150  y   150 x2 h = 110 yd. 220  220  yd, 21. . 1500 1500 , C  ( x)  12 x  2 x x 1500 C ( x)  0  12 x  2  0  x  5 x The optimal dimensions are 5 ft  5 ft  6 ft. C ( x)  6 x 2  18. Let x be the length of the front of the building and let y be the other dimension. The objective is C = 70x + 2 · 50y + 50x = 120x + 100y and 12, 000 . the constraint is xy = 12,000 y  x 1, 200, 000 , So C ( x )  120 x  x 1, 200, 000 C  ( x )  120  , C  (100)  0 . x2 The optimal dimensions are x = 100 ft, y = 120 ft. 19. Let x be the length of the square end and let h be the other dimension. The objective is V  x 2 h and the constraint is 2x + h = 120 h = 120 – 2x. V ( x)  120 x 2  2 x 3 , V ( x)  240 x  6 x 2  V  ( x)  0  240 x  6 x 2  0  . 6 x  40  x   0  x  0 or x  40 The maximum value of V for x > 0 occurs at x = 40 cm, h = 40 cm. The optimal dimensions are 40 cm  40 cm  40 cm. 20. The objective equation is V  w 2 x and the constraint is w  2 x  16  w  16  2 x. V ( x)  (16  2 x) 2 x  4 x 3  64 x 2  256 x V  ( x)  12 x 2  128 x  256 V  ( x)  0  12 x 2  128 x  256  0  3 4  x  83x  8  0  x  8 or x  8 8 V     0, V  (8)  0 3 The maximum value of V for x between 0 and 8 8 occurs at x  in. 3 22. Let x be the width of the base and let h be the other dimension. The objective is V  2 x 2 h and the constraint is 2(2 x 2 )  2 xh  2(2 xh)  27 , or 4 x 2  6 xh  27  h  V 27  4 x 2 . Thus, 6x 2 x 2 (27  4 x 2 ) 4  9x  x3 . 6x 3 V  ( x)  9  4 x 2  9  4 x 2  0  x 2  x 3 2 The optimal values are x  dimensions should be Copyright © 2018 Pearson Education Inc. 9  4 3 , h = 2. The 2 3 ft  3 ft  2 ft . 2 Section 2.6 Further Optimization Problems 23. We want to find the maximum value of f (t ) . f  (t )  f  (t )  10 (t  10) 20 2 (t  10) 3 f  (t )  0 gives 20   200 (t  10) 3 600 (t  10) 4 27. ; . Setting A  x 2  5 xh (Area: where x is the length of the square base and h is the height.) 400 V  x 2 h  400  h  2 , so x dA 2000 2000 , and  2x  2 . A  x2  dx x x dA Setting  0 gives 2 x 3  2000 or x = 10 dx which in turn yields h = 4 in. The dimensions should be 10 in.  10 in.  4 in. 600 600  20   t  20. (t  10) (t  10) 4 60 2400 f  (t )   ; f  (20)  0 , so 4 (t  10) (t  10) 5 t = 20 is the maximum value of f (t ) . Oxygen content is increasing fastest after 20 days. (t  10) 3 107  24. We want to find the maximum value of 1 f (t )  40  2t  t 2 . 5 2 2 f  (t )  2  t  2  t  0  t  5 . 5 5 The maximum rate of output occurs at t = 5. The maximum output rate is f  (5)  45 tons/hour. 28. Since f  ( x) is negative on the interval 0  x  5, f(x) is decreasing on the interval 0. Therefore f(x) has its greatest value at zero. 29. 25. Let (x, y) be the top right-hand corner of the window. The objective is A = 2xy and the constraint is y  9  x 2 . Thus, A( x)  2 x(9  x 2 )  18 x  2 x 3 , A ( x)  18  6 x 2 . A ( x)  0  18  6 x 2  0  x  3 The maximum value of A for x > 0 occurs at x  3 . Thus, the window should be 6 units high and 2 3 units wide. 26. We want to find the minimum value of 1000 8000 f  (t )  ;  (t  8) 2 (t  8) 3 2000 24, 000 . Setting f  (t )  0 f  (t )   3 (t  8) (t  8) 4 gives 2000 24, 000 24, 000   2000   t  4. t 8 (t  8) 3 (t  8) 4 6000 96, 000 , f  (4)  0 , so t = (t  8) (t  8) 5 4 gives the minimum value of f (t ) . Sales fall the fastest after 4 weeks. f  (t )  4  Let V = volume of box, and let l and w represent the dimensions of the base of the box. Objective: V = lwx 40  3x , w = 20 – 2x Constraints: l  2 Substituting, the volume of the box is given by  40  3x  V  (20  2 x) x  3 x 3  70 x 2  400 x .  2  [0, 10] by [0, 700] Since we require the dimensions of the box to be positive, the appropriate domain is 0 < x 0 occurs at x = 5. Copyright © 2018 Pearson Education Inc. Section 2.7 Applications of Derivatives to Business and Economics 6. The revenue function is R(x) = 3.5x. Thus, the profit function is P(x) = R(x) – C(x) P ( x)  R( x)  C ( x)  3.5 x  (.0006 x 3  .03x 2  2 x  20)  .0006 x 3  .03 x 2  1.5 x  20 P  ( x)  .0018 x 2  .06 x  1.5 P  ( x)  0  .0018 x 2  .06 x  1.5  0  50 x  50 or x   3 Thus, the maximum value of P(x) for x > 0 occurs at x = 50. 7. The revenue function is 1  R ( x )  x  x 2  10 x  300   12  1 3 2  x  10 x  300 x 12 1 R  ( x)  x 2  20 x  300  R  ( x)  0  4 1 2 x  20 x  300  0  x 2  80 x  1200  0  4  x  60 x  20  0  x  60 or x  20 1 x  20 2 R  (20)  0, R  (60)  0 The maximum value of R(x) occurs at x = 20. 1 The corresponding price is $133 or $133.33 . 3 R  ( x)  8. The revenue function is R ( x )  x(2  .001x)  2 x  .001x 2 . R  ( x)  2  .002 x R  ( x)  0  2  .002 x  0  x  1000 The maximum value of R(x) occurs at x = 1000. The corresponding price is p = 2 – .001(1000) = $1. 9. The revenue function is R ( x)  x (256  50 x)  256 x  50 x 2 . Thus, the profit function is P ( x)  R ( x )  C ( x)  256 x  50 x 2  182  56 x  50 x 2  200 x  182 P  ( x)  100 x  200 P  ( x)  0  100 x  200  0  x  2 The maximum profit occurs at x = 2 (million tons). The corresponding price is 256 – 50(2) = 156 dollars per ton. 109 10. The objective is A = xy and the constraint is y = 30 – x. A(x) = x(30 – x)  30x  x 2 , A ( x)  30  2 x A ( x)  0  30  2 x  0  x  15 The maximum value of A(x) occurs at x = 15. Thus, the optimal values are a = 15, b = 15. If y = 30 – x is a demand curve, then A(x) above corresponds to the revenue function R(x) and the optimal values a, b correspond to the revenue-maximizing quantity and price, respectively. 11. a. Let p stand for the price of hamburgers and let x be the quantity. Using the pointslope equation, 4.4  4 p4 ( x  10, 000) or 8000  10, 000 p = –.0002x + 6. Thus, the revenue function is R ( x )  x(.0002 x  6)  .0002 x 2  6 x. R  ( x)  .0004 x  6 R  ( x)  0  .0004 x  6  0  x  15, 000 The maximum value of R(x) occurs at x = 15,000. The optimal price is thus –.0002(15,000) + 6 = $3.00 b. The cost function is C(x) = 1000 + .6x, so the profit function is P(x) = R(x) – C(x) P ( x)  R( x)  C ( x)  .0002 x 2  6 x  1000  .6 x   .0002 x 2  5.4 x  1000 P  ( x)  .0004 x  5.4 P  ( x)  0  .0004 x  5.4  0  x  13, 500 The maximum value of P(x) occurs at x = 13,500. The optimal price is –.0002(13,500) + 6 = $3.30. 12. Let 50 + x denote the ticket price and y the attendance. Since a $2 increase in price lowers the attendance by 200, we have y = 4000 – 100x. We now have Revenue  R  price  attendance  (50  x)(4000  100 x)  100 x 2  1000 x  200, 000 R  ( x)  200 x  1000 R  ( x)  0  200 x  1000  0  x  5 R = (50 – 5)(4000 – 100(–5)) = 202,500 Answer: Charge $45 per ticket. Revenue = $202,500 Copyright © 2018 Pearson Education Inc. 110 Chapter 2 Applications of the Derivative The maximum value of P(x) occurs at 13. Let x be the number of prints the artist sells. Then his revenue = [price] · [quantity]. (400  5( x  50)) x if x  50 400 x if x  50 x  1.5  10 5 (thousand kilowatt-hours). The corresponding price is  For x > 50, r ( x)  5 x 2  650 x, r  ( x)  10 x  650 r  ( x)  0  10 x  650  0  x  65 The maximum value of r(x) occurs at x = 65. The artist should sell 65 prints. 14. Let x be the number of memberships the club sells. Then their revenue is 200 x if x  100 r ( x)  (200  ( x  100)) x if 100  x  160 if x  100 200 x  2 if 100  x  160  x  300 x   This represents $45/thousand kilowatthours. b. The new profit function is P1 ( x)  R ( x)  C1 ( x)  60 x  10 5 x 2  7  10 6  40 x  10 5 x 2  20 x  7  10 6 P1( x)  2  10 5 x  20 . P1( x)  0  2  10 5 x  20  0  x  10 6 The maximum value of P1 ( x) occurs at x  10 6 (thousand kilowatt-hours). The corresponding price is For 100  x  160, r  ( x)  2 x  300 r ( x)  0  2 x  300  0  x  150 The maximum value of r(x) occurs at x = 150. The club should try to sell 150 memberships. p  60  10 5 (10 6 )  50 , representing $50/thousand kilowatt-hours. The maximum profit will be obtained by charging $50/thousand kilowatt-hours. Since this represents an increase of only $5/thousand kilowatt-hours over the answer to part (a), the utility company should not pass all of the increase on to consumers. 15. Let P(x) be the profit from x tables. Then P ( x )  (10  ( x  12)(.5) x  .5 x 2  16 x For x ≥ 12, P  ( x)  16  x P  ( x)  0  16  x  0  x  16 The maximum value of P(x) occurs at x = 16. The cafe should provide 16 tables. 16. The revenue function is R ( x)  x 36, 000  300  x  100  300 x 2  66, 000 x where x is the price in cents and x ≥ 100. R  ( x)  66, 000  600 x R  ( x)  0  66, 000  600 x  0  x  110 The maximum value occurs at x = 110. The toll should be $1.10. 17. a.   R ( x )  x 60  10 5 x  60 x  10 5 x 2 ; so the profit function is P(x) = R(x) – C(x) P ( x)  R( x)  C ( x)     60 x  10 5 x 2  7  10 6  30 x  10 5 2 x  30 x  7  10 6 P  ( x)  –2  10 5 x  30 P  ( x)  0  –2  10 5 x  30  0  x  15  10 5   p  60  10 5 15  10 5  45. 18. a. R ( x)  x(200  3x)  200 x  3 x 2 , so the profit function is P ( x)  C ( x)  R( x)  200 x  3 x 2  (75  80 x  x 2 )  2 x 2  120 x  75 P  ( x)  4 x  120 P  ( x)  0  4 x  120  0  x  30 The corresponding price is p = 200 – 3(30) = 110. Thus, x = 30 and the price is $110. b. The tax increases the cost function by 4x, so the new cost function is C ( x)  75  84 x  x 2 and the profit function is now P(x) = R(x) – C(x) P ( x)  R( x)  C ( x)  200 x  3 x 2  (75  84 x  x 2 )  2 x 2  116 x  75 P  ( x)  4 x  116 P  ( x)  0  4 x  116  0  x  29 The corresponding price is p = 200 – 3(29) = 113, or $113. Copyright © 2018 Pearson Education Inc. Chapter 2 Fundamental Concept Check Exercises The profit function is now P ( x)  R( x)  C ( x)  200 x  3x 2  75  (80  T ) x  x 2   2 x 2  (120  T ) x  75 P  ( x)  4 x  (120  T ) T P  ( x)  0  4 x  (120  T )  x  30  4 T The new value of x is 30  . 4 The government’s tax revenue is given by T 1  G (T )  Tx  T  30    30T  T 2 ,   4 4 1 G  (T )  30  T 2 . 1 G  (T )  0  30  T  0  T  60 2 The maximum value of G(T) occurs at T = 60. Thus a tax of $60/unit will maximize the government’s tax revenue. 19. Let r be the percentage rate of interest (r = 4 represents a 4% interest rate). Total deposit is $1,000,000r. Total interest c. paid out in one year is 10, 000r 2 . Total interest received on the loans of 1,000,000r is 100,000r. P  100, 000r  10, 000r 2 dP  100, 000  20, 000r dr dP  0 and solve for r: Set dr 100, 000  20, 000r  0  r  5 An interest rate of 5% generates the greatest profit. 20. a. P(0) is the profit with no advertising budget. b. As money is spent on advertising, the marginal profit initially increases. However, at some point the marginal profit begins to decrease. c. Additional money spent on advertising is most advantageous at the inflection point. 21. a. Since R(40) = 75, the revenue is $75,000. b. Since R  (17.5)  3.2 , the marginal revenue is about $3200 per unit. c. Since the solution of R(x) = 45 is x = 15, the production level in 15 units. 111 d. Since the solution of R  ( x)  .8 is x = 32.5, the production level is 32.5 units. e. Looking at the graph of y = R(x), the revenue appears to be greatest at x ≈ 35. To confirm, observe that the graph of y  R  ( x) crosses the x-axis at x = 35. The revenue is greatest at a production level of 35 units. 22. a. Since C(60) = 1100, the cost is $1100. b. Since C  (40)  12.5 , the marginal cost is $12.50. c. Since the solution of C(x) = 1200 is x = 100, the production level is 100 units. d. Since the solutions of C ( x)  22.5 are x = 20 and x = 140, the production levels are 20 units and 140 units. e. Looking at the graph of y  C  ( x) , the marginal cost appears to be least at x ≈ 80. The production level is 80 units, and the marginal cost is $5. Chapter 2 Fundamental Concept Check Exercises 1. Increasing and decreasing functions relative maximum and minimum points absolute maximum and minimum points concave up and concave down inflection point, intercepts, asymptotes 2. A point is a relative maximum at x = 2 if the function attains a maximum at x = 2 relative to nearby points on the graph. The function has an absolute maximum at x = 2 if it attains its largest value at x = 2. 3. Concave up at x = 2: The graph “opens” up as it passes through the point at x = 2; there is an open interval containing x = 2 throughout which the graph lies above its tangent line; the slope of the tangent line increases as we move from left to right through the point at x = 2. Concave down at x = 2: The graph “opens” down as it passes through the point at x = 2; there is an open interval containing x = 2 throughout which the graph lies below its tangent line; the slope of the tangent line decreases as we move from left to right through the point at x = 2. Copyright © 2018 Pearson Education Inc. 112 Chapter 2 Applications of the Derivative 4. f  x  has an inflection point at x = 2 if the concavity of the graph changes at the point 2, f 2 . 5. The x-coordinate of the x-intercept is a zero of the function. 6. To determine the y-intercept, set x = 0 and compute f 0 . 7. An asymptote is a line that a curve approaches as the curve approaches infinity. There are three types of asymptotes: horizontal, vertical, and oblique (or slant) asymptotes. Note that the distance between the curve and the asymptote approaches zero. For example, in the figure y = 2 is a vertical asymptote and x = 2 is a horizontal asymptote. 11. Solve f   x   0. Let a solution be represented by a. If f   a   0 and f   x  changes sign as we move from left to right through x = a, then there is an inflection point at x = a. 12. See pages 161−162 in section 2.4 for more detail. 1. Compute f   x  and f   x  . 2. Find all relative extreme points. a. Apply the first and second derivative tests to find the relative extreme points. Set f   x   0, and solve for x to find the critical value x = a. (i) If f  a   0, the curve has a (ii) relative minimum at x = a. If f  a   0, the curve has a relative maximum at x = a. (iii) If f  a   0, there is an b. 8. First derivative rule: If f   a   0, then f is increasing at x = a. If f   a   0, then f is inflection point at x = a. Repeat the preceding steps for each solution to f   x   0. 3. Find all the inflection points of f  x  4. using the second derivative test. Consider other properties of the function and complete the sketch. decreasing at x = a. Second derivative rule: If f   a   0, then f is 13. In an optimization problem, the quantity to be optimized (maximized or minimized) is given by the objective equation. concave down at x = a. 14. A constraint equation is an equation that places a limit, or a constraint, on the variables in an optimization problem. concave up at x = a. If f   a   0, then f is 9. We can think of the derivative of f  x  as a “slope function” for f  x  . The y-values on the graph of y  f   x  are the slopes of the corresponding points on the graph of y  f  x  . Thus, on an interval where 15. 1. 2. 3. f   x   0, f is increasing. On an interval 4. 10. Solve f   x   0. Let a solution be represented 5. where f   x  is increasing, f is concave up. by a. If f  changes from positive to negative at x = a, then f has a local maximum at a. If f  changes from negative to positive at x = a, then f has a local minimum at a. If f  does not change sign at a (that is, f  is either positive on both sides of a or negative on both sides of a, then f has no local extremum at a. 6. Draw a picture, if possible. Decide what quantity Q is to be maximized or minimized. Assign variables to other quantities in the problem. Determine the objective equation that expresses Q as a function of the variables assigned in step 3. Find the constraint equation that relates the variable to each other and to any constants that are given in the problem. Use the constraint equation to simplify the objective equation in such a way that Q becomes a function of only one variable. Determine the domain of this function. Copyright © 2018 Pearson Education Inc. (continued on next page) Chapter 2 Review Exercises (continued) 7. 113 6. Sketch the graph of the function obtained in step 6 and use this graph to solve the optimization problem. Alternatively, use the second derivative test. 16. P  x   R  x   C  x  Chapter 2 Review Exercises 1. a. The graph of f(x) is increasing when f ( x)  0 : –3 < x 5. The graph of f(x) is decreasing when f ( x)  0 : x < –3, 1 < x < 5. b. The graph of f(x) is concave up when f  ( x) is increasing: x 3. The graph of f(x) is concave down when f  ( x) is decreasing: –1 < x < 3. 2. a. f(3) = 2 b. The tangent line has slope f (3)  c. 3. 1 , so 2 1 . 2 7. (d), (e) 8. (b) 9. (c), (d) 10. (a) 11. (e) 12. (b) 13. Graph goes through (1, 2), increasing at x = 1. 14. Graph goes through (1, 5), decreasing at x = 1. 15. Increasing and concave up at x = 3. 16. Decreasing and concave down at x = 2. 17. (10, 2) is a relative minimum point. 18. Graph goes through (4, –2), increasing and concave down at x = 4. 19. Graph goes through (5, –1), decreasing at x = 5. 20. (0, 0) is a relative minimum point. Since the point (3, 2) appears to be an inflection point, f (3)  0 . 21. a. f(t) = 1 at t = 2, after 2 hours. b. f(5) = .8 c. f  (t )  .08 at t = 3, after 3 hours. d. Since f (8)  .02, the rate of change is –.02 unit per hour. 22. a. 4. b. Since f (50)  35, the rate of change was 35 trillion kilowatt-hours per year. c. 5. Since f(50) = 400, the amount of energy produced was 400 trillion kilowatt-hours. Since f(t) = 3000 at t = 95, the production level reached 300 trillion kilowatt-hours in 1995. d. Since f (t )  10 at t = 35, the production level was rising at the rate of 10 trillion kilowatt-hours per year in 1935. e. Looking at the graph of y  f  (t ), the value of f (t ) appears to be greatest at t = 70. To confirm, observe that the graph of y  f  (t ) crosses the t-axis at t = 70. Energy production was growing at the greatest rate in 1970. Since f(70) = 1600, the production level at that time was 1600 trillion kilowatt-hours. Copyright © 2018 Pearson Education Inc. 114 23. Chapter 2 Applications of the Derivative y  3  x2 y   2 x y   2 y   0 if x  0 If x = 0, y = 3, so (0, 3) is a critical point and the y-intercept. y   0, so (0, 3) is a relative maximum. 0  3  x 2  x   3, so the x-intercepts are 26.  3, 0. y  4  3x  x 2 y  3  2x y   2 3 y   0 if x  2 3 so If x  2 , y   49 4  32 , 254  is a critical is a relative point. y   0, so  32 , 25 4  24. maximum. 0  4  3 x  x 2  x  1 or x  4, so the x-intercepts are (–1, 0) and (4, 0). The y-intercept is (0, 4). y  7  6 x  x2 y  6  2x y   2 y   0 if x  3 If x = 3, y = 16, so (3, 16) is a critical point. y   0, so (3, 16) is a relative maximum. 0  7  6 x  x 2  x  1 or x  7, so the x-intercepts are (–1, 0) and (7, 0). The y-intercept is (0, 7). 27. y  2 x 2  10 x  10 y   4 x  10 y   4 5 y   0 if x  2 5 If x  2 , y  52 so 52 , 52 is a critical point.  25. y  x 2  3x  10 y  2x  3 y   2 3 y   0 if x   2 3 so  32 ,  49 is a critical If x   2 , y   49 4 4     y   0, so   52 , 52  is a relative maximum. 0  2 x 2  10 x  10  x  5 2 5 , so the x-intercepts are  5 5 ,0 2  and  The y-intercept is (0, –10). is a relative point. y   0, so  32 ,  49 4 minimum. x 2  3 x  10  0  x  5 or x  2, so the x-intercepts are (–5, 0) and (2, 0). The y-intercept is (0, –10). Copyright © 2018 Pearson Education Inc.  5 5 ,0 . 2 Chapter 2 Review Exercises 28. 115 y  x 2  9 x  19 y  2x  9 y   2 9 y   0 if x  2 9 If x  2 , y   54 so  92 ,  54  is a critical point. y   0, so  92 ,  54  is a relative 31. minimum. 0  x 2  9 x  19  x  9 2 5 , so the x-intercepts are  9 5 ,0 2  and   9 5 ,0 . 2 The y-intercept is (0, 19). y   x 2  20 x  90 y   2 x  20 y   2 y   0 if x  10 If x = 10, y =10, so (10, 10) is a critical point. y   0, so (10, 10) is a relative maximum. 0   x 2  20 x  90  x  10  10, so the   x-intercepts are 10  10, 0 and 10  10, 0 . The y-intercept is (0, –13). 29. y  x2  3x  2 y  2x  3 y   2 3 y   0 if x   2 3 If x   2 , y   14 so  32 ,  14 is a critical     point. y   0, so  32 ,  14 is a relative minimum. 0  x 2  3x  2  x  2 or x  1, so the x-intercepts are (–2, 0) and (–1, 0). The y-intercept is (0, 2). 32. y  2 x2  x  1 y  4x  1 y   4 1 y   0 if x   4  If x   14 , y   89 so  14 ,  89  point. y   0, so  14 ,  89  is a critical  is a relative minimum. 0  2 x 2  x  1  x  1 or x  12 , so the x-intercepts are (–1, 0) and The y-intercept is (0, –1). 30. y   x 2  8 x  13 y   2 x  8 y   2 y   0 if x  4 If x = 4, y = 3, so (4, 3) is a critical point. y   0, so (4, 3) is a relative maximum. 0   x 2  8 x  13  x  4  3, so the     x-intercepts are 4  3, 0 and 4  3, 0 . The y-intercept is (0, –13). Copyright © 2018 Pearson Education Inc.  12 , 0. 116 33. Chapter 2 Applications of the Derivative f  x   2 x3  3 x 2  1 f   x  6x  6 x f   x   12 x  6 f   x   0 if x  0 or x  1 f 0  1  0, 1 is a critical pt. f 1  2   1, 2 is a critical pt. 35. f  x   x 3  3x 2  3x  2 f   x   3x 2  6 x  3 f   x   6 x  6 2 To find possible extrema, set f ( x)  0 and solve for x. 3x 2  6 x  3  0  x  1 f 1  1, so (1, –1) is a critical point. f  0  6  0, so the graph is concave up at x = 0, and (0, 1) is a relative minimum. f   1  6  0, so the graph is concave down Since f  ( x)  0 for all x, the graph is always increasing, and (1, –1) is neither a relative maximum nor a relative minimum. To find possible inflection points, set f  ( x)  0 and solve for x. 6x  6  0  x  1 Since f   x   0 for x 1 (meaning the graph is concave up), the point (1, –1) is an inflection point. The y-intercept is (0, –2). 34. 3 2 x  6x 2 f   x   3x 2  3x  6 f   x   6 x  3 f   x   0 if x  1 or x  2 7 7  f  1    1,  is a critical pt.  2 2 f  2  10   2,  10 is a critical pt. f  x   x3  f   1  9  0, so the graph is concave down   at x = –1, and 1, 72 is a relative maximum. f   2  9  0, so the graph is concave up at x = 2, and (2, –10) is a relative minimum. 1 f   x   0 when x  . 2 13 13  1 1 f       ,   is an inflection pt. 2 2 4 4 The y-intercept is (0, 1). 36. f  x   100  36 x  6 x 2  x 3 f   x   36  12 x  3 x 2 f   x   12  6 x f   x   0 if x  6 or x  2 f  6  116   6,  116 is a critical pt. f  2  140   2, 140 is a critical pt. f   6  24  0, so the graph is concave up at x = –6, and (–6, –116) is a relative minimum. f   2  24  0, so the graph is concave down at x = –1, and (2, 140) is a relative maximum. f   x   0 when x  2. f  2  12   2, 12 is an inflection pt. The y-intercept is (0, 100). Copyright © 2018 Pearson Education Inc. Chapter 2 Review Exercises 37. 11 1  3x  x 2  x3 3 3 f   x  3  2 x  x2 f   x   2  2 x f   x   0 if x  3 or x  1 16 16   f  3     3,   is a critical pt.  3 3 16  16  f 1   1,  is a critical pt.  3 3 f  x  39.  x 2  4 x  5  0  no real solution Thus, there are no extrema. Since f  ( x)  0 for all x, the graph is always decreasing. To find possible inflection points, set f  ( x)  0 and solve for x. 2 x  4  0  x  2 14 f  2  3 Since f   x   0 for x  143 (meaning the  x = –3, and 3,  163 is a relative minimum. f  1  4  0, so the graph is concave down   at x = 1, and 1, 163 is a relative maximum. f   x   0 when x  1. f  1  0  1, 0 is an inflection pt.  1 f  x   x 3  2x 2  5x 3 f   x   x 2  4 x  5 f   x   2 x  4 To find possible extrema, set f ( x)  0 and solve for x. f   3  4  0, so the graph is concave up at  117 graph is concave up) and f   x   0 for  The y-intercept is 0, 11 . 3 x  143 (meaning the graph is concave down),   the point 2, 143 is an inflection point. f 0  6, so the y-intercept is (0, 0). 38. f  x   x3  3 x 2  9 x  7 f   x   3x 2  6 x  9 f   x   6 x  6 f   x   0 if x  1 or x  3 f  1  12   1, 12 is a critical pt. f 3  10  3,  20 is a critical pt. f   1  12  0, so the graph is concave down at x = –1, and (–1, 12) is a relative maximum. f  3  12  0, so the graph is concave up at x = 3, and (3, –20) is a relative minimum. f   x   0 when x  1. f 1  4  1,  4 is an inflection pt. The y-intercept is (0, 7). 40. y  x3  6 x 2  15 x  50 y   3x 2  12 x  15 y   6 x  12 y   0 if x  1 or x  5 If x = –1, y = 58. If x = 5, y = –50. So, (–1, 58) and (5, –50) are critical points. If x = –1, y   18  0, so the graph is concave down and (–1, 58) is a relative maximum. If x = 5, y   18  0, so the graph is concave up and (5, –50) is a relative minimum. y   0 when x = 2. If x = 2, y = 4, so (2, 4) is an inflection point. The y-intercept is (0, 50). Copyright © 2018 Pearson Education Inc. 118 Chapter 2 Applications of the Derivative 41. y  x4  2×2 y   4 x3  4 x y   12 x 2  4 y   0 if x  0, x  1, or x  1 If x = –1, y = –1. If x = 0, y = 0. If x = 1, y = –1. So, (–1, –1), (0, 0), and (1, –1) are critical points. If x = –1, y   8  0, so the graph is concave up and (–1, –1) is a relative minimum. If x = 1, y = –1, y   8  0, so the graph is concave up and (1, –1) is a relative minimum. If x = 0, y   0, so we must use the first derivative test. Since y   0 when x 0, (0, 0) is a relative maximum. Thus, (0, 0) is neither a relative maximum nor a relative minimum. It may be an inflection point. Verify by using the second derivative test. y   0 when x  0 or x  2. Critical Points, Intervals x0 0 x2 2 x3 3 x 12x – + + + x2 − − + + y + – + + Concavity up down up up If x = 0, y = 0 so (0, 0) is an inflection point. If x = 2, y = –16 so (2, –16) is an inflection point. The y-intercept is (0, 0). y   0 when x   1 . If x   1 , y   59 ,  so  3 3  is an inflection point. , y   , so  ,   is an 1 ,  59 3 If x  1 3 5 9 1 3 5 9 inflection point. The y-intercept is (0, 0). 43. y  x 4  4 x3 y   4 x3  12 x 2  4 x 2  x  3 42. y   12 x 2  24 x y   0 if x  0 or x  3 If x = 0, y = 0. If x = 0, y = 0. If x =3, y = –27. So, (0, 0), and (3, –27) are critical points. If x = 3, y   36  0, so the graph is concave up and (3, –27) is a relative minimum. If x = 0, y   0, so we must use the first derivative test. Critical Points, Intervals x0 4x 2 + + + x3 − − + y – – + y Decreasing on  , 0 Decreasing on 0, 3 Increasing on 3,   0 x3 3 x x 20   3. x  0 5 x 1 20 y   2 5 x 40 y   3 x y   0 if x  10 Note that we need to consider the positive solution only because the function is defined only for x > 0. When x = 10, y = 7, and 1 y   25  0, so the graph is concave up and y (10, 7) is a relative minimum. Since y can never be zero, there are no 20 tells us that the x y-axis is an asymptote. As x  , the graph inflection points. The term approaches y  4x  3, so this is also an asymptote of the graph. Copyright © 2018 Pearson Education Inc. Chapter 2 Review Exercises 44. 1  2 x  1, x  0 2x 1 y   2  2 2x 1 y   3 x 1 y   0 if x   2 Note that we need to consider the positive solution only because the function is defined only for x > 0. When x  12 , y = 3, and y 48. 50. A – c, B – e, C – f, D – b, E – a, F – d 51. a.  12 , 3 is a relative minimum.     1/ 2 1/ 2 3 45. f ( x)  x 2  2 2 x   3x x 2  2 2 Since f  (0)  0, f has a possible extreme value at x = 0. 46. f ( x)      1/ 2 1/ 2 3 2x 2  3 4 x   6 x 2 x 2  3 2 2 x 2  3  0 for all x, so the sign of f  ( x) is determined by the sign of 4x. Therefore, f ( x)  0 if x > 0, f ( x)  0 if x < 0. This means that f(x) is decreasing for x 0. 47. f  ( x)  The number of people living between 10 + h and 10 miles from the center of the city. b. If so, f(x) would be decreasing at x = 10, which is not possible. Since y can never be zero, there are no 1 tells us that the 2x y-axis is an asymptote. As x  , the graph approaches y  2 x  1, so this is also an asymptote of the graph.  1/ 2 1 5x 5x 2  1 , so 10 x   2 5x 2  1 f  (0)  0. Since f ( x)  0 for all x, f ( x) is positive for x > 0 and negative for x 0, f(x) is decreasing on the interval [0, 5]. Thus, the maximum value occurs at x = 0. The maximum value is f(0) = 2. 54. g (t )  t 2  6t  9 (1 ≤ t ≤ 6) g  (t )  2t  6 g  (t )  0  2t  6  0  t  3 g  (t )  2 The minimum value of g(t) is g(3) = 0. 2 x , so f  (0)  0. Since (1  x 2 ) 2 f ( x)  0 for all x, it follows that 0 must be an inflection point. Copyright © 2018 Pearson Education Inc. 120 Chapter 2 Applications of the Derivative 55. Let x be the width and h be the height. The objective is S = 2xh + 4x + 8h and the 50 constraint is 4 xh  200  h  . x Thus, 400 400  50   100  4 x  S ( x)  2 x    4 x  .  x x x 400 S  ( x)  4  2 x 400 S  ( x)  0  4  2  0  x  10 x 50 50  5 h x 10 The minimum value of S(x) for x > 0 occurs at x = 10. Thus, the dimensions of the box should be 10 ft  4 ft  5 ft. 56. Let x be the length of the base of the box and let y be the other dimension. The objective is 2 V  x y and the constraint is 3x 2  x 2  4 xy  48 y 12  x 2  12 x  x 3 x V  ( x)  12  3x 2 V  ( x)  0  12  3 x 2  0  x  2 V  ( x)  6 x; V (2)  0 . The maximum value for x for x > 0 occurs at x = 2. The optimal dimensions are thus 2 ft  2 ft  4 ft. 57. Let x be the number of inches turned up on each side of the gutter. The objective is A(x) = (30 – 2x)x (A is the cross-sectional area of the gutter— maximizing this will maximize the volume). A ( x)  30  4 x 15 A ( x)  0  30  4 x  0  x  2  15  A ( x)  4, A    0 2 x f  ( x)  1; f  (45)  0 . The maximum value of f(x) occurs at x = 45. Thus, 45 trees should be planted. 59. Let r be the number of production runs and let x be the lot size. Then the objective is x C  1000r  .5   and the constraint is 2 rx  400, 000  r  so C ( x)  400, 000 x 4  108 x  4 x 4  108 1 4 x 4  108 1 C ( x)  0    0  x  4  10 4 4 x2 8 8  10 C ( x)  ; C  (4  10 4 )  0 . 3 x The minimum value of C(x) for x > 0 occurs at C ( x)  48  4 x 2 12  x 2  4x x V ( x)  x 2  58. Let x be the number of trees planted. The 1   objective is f ( x)   25  ( x  40)  x (x ≥   2 1 40). f ( x)  45 x  x 2 2 f  ( x )  45  x f  ( x )  0  45  x  0  x  45 2  x  4  10 4  40, 000. Thus the economic lot size is 40,000 books/run. 60. The revenue function is R ( x)  (150  .02 x) x  150 x  .02 x 2 . Thus, the profit function is P ( x)  (150 x  .02 x 2 )  (10 x  300)  .02 x 2  140 x  300 P  ( x)  .04 x  140 P  ( x)  0  .04 x  140  0  x  3500 P  ( x)  .04; P  (3500)  0 . The maximum value of P(x) occurs at x = 3500. 15 inches gives the maximum value for A. 2 Copyright © 2018 Pearson Education Inc. Chapter 2 Review Exercises 121 61. The distance from point A to point P is 25  x 2 and the distance from point P to point B is 15  x. The time 25  x 2 and the time it takes to travel from point P to point B is 8 12 15  x 1 1 . Therefore, the total trip takes T ( x)  25  x 2  15  x  hours. 17 8 17 1 1 2 1 2 T ( x)  (25  x ) (2 x)  16 17 1 1 8 2 1 2 0 x T ( x)  0  (25  x ) (2 x)  16 17 3 1 1 T  ( x)  (25  x 2 ) 1 2  x 2 (25  x 2 )  3 2 8 8 1 2 2 2 2 3 2 1 8  675 8 1  8  8  T      25         25      0 3 8  3      8 3  3  39304 it takes to travel from point A to point P is   The minimum value for T ( x) occurs at x  83 . Thus, Jane should drive from point A to point P, 83 miles from point C, then down to point B. 62. Let 12  x  25 be the size of the tour group. Then, the revenue generated from a group of x people, R(x), is R ( x)  800  20( x  12)  x. To maximize revenue: R  ( x)  1040  40 x R  ( x)  0  1040  40 x  0  R  26 R  ( x)  40; R  (26)  0 Revenue is maximized for a group of 26 people, which exceeds the maximum allowed. Although, R(x) is an increasing function on [12, 25], therefore R(x) reaches its maximum at x = 25 on the interval 12  x  25 . The tour group that produces the greatest revenue is size 25. Copyright © 2018 Pearson Education Inc.