applications of the derivative - grosse pointe...

Chapter 6

APPLICATIONS OF THE DERIVATIVE

6.1 Absolute Extrema

1. As shown on the graph, the absolute maximum oc-curs at x3; there is no absolute minimum. (Thereis no functional value that is less than all others.)

2. As shown on the graph, the absolute minimum oc-curs at x1; there is no absolute maximum. (Thereis no functional value that is greater than all oth-ers.)

3. As shown on the graph, there are no absolute ex-trema.

4. As shown on the graph, there are no absolute ex-trema.

5. As shown on the graph, the absolute minimumoccurs at x1; there is no absolute maximum.

6. As shown on the graph, the absolute maximumoccurs at x1; there is no absolute minimum.

7. As shown on the graph, the absolute maximumoccurs at x1; the absolute minimum occurs at x2:

8. As shown on the graph, the absolute maximumoccurs at x2; the absolute minimum occurs at x1:

10. f(x) = x3 ¡ 3x2 ¡ 24x+ 5; [¡3; 6]

Find critical numbers:

f 0(x) = 3x2 ¡ 6x¡ 24 = 03(x2 ¡ 2x¡ 8) = 03(x+ 2)(x¡ 4) = 0x = ¡2 or x = 4

x f(x)

¡3 23

¡2 33 Absolute maximum4 ¡75 Absolute minimum6 ¡31

11. f(x) = x3 ¡ 6x2 + 9x¡ 8; [0; 5]Find critical numbers:

f 0(x) =3x2 ¡ 12x+ 9 = 0x2 ¡ 4x+ 3 = 0

(x¡ 3)(x¡ 1) = 0x = 1 or x = 3

x f(x)

0 ¡8 Absolute minimum1 ¡43 ¡8 Absolute minimum5 12 Absolute maximum

12. f(x) =1

3x3 ¡ 1

2x2 ¡ 6x+ 3; [¡4; 4]


f 0(x) = x2 ¡ x¡ 6 = 0(x+ 2)(x¡ 3) = 0x = ¡2 or x = 3

x f(x)

¡4 ¡73¼ ¡2:3

¡2 31

3¼ 10:3 Absolute maximum

3 ¡212¼ ¡10:5 Absolute minimum

4 ¡233¼ ¡7:7

13. f(x) =1

3x3 +

3

2x2 ¡ 4x+ 1; [¡5; 2]


f 0(x) = x2 + 3x¡ 4 = 0(x+ 4)(x¡ 1) = 0x = ¡4 or x = 1

x f(x)

¡4 59


1 ¡76¼¡1:17 Absolute minimum

¡5 101

6¼ 16:83

25

3¼ 1:67

376

Section 6.1 Absolute Extrema 377

14. f(x) = x4 ¡ 32x2 ¡ 7; [¡5; 6]f 0(x) = 4x3 ¡ 64x = 0

4x(x2 ¡ 16) = 04x(x¡ 4)(x+ 4) = 0x = 0 or x = 4 or x = ¡4x f(x)

¡5 ¡182¡4 ¡263 Absolute minimum0 ¡74 ¡263 Absolute minimum6 137 Absolute maximum

15. f(x) = x4 ¡ 18x2 + 1; [¡4; 4]f 0(x) = 4x3 ¡ 36x = 0

4x(x2 ¡ 9) = 04x(x+ 3)(x¡ 3) = 0

x = 0 or x = ¡3 or x = 3

x f(x)

¡4 ¡31¡3 ¡80 Absolute minimum0 1 Absolute maximum3 ¡80 Absolute minimum4 ¡31

16. f(x) =8 + x

8¡ x ; [4; 6]

f 0(x) =(8¡ x)(1)¡ (8 + x)(¡1)

(8¡ x)2

=16

(8¡ x)2

f 0(x) is never zero. Although f 0(x) fails to existif x = 8; 8 is not in the given interval.

x f(x)

4 3 Absolute minimum6 7 Absolute maximum

17. f(x) =1¡ x3 + x

; [0; 3]

f 0(x) =¡4

(3 + x)2

No critical numbers

x f(x)

01

3Absolute maximum

3 ¡13

Absolute minimum

18. f(x) =x

x2 + 2; [0; 4]

f 0(x) =(x2 + 2)1¡ x(2x)

(x2 + 2)2

=¡x2 + 2(x2 + 2)2

= 0

¡x2 + 2 = 0x2 = 2

x =p2 or x = ¡p2; but ¡p2 is not in [0; 4]:

f 0(x) is de…ned for all x:

x f(x)

0 0 Absolute minimum

p2

p2


42

9¼ 0:22

19. f(x) =x¡ 1x2 + 1

; [1; 5]

f 0(x) =¡x2 + 2x+ 1(x2 + 1)2

f 0(x) = 0 when

¡x2 + 2x+ 1 = 0x = 1§p2;

but 1¡p2 is not in [1; 5]:x f(x)

1 0 Absolute minimum

52

13¼ 0:15

1 +p2

p2¡ 12

¼ 0:21 Absolute maximum

20. f(x) = (x2 ¡ 16)2=3; [¡5; 8]

f 0(x) =2

3(x2 ¡ 16)¡1=3(2x) = 4x

3(x2 ¡ 16)1=3f 0(x) = 0 when 4x = 0

x = 0

f 0(x) is unde…ned at x = ¡4 and x = 4; butf(x) is de…ned there, so ¡4 and 4 are also criticalnumbers.

x f(x)

¡5 92=3 ¼ 4:327¡4 0 Absolute minimum0 (¡16)2=3 ¼ 6:3504 0 Absolute minimum8 482=3 ¼ 13:21 Absolute maximum

378 Chapter 6 APPLICATIONS OF THE DERIVATIVE

21. f(x) = (x2 ¡ 4)1=3; [¡2; 3]

f 0(x) =1

3(x2 ¡ 4)¡2=3(2x)

=2x

3(x2 ¡ 4)2=3

f 0(x) = 0 when 2x = 0x = 0

f 0(x) is unde…ned at x = ¡2 and x = 2, butf(x) is de…ned there, so ¡2 and 2 are also criticalnumbers.

x f(x)

¡2 0

0 (¡4)1=3 ¼ ¡1:587 Absolute minimum2 03 51=3 ¼ 1:710 Absolute maximum

22. f(x) = x+ 3x2=3; [¡10; 1]f 0(x) = 1 + 2x¡1=3

= 1 +23px

=3px+ 23px

f 0(x) = 0 when 3px+ 2 = 0

3px = ¡2x = ¡8

f 0(x) is unde…ned at x = 0; but f(x) is de…ned atx = 0; so 0 is also a critical number.

x f(x)

¡10 3:925

¡8 4 Absolute maximum0 0 Absolute minimum1 4 Absolute maximum

23. f(x) = 5x2=3 + 2x5=3; [¡2; 1]

f 0(x) =10

3x¡1=3 +

10

3x2=3

=10

3x1=3+10x2=3

3

=10x+ 10

3x1=3

=10(x+ 1)

3 3px

f 0(x) = 0 when 10(x+ 1) = 0x+ 1 = 0

x = ¡1:

f 0(x) is unde…ned at x = 0; but f(x) is de…ned atx = 0; so 0 is also a critical number.

x f(x)

¡2 1:587

¡1 3

0 0 Absolute minimum1 7 Absolute maximum

24. f(x) =lnx

x2; [1; 4]

f 0(x) =x2 ¢ 1x ¡ lnx ¢ 2x

x4

=x¡ 2x lnx

x4

=x(1¡ 2 lnx)

x4

=1¡ 2 lnxx3

f 0(x) = 0 when 1¡ 2 lnx = 0¡2 lnx = ¡1

lnx =1

2

x = e1=2

Although f 0(x) fails to exist at x = 0; 0 is not inthe speci…ed domain for f(x); so 0 is not a criticalnumber.

x f(x)

0 0 Absolute minimume1=2 0:1839 Absolute maximum4 0:0866

25. f(x) = x2 ¡ 8 lnx; [1; 4]

f 0(x) = 2x¡ 8

x

f 0(x) = 0 when 2x¡ 8

x= 0

2x =8

x

2x2 = 8

x2 = 4

x = ¡2 or x = 2

but x = ¡2 is not in the given interval.


Although f 0(x) fails to exist at x = 0; 0 is not inthe speci…ed domain for f(x); so 0 is not a criticalnumber.

x f(x)

1 1

2 ¡1:545 Absolute minimum4 4:910 Absolute maximum

26. f(x) = x2e¡0:5x; [2; 5]

f 0(x) = x2μ¡12e¡0:5x

¶+ 2xe¡0:5x

= e¡0:5xμ2x¡ 1

2x2¶

f 0(x) = 0when e¡0:5xμ2x¡ 1

2x2¶= 0

x = 0 or x = 4

but x = 0 is not in the given interval.

x f(x)

2 1:472 Absolute minimum4 2:165 Absolute maximum5 2:052

27. f(x) = x+ e¡3x; [¡1; 3]f 0(x) = 1¡ 3e¡3x

f 0(x) = 0 when 1¡ 3e¡3x = 0¡3e¡3x = ¡1

e¡3x =1

3

¡3x = ln 13

x =ln 3

3

x f(x)

¡1 19:09 Absolute maximumln 3

30:6995 Absolute minimum

3 3:000

28. f(x) =x3 + 2x+ 5

x4 + 3x3 + 10; [¡3; 0]

The indicated domain tells us the x-values to usefor the viewing window, but we must experimentto …nd a suitable range for the y-values. In orderto show the absolute extrema on [¡3; 0]; we …ndthat a suitable window is [¡3; 0] by [¡9; 1]:From the graph, we see that on [¡3; 0]; f has anabsolute maximum of 0.5 at 0 and an absoluteminimum of 8.10 at about ¡2:35:

29. f(x) =¡5x4 + 2x3 + 3x2 + 9x4 ¡ x3 + x2 + 7 ; [¡1; 1]

The indicated domain tells us the x-values to usefor the viewing window, but we must experimentto …nd a suitable range for the y-values. In orderto show the absolute extrema on [¡1; 1]; we …ndthat a suitable window is [¡1; 1] by [0; 1:5] withXscl = 0.1, Yscl = 0.1.

From the graph, we se that on [¡1; 1]; f has anabsolute maximum of 1.356 at about 0.6085 andan absolute minimum of 0.5 at ¡1:

30. f(x) = 12¡ x¡ 9

x; x > 0

f 0(x) = ¡1 + 9

x2

=9¡ x2x2

=(3 + x)(3¡ x)

x2

f 0(x) = 0 when x = ¡3 or x = 3; and f 0(x)does not exist when x = 0: However, the speci…eddomain for f is (0;1): Since ¡3 and 0 are not inthe domain of f; the only critical number is 3.

x f(x)

3 6

There is an absolute maximum at x = 3: There isno absolute minimum, as can be seen by lookingat the graph of f:

31. f(x) = 2x+8

x2+ 1; x > 0

f 0(x) = 2¡ 16x3

=2x3 ¡ 16x3

=2(x¡ 2)(x2 + 2x+ 4)

x3

Since the speci…ed domain is (0;1), a criticalnumber is x = 2:

x f(x)

2 7

There is an absolute minimum at x = 2; there isno absolute maximum, as can be seen by lookingat the graph of f:


32. f(x) = x4 ¡ 4x3 + 4x2 + 1f 0(x) = 4x3 ¡ 12x2 + 8x

= 4x(x2 ¡ 3x+ 2)= 4x(x¡ 2)(x¡ 1)

The critical numbers are 0, 1, and 2.

x f(x)

0 1

1 2

2 1

There is no absolute maximum, as can be seen bylooking at the graph of f: There is an absoluteminimum at x = 0 and x = 2:

33. f(x) = ¡3x4 + 8x3 + 18x2 + 2f 0(x) = ¡12x3 + 24x2 + 36x

= ¡12x(x2 ¡ 2x¡ 3)= ¡12x(x¡ 3)(x+ 1)

Critical numbers are 0; 3; and ¡1:

x f(x)

¡1 9

0 2

3 137

There is an absolute maximum at x = 3; there isno absolute minimum, as can be seen by lookingat the graph of f:

34. f(x) =x

x2 + 1

f 0(x) =(x2 + 1)¡ x(2x)

(x2 + 1)2

=x2 + 1¡ 2x2(x2 + 1)2

=1¡ x2(x2 + 1)2

=(1 + x)(1¡ x)(x2 + 1)2

The critical numbers are ¡1 and 1.

x f(x)

¡1 ¡0:51 0:5

There is an absolute maximum of 0.5 at x = 1 andan absolute minimum of ¡0:5 at x = ¡1: This canbe veri…ed by looking at the graph of f:

35. f(x) =x¡ 1

x2 + 2x+ 6

f 0(x) =(x2 + 2x+ 6)(1)¡ (x¡ 1)(2x+ 2)

(x2 + 2x+ 6)2

=x2 + 2x+ 6¡ 2x2 + 2

(x2 + 2x+ 6)2

=¡x2 + 2x+ 8(x2 + 2x+ 6)2

=¡(x2 ¡ 2x¡ 8)(x2 + 2x+ 6)2

=¡(x¡ 4)(x+ 2)(x2 + 2x+ 6)2

Critical numbers are 4 and ¡2:

x f(x)

¡2 ¡12

4 0:1

There is an absolute maximum at x = 4 and anabsolute minimum at x = ¡2: This can be veri…edby looking at the graph of f:

36. f(x) = x lnx

f 0(x) = x ¢ 1x+ 1 ¢ lnx

= 1 + lnx

f 0(x) = 0 when x = e¡1, and f 0(x) does not existwhen x · 0: The only critical number is e¡1:

x f(x)

e¡1 ¡e¡1 ¼ ¡0:3679

There is an absolute minimum of ¡0:3679 at x =e¡1. There is no absolute maximum, as can beseen by looking at the graph of f:

37. f(x) =lnx

x3

f 0(x) =x3 ¢ 1x ¡ 3x2 lnx

x6

=x2 ¡ 3x2 lnx

x6

=x2(1¡ 3 lnx)

x6

=1¡ 3 lnxx4


f 0(x) = 0 when x = e1=3, and f 0(x) does not existwhen x · 0. The only critical number is e1=3.

x f(x)

e1=31

3e¡1 ¼ 0:1226

There is an absolute maximum of 0.1226 at x =e1=3. There is no absolute minimum, as can beseen by looking at the graph of f:

38. f(x) = 2x¡ 3x2=3

f 0(x) = 2¡ 2x¡1=3 = 2¡ 23px=2 3px¡ 23px

f 0(x) = 0 when 2 3px¡ 2 = 02 3px = 2

3px = 1

x = 1

f 0(x) is unde…ned at x = 0; but f(x) is de…ned atx = 0: So the critical numbers are 0 and 1.

(a) On [¡1; 0:5]x f(x)

¡1 ¡50 01 ¡10:5 ¡0:88988

Absolute minimum of ¡5 at x = ¡1; absolutemaximum of 0 at x = 0

(b) On [0:5; 2]

x f(x)

0:5 ¡0:889881 ¡12 ¡0:7622

Absolute maximum of about ¡0:76 at x = 2; ab-solute minimum of ¡1 at x = 1:

39. Let P (x) be the perimeter of the rectangle withvertices (0; 0); (x; 0); (x; f(x)); and (0; f(x)) for x >0 when f(x) = e¡2x:

The length of the rectangle is x and the widthis given by e¡2x. Therefore, an equation for theperimeter is

P (x) = x+ e¡2x + x+ e¡2x = 2(x+ e¡2x):

P 0(x) = 2¡ 4e¡2xP 0(x) = 0 when 2¡ 4e¡2x = 0

¡4e¡2x = ¡2

e¡2x =1

2

e2x = 2

2x = ln 2

x =ln 2

2x P (x)

ln 2

21 + ln 2 ¼ 1:693

There is an absolute minimum of 1.693 at x = ln 22 .

There is no absolute maximum, as can be seen bylooking at the graph of P . Therefore, the correctstatement is a.

40. (a) By looking at the graph, there are relativemaxima of 8046 in 1996, 8496 in 2001, and 7556in 2004. There are relative minima of 6599 in 1999and 7465 in 2003.

(b) Annual bank robberies reached an absolutemaximum of 8496 in 2001 and an absolute mini-mum of 6599 in 1999.

41. (a) By looking at the graph, there are relativemaxima of 413 in 1997, 341 in 2000, and 134 in2004. There are relative minima of 290 in 1996,313 in 1998, and 131 in 2003.

(b) Annual bank burglaries reached an absolutemaximum of 413 in 1997 and an absolute mini-mum of 131 in 2003.

42. P (x) = ¡0:02x3 + 600x¡ 20,000; [50; 150]P 0(x) = ¡0:06x2 + 600

= ¡0:06(x2 ¡ 10,000)= ¡0:06(x+ 100)(x¡ 100)

There are critical numbers at x = ¡100 andx = 100: Only x = 100 is in the domain of P (x):

x P (x)

50 7500100 20,000150 2500


The maximum pro…t is $20,000, which occurs when100 units per week are made.

43. P (x) = ¡x3 + 9x2 + 120x¡ 400; x ¸ 5P 0(x) = ¡3x2 + 18x+ 120

= ¡3(x2 ¡ 6x¡ 40)= ¡3(x¡ 10)(x+ 4) = 0x = 10 or x = ¡4

¡4 is not relevant since x ¸ 5; so the only criticalnumber is 10:

The graph of P 0(x) is a parabola that opens down-ward, so P 0(x) > 0 on the interval [5; 10) andP 0(x) < 0 on the interval (10;1): Thus, P (x)is a maximum at x = 10:Since x is measured in hundred thousands, 10 hun-dred thousand or 1,000,000 tires must be sold tomaximize pro…t.Also,

P (10) = ¡(10)3 + 9(10)2 + 120(10)¡ 400= 700:

The maximum pro…t is $700 thousand or $700,000.

44. C(x) = 81x2 + 17x+ 324

(a) 1 · x · 10

C(x) =C(x)

x=81x2 + 17x+ 324

x

= 81x+ 17 +324

x

C0(x) = 81¡ 324

x2= 0 when

81 =324

x2

x2 = 4

x = §2:

¡2 is not meaningful is this application and is notin the given domain.Test 2 for minimum.

C0(1) = ¡243 < 0

C0(3) = 45 > 0

C(x) is a minimum when x = 2:

C(2) = 81(2) + 17 +324

2= 341

The minimum on the interval 1 · x · 10 is 341.

(b) 10 · x · 20There are no critical numbers in this interval. Testthe endpoints.

C(10) = 859:4

C(20) = 1620 + 17 + 16:2

= 1653:2

The minimum on the interval 10 · x · 20 is859.4.

45. C(x) = x3 + 37x+ 250

(a) 1 · x · 10

C(x) =C(x)

x=x3 + 37x+ 250

x

= x2 + 37 +250

x

C0(x) = 2x¡ 250

x2

=2x3 ¡ 250

x2= 0 when

2x3 = 250

x3 = 125

x = 5:

Test for relative minimum.

C0(4) = ¡7:625 < 0

C0(6) ¼ 5:0556 > 0

C(5) = 112

C(1) = 1 + 37 + 250 = 288

C(10) = 100 + 37 + 25 = 162

The minimum on the interval 1 · x · 10 is 112.(b) 10 · x · 20There are no critical values in this interval. Checkthe endpoints.

C(10) = 162

C(20) = 400 + 37 + 12:5 = 449:5

The minimum on the interval 10 · x · 20 is 162.46. The value x = 20 minimizes f(x)

x because this isthe point where the line from the origin to thecurve is tangent to the curve.A production level of 20 units results in the min-imum cost per unit.

47. The value x = 11 minimizes f(x)x because this is

the point where the line from the origin to thecurve is tangent to the curve.A production level of 11 units results in the min-imum cost per unit.


48. The value x = 300 maximizes f(x)x because this

is the point where the line from the origin to thecurve is tangent to the curve.A production level of 300 units results in the max-imum pro…t per item produced.

49. The value x = 100 maximizes f(x)x because this

is the point where the line from the origin to thecurve is tangent to the curve.A production level of 100 units results in the max-imum pro…t per item produced.

50. f(x) =x2 + 36

2x; 1 · x · 12

f 0(x) =2x(2x)¡ (x2 + 36)(2)

(2x)2=4x2 ¡ 2x2 ¡ 72

4x2

=2x2 ¡ 724x2

=2(x2 ¡ 36)4x2

=(x+ 6)(x¡ 6)

2x2

f 0(x) = 0 when x = 6 and when x = ¡6: Only 6is in the interval 1 · x · 12:Test for relative maximum or minimum.

f 0(5) =(11)(¡1)50

< 0

f 0(7) =(13)(1)

98> 0

The minimum occurs at x = 6; or at 6 months.Since f(6) = 6; f(1) = 18:5, and f(12) = 7:5, theminimum percent is 6%.

51. S(x) = ¡x3 + 3x2 + 360x+ 5000; 6 · x · 20

S0(x) = ¡3x2 + 6x+ 360= ¡3(x2 ¡ 2x¡ 120)

S0(x) = ¡3(x¡ 12)(x+ 10) = 0x = 12 or x = ¡10 (not in the

interval)

x f(x)

6 7052

12 8024

10 7900

12± is the temperature that produces the maxi-mum number of salmon.

52. Since we are only interested in the length duringweeks 22 through 28, the domain of the functionfor this problem is [22; 28]: We now look for anycritical numbers in this interval. We …nd

L0(t) = 0:788¡ 0:02t

There is a critical number at t = 0:7880:02 = 39:4;

which is not in the interval. Thus, the maximumvalue will occur at one of the endpoints.

t L(t)

22 5:4

28 7:2

The maximum length is about 7.2 millimeters.

53. The function is de…ned on the interval [15; 46]:Welook …rst for critical numbers in the interval. We…nd

R0(T ) = ¡0:00021T 2 + 0:0802T ¡ 1:6572

Using our graphing calculator, we …nd one criticalnumber in the interval at about 21.92

T R(T )

15 81.0121:92 79.2946 98.89

The relative humidity is minimized at about 21.92±C.

54. M(x) = ¡ 1

45x2 + 2x¡ 20; 30 · x · 65

M 0(x) = ¡ 1

45(2x) + 2 = ¡2x

45+ 2

When M 0(x) = 0;

¡2x45+ 2 = 0

2 =2x

45

45 = x:

x M(x)

30 2045 25

65145

9¼ 16:1

The absolute maximum miles per gallon is 25 andthe absolute minimum miles per gallon is about16.1.


55. M(x) = ¡0:015x2 + 1:31x¡ 7:3; 30 · x · 60M 0(x) = ¡0:03x+ 1:31 = 0

x ¼ 43:7x M(x)

30 18.543:7 21.3060 17:3

The absolute maximum of 21.30 mpg occurs at43.7 mph. The absolute minimum of 17.3 mpgoccurs at 60 mph.

56. Total area A(x) = ¼³ x2¼

´2+

μ12¡ x4

¶2

=x2

4¼+(12¡ x)216

A0(x) =x

2¼¡ 12¡ x

8= 0

4x¡ ¼(12¡ x)8¼

= 0

x =12¼

4 + ¼¼ 5:28

x Area0 9

5:28 5:04

12 11:46

The total area is minimized when the piece usedto form the circle is 12¼

4+¼ feet, or about 5.28 feetlong.

57. Total area = A(x)

= ¼³ x2¼

´2+

μ12¡ x4

¶2

=x2

4¼+(12¡ x)216

A0(x) =x

2¼¡ 12¡ x

8= 0

4x¡ ¼(12¡ x)8¼

= 0

x =12¼

4 + ¼¼ 5:28

x Area0 9

5:28 5:04

12 11:46

The total area is maximized when all 12 feet ofwire are used to form the circle.

58. For the solution to Exercise 56, the piece of lengthx used to form the circle is 12¼

4+¼ feet. The circlecan be inscribed inside the square if the side ofthe square equals the diameter of the circle (thatis, twice the radius).

side of the square = 2(radius)

12¡ x4

= 2³ x2¼

´12¡ x4

=x

¼

4x = 12¼ ¡ ¼xx(4 + ¼) = 12¼

x =12¼

4 + ¼

Therefore, the circle formed by piece of length x =12¼4+¼ can be inscribed inside the square.

59. (a) I(p) = ¡p ln p¡ (1¡ p) ln (1¡ p)

I 0(p) = ¡pμ1

p

¶+ (ln p)(¡1)

¡·(1¡ p) ¡1

1¡ p + [ln (1¡ p)](¡1)¸

= ¡1¡ ln p+ 1+ ln (1¡ p)= ¡ ln p+ ln (1¡ p)

(b) ¡ ln p+ ln (1¡ p) = 0ln (1¡ p) = ln p

1¡ p = p1 = 2p

1

2= p

I 0(0:25) = 1:0986I 0(0:75) = ¡1:099

There is a relative maximum of 0.693 at p = 12 :

Section 6.2 Applications of Extrema 385

6.2 Applications of Extrema

1. x+ y = 180; P = xy

(a) y = 180¡ x

(b) P = xy = x(180¡ x)

(c) Since y = 180¡x and x and y are nonnegativenumbers, x ¸ 0 and 180¡ x ¸ 0 or x · 180. Thedomain of P is [0, 180].

(d) P 0(x) = 180¡ 2x180¡ 2x = 02(90¡ x) = 0

x = 90

(e) x P

0 090 8100180 0

(f) From the chart, the maximum value of P is8100; this occurs when x = 90 and y = 90:

2. x+ y = 140

Minimize x2 + y2:

(a) y = 140¡ x

(b) Let P = x2 + y2 = x2 + (140¡ x)2= x2 + 19,600 ¡ 280x+ x2= 2x2 ¡ 280x+ 19,600.

(c) Since y = 140¡x and x and y are nonnegativenumbers, the domain of P is [0; 140]:

(d) P 0 = 4x¡ 2804x¡ 280 = 0

4x = 280

x = 70

(e) x P

0 19,60070 9800140 19,600

(f) The minimum value of x2 + y2 occurs whenx = 70 and y = 140 ¡ x = 140 ¡ 70 = 70: Theminimum value is 9800.

3. x+ y = 90

Minimize x2y:

(a) y = 90¡ x

(b) Let P = x2y = x2(90¡ x)= 90x2 ¡ x3:


(d) P 0 = 180x¡ 3x2

180x¡ 3x2 = 03x(60¡ x) = 0

x = 0 or x = 60(e) x P

0 0

60 108; 000

90 0

(f) The maximum value of x2y occurs when x =60 and y = 30: The maximum value is 108,000.

4. x+ y = 105

Maximize xy2:

(a) y = 105¡ x

(b) Let P = xy2 = x(105¡ x)2= x(11,025¡ 210x+ x2)= 11,025x¡ 210x2 + x3:


(d) P 0 = 11,025¡ 420x+ 3x2

3x2 ¡ 420x+ 11,025 = 03(x2 ¡ 140x+ 3675) = 03(x¡ 35)(x¡ 105) = 0x = 35 or x = 105

(e) x P

0 0

35 171,500105 0

(f) The maximum value of xy2 occurs when x =35 and y = 70: The maximum value is 171,500.


5. C(x) =1

2x3 + 2x2 ¡ 3x+ 35

The average cost function is

A(x) = C(x) =C(x)

x

=12x

3 + 2x2 ¡ 3x+ 35x

=1

2x2 + 2x¡ 3 + 35

x

or1

2x2 + 2x¡ 3 + 35x¡1:

ThenA0(x) = x+ 2¡ 35x¡2

or x+ 2¡ 35x2:

Graph y = A0(x) on a graphing calculator. Asuitable choice for the viewing window is [0; 10] by[¡10; 10]: (Negative values of x are not meaningfulin this application.) Using the calculator, we seethat the graph has an x-intercept or “zero” at x ¼2:722: Thus, 2.722 is a critical number.Now graph y = A(x) and use this graph to con…rmthat a minimum occurs at x ¼ 2:722:Thus, the average cost is smallest at x ¼ 2:722:

6. C(x) = 10 + 20x1=2 + 16x3=2

The average cost function is

A(x) = C(x) =C(x)

x

=10 + 20x1=2 + 16x3=2

x

=10

x+ 20x¡1=2 + 16x1=2

or 10x¡1 + 20x¡1=2 + 16x1=2:

Then

A0(x) = ¡10x¡2 ¡ 10x¡3=2 + 8x¡1=2:Graph y = A0(x) on a graphing calculator. Asuitable choice for the viewing window is [0; 10] by[¡10; 10]: (Negative values of x are not meaningfulin this application.) We see that this graph hasone x-intercept or “zero.” Using the calculator, we…nd that this x-value is about 2.110, which showsthat 2.110 is the only critical number of A.

Now graph y = A(x) and use this graph to con…rmthat a minimum occurs at x ¼ 2:110: Thus, theaverage cost is smallest at x ¼ 2:110:

7. p(x) = 160¡ x

10

(a) Revenue from sale of x thousand candy bars:

R(x) = 1000xp

= 1000x³160¡ x

10

´= 160,000x¡ 100x2

(b) R0(x) = 160,000¡ 200x

160,000¡ 200x = 0160,000 = 200x

800 = x

The maximum revenue occurs when 800 thousandbars are sold.

(c) R(800) = 160,000(800)¡ 100(800)2= 64,000,000

The maximum revenue is 64,000,000 cents.

8. p(x) = 12¡ x8

(a) Revenue from x thousand compact discs:

R(x) = 1000xp

= 1000x³12¡ x

8

´= 12,000x¡ 125x2

(b) R0(x) = 12,000¡ 250x

12,000¡ 250x = 012,000 = 250x

48 = x

The maximum revenue occurs when 48 thousandcompact discs are sold.

(c) R(48) = 12,000(48)¡ 125(48)2= 288,000

The maximum revenue is $288,000.

9. Let x = the widthand y = the length.

(a) The perimeter is

P = 2x+ y

= 1400;

soy = 1400¡ 2x:


(b) Area = xy = x(1400¡ 2x)A(x) = 1400x¡ 2x2

(c) A0 = 1400¡ 4x1400¡ 4x = 0

1400 = 4x

350 = x

A00 = ¡4; which implies that x = 350 m leads tothe maximum area.

(d) If x = 350;

y = 1400¡ 2(350) = 700:The maximum area is (350)(700) = 245,000 m2:

10. Let x = length of …eldy = width of …eld.

Perimeter:

P = 2x+ 2y = 300

x+ y = 150

y = 150¡ xArea:

A = xy

= x(150¡ x)= 150x¡ x2

Thus,

A(x) = 150x¡ x2A0(x) = 150¡ 2x:

A0(x) = 0 when

150¡ 2x = 0x = 75:

A00(x) = ¡2; so A00(75) = ¡2 < 0; which con…rmsthat a maximum value occurs at x = 75:If x = 75;

y = 150¡ x = 150¡ 75 = 75:A maximum area occurs when the length is 75 mand the width is 75 m.

11. Let x = the width of the rectangley = the total length of the

rectangle.

An equation for the fencing is

3600 = 4x+ 2y

2y = 3600¡ 4xy = 1800¡ 2x:

Area = xy = x(1800¡ 2x)A(x) = 1800x¡ 2x2

A0 = 1800¡ 4x

1800¡ 4x = 01800 = 4x

450 = x

A00 = ¡4; which implies that x = 450 is the loca-tion of a maximum.If x = 450; y = 1800¡ 2(450) = 900:The maximum area is

(450)(900) = 405,000 m2:

12. Let x = the length at $2.50 per footy = the width at $3.20 per foot.

xy = 20,000

y =20,000x

Perimeter = 2x+ 2y = 2x+40,000x

Cost = C(x) = 2x(2:5) +40,000x

(3:2) = 5x+128,000x

Minimize cost:

C0(x) = 5¡ 128,000x2

5¡ 128,000x2

= 0

5 =128,000x2

5x2 = 128,000x2 = 25,600x = 160

y =20,000160

= 125

320 ft at $2.50 per foot will cost $800. 250 ft at$3.20 per foot will cost $800. The entire cost willbe $1600.


13. Let x = length at $1.50 per metery = width at $3 per meter.

xy = 25,600

y =25,600x

Perimeter = x+ 2y = x+51,200x

Cost = C(x) = x(1:5) +51,200x

(3)

= 1:5x+153,600x

Minimize cost:

C0(x) =1:5¡ 153,600x2

1:5¡ 153,600x2

= 0

1:5 =153,600x2

1:5x2 = 153,600x2 = 102,400x = 320

y =25,600320

= 80

320 m at $1.50 per meter will cost $480. 160 mat $3 per meter will cost $480. The total cost willbe $960.

14. Let x = the number of seats.Pro…t is 6 dollars per seat for 0 · x · 50.Pro…t (in dollars) is 6¡ 0:10(x¡ 50) per seat forx > 50.We expect that the number of seats which makesthe total pro…t a maximum will be greater than50 because after 50 the pro…t is still increasing,though at a slower rate. (Thus we know the func-tion is concave down and its one extremum willbe a maximum.)

(a) The total pro…t for x seats is

P (x) = [6¡ 0:1(x¡ 50)]x= (6¡ 0:1x+ 5)x= (11¡ 0:1x)x= 11x¡ 0:1x2:

P 0(x) = 11¡ 0:2x11¡ 0:2x = 0

11 = 0:2x

x = 55

55 seats will produce maximum pro…t.

(b) P (55) = 11(55)¡ 0:10(552)= 605¡ 0:10(3025)= 302:5

The maximum pro…t is $302.50.

15. Let x = the number of days to wait.

12; 000

100= 120 = the number of 100-lb groups

collected already.Then 7:5¡ 0:15x = the price per 100 lb;

4x = the number of 100-lb groupscollected per day;

120 + 4x = total number of 100-lb groupscollected.

Revenue = R(x)= (7:5¡ 0:15x)(120 + 4x)= 900 + 12x¡ 0:6x2

R0(x) = 12¡ 1:2x = 0x = 10

R00(x) = ¡1:2 < 0 so R(x) is maximized atx = 10:

The scouts should wait 10 days at which time theirincome will be maximized at

R(10) = 900 + 12(10)¡ 0:6(10)2 = $960:

16. Let x = a side of the baseh = the height of the box.

An equation for the volume of the box is

V = x2h;

so 32 = x2h

h =32

x2:

The box is open at the top so the area of thesurface material m(x) in square inches is the areaof the base plus the area of the four sides.


m(x) = x2 + 4xh

= x2 + 4x

μ32

x2

¶

= x2 +128

x

m0(x) = 2x¡ 128x2

2x3 ¡ 128x2

= 0

2x3 ¡ 128 = 02(x3 ¡ 64) = 0

x = 4

m0(x) = 2 + 256x3 > 0 since x > 0:

So, x = 4 minimizes the surface material.

If x = 4;

h =32

x2=32

16= 2:

The dimensions that will minimize the surface ma-terial are 4 in by 4 in by 2 in.

17. Let x = the number of refunds.Then 535¡ 5x = the cost per passengerand 85 + x = the number of passengers.

(a) Revenue = R(x) = (535¡ 5x)(85 + x)= 45,475+ 110x¡ 5x2

R0(x) = 110¡ 10x = 0x = 11

R00(x) = ¡10 < 0; so R(x) is maximized whenx = 11:

Thus, the number of passengers that will maxi-mize revenue is 85 + 11 = 96.

(b) R(11) = 45,475 + 110(11)¡ 5(11)2= 46,080

The maximum revenue is $46,080.

18. Let x = the width.Then 2x = the lengthand h = the height.

An equation for volume is

V = (2x)(x)h = 2x2h

36 = 2x2h:

So, h = 18x2 :

The surface area S(x) is the sum of the areas ofthe base and the four sides.

S(x) = (2x)(x) + 2xh+ 2(2x)h

= 2x2 + 6xh

= 2x2 + 6x

μ18

x2

¶

= 2x2 +108

x

S0(x) = 4x¡ 108x2

4x3 ¡ 108x2

= 0

4(x3 ¡ 27) = 0x = 3

S00(x) = 4 +108(2)

x3

= 4 +216

x3> 0 since x > 0

So x = 3 minimizes the surface material.

If x = 3;

h =18

x2=18

9= 2:

The dimensions are 3 ft by 6 ft by 2 ft.

19. Let x = the length of a side ofthe top and bottom.

Then x2 = the area of the top andbottom

and (3)(2x2) = the cost for the top andbottom.

Let y = depth of box.Then xy = the area of one side,

4xy = the total area of thesides,

and (1:50)(4xy) = the cost of the sides.

The total cost is

C(x) = (3)(2x2) + (1:50)(4xy) = 6x2 + 6xy:

The volume is

V = 16; 000 = x2y:

y =16,000x2

C(x) = 6x2 + 6x

μ16,000x2

¶= 6x2 +

96,000x

C 0(x) =12x¡ 96,000x2

= 0

x3 = 8000

x = 20


C00(x) = 12 +192,000x3

> 0 at x = 20; which im-

plies that C(x) is minimized when x = 20:

y =16,000(20)2

= 40

So the dimensions of the box are x by x by y; or20 cm by 20 cm by 40 cm.

C(20) = 6(20)2 +96; 000

20= 7200

The minimum total cost is $7200.

20. 120 centimeters of ribbon are available; it willcover 4 heights and 8 radii.

4h+ 8r = 120

h+ 2r = 30

h = 30¡ 2r

V = ¼r2h

V = ¼r2(30¡ 2r)= 30¼r2 ¡ 2¼r3

Maximize volume.

V 0 = 60¼r ¡ 6¼r260¼r ¡ 6¼r2 = 06¼r(10¡ r) = 0

r = 0 or r = 10

If r = 0; there is no box, so we discard this value.V 00 = 60¼ ¡ 12¼r < 0 for r = 10; which impliesthat r = 10 gives maximum volume.When r = 10; h = 30¡ 2(10) = 10:The volume is maximized when the radius andheight are both 10 cm.

21. (a) S = 2¼r2 + 2¼rh; V = ¼r2h

S = 2¼r2 +2V

r

Treat V as a constant.

S0 = 4¼r ¡ 2Vr2

4¼r ¡ 2Vr2

= 0

4¼r3 ¡ 2Vr2

= 0

4¼r3 ¡ 2V = 02¼r3 ¡ V = 0

2¼r3 = V

2¼r3 = ¼r2h

2r = h

22. V = ¼r2h = 16

h =16

¼r2

The total cost is the sum of the cost of the topand bottom and the cost of the sides.

C = 2(2)(¼r2) + 1(2¼rh)

= 4(¼r2) + 1(2¼r)

μ16

¼r2

¶

= 4¼r2 +32

r

Minimize cost.

C0 = 8¼r ¡ 32r2

8¼r ¡ 32r2= 0

8¼r3 = 32

¼r3 = 4

r = 3

r4

¼

¼ 1:08

h =16

¼(1:08)2¼ 4:34

The radius should be 1.08 ft and the height shouldbe 4.34 ft. If these rounded values for the heightand radius are used, the cost is

$2(2)(¼r2) + $1(2¼rh)

= 4¼(1:08)2 + 2¼(1:08)(4:34)

= $44:11:

23. Let x = the length of the sideof the cutout square.

Then 3¡ 2x = the width of the boxand 8¡ 2x = the length of the box.

V (x) = x(3¡ 2x)(8¡ 2x)= 4x3 ¡ 22x2 + 24x

The domain of V is¡0; 32

¢:

Maximize the volume.

V 0(x) = 12x2 ¡ 44x+ 2412x2 ¡ 44x+ 24 = 04(3x2 ¡ 11x+ 6) = 04(3x¡ 2)(x¡ 3) = 0x =

2

3or x = 3


3 is not in the domain of V:

V 00(x) = 24x¡ 44

V 00μ2

3

¶= ¡28 < 0

This implies that V is maximized when x = 23 :

The box will have maximum volume when x = 23

ft or 8 in.

24. (a) From Example 3, the area of the baseis (12¡ 2x)(12¡ 2x) = 4x2 ¡ 48x+ 144and the total area of all four walls is4x(12¡ 2x) = ¡8x2 + 48x. Since the box hasmaximum volume when x = 2, the area ofthe base is 4(2)2 ¡ 48(2) + 144 = 64 squareinches and the total area of all four walls is¡8(2)2 + 48(2) = 64 square inches. So, bothare 64 square inches.

(b) From Exercise 23, the area of the baseis (3¡ 2x)(8¡ 2x) = 4x2 ¡ 22x+ 24 andthe total area of all four walls is2x(3¡ 2x) + 2x(8¡ 2x) = ¡8x2 + 22x.Since the box has maximum volume whenx = 2

3 , the area of the base is

4¡23

¢2 ¡ 22 ¡23¢+ 24 = 1009 square feet

and the total area of all four walls is¡8 ¡23¢2 + 22 ¡23¢ = 100

9 square feet. So, both are1009 square feet.

(c) Based on the results from parts (a) and (b),it appears that the area of the base and the to-tal area of the walls for the box with maximumvolume are equal. (This conjecture is true.)

25. Let x = the width of printed materialand y = the length of printed material.

Then, the area of the printed material is

xy = 36;

so y =36

x:

Also, x+ 2 = the width of a pageand y + 3 = the length of a page.

The area of a page is

A = (x+ 2)(y + 3)

= xy + 2y + 3x+ 6

= 36 + 2

μ36

x

¶+ 3x+ 6

= 42 +72

x+ 3x:

A0 =¡72x2+ 3 = 0

x2 = 24

x =p24

= 2p6

(We discard x = ¡2p6 once we must have x > 0:)A00 = 216

x3 > 0 when x = 2p6; which implies that

A is minimized when x = 2p6:

y =36

x=

36

2p6=18p6=18p6

6= 3

p6

The width of a page is

x+ 2 = 2p6 + 2

¼ 6:9 in.

The length of a page is

y + 3 = 3p6 + 3

¼ 10:3 in.

26. Distance on shore: 9¡ x milesCost on shore: $400 per mileDistance underwater:

px2 + 36

Cost underwater: $500 per mileFind the distance from A, that is, (9¡x); to min-imize cost, C(x):

C(x) = (9¡ x)(400) + (px2 + 36)(500)= 3600¡ 400x+ 500(x2 + 36)1=2

C 0(x) = ¡400 + 500μ1

2

¶(x2 + 36)¡1=2(2x)

= ¡400 + 500xpx2 + 36


If C0(x) = 0;

500xpx2 + 36

= 400

5x

4=px2 + 36

25

16x2 = x2 + 36

9

16x2 = 36

x2 =36 ¢ 169

x =6 ¢ 43= 8:

(Discard the negative solution.)Then the distance should be

9¡ x = 9¡ 8= 1 mile from point A.

27. Distance on shore: 7¡ x milesCost on shore: $400 per mileDistance underwater:

px2 + 36

Cost underwater: $500 per mileFind the distance from A, that is, 7¡ x; to mini-mize cost, C(x):

C(x) = (7¡ x)(400) + (px2 + 36)(500)= 2800¡ 400x+ 500(x2 + 36)1=2

C0(x) = ¡400 + 500μ1

2

¶(x2 + 36)¡1=2(2x)

= ¡400 + 500xpx2 + 36

If C0(x) = 0;

500xpx2 + 36

= 400

5x

4=px2 + 36

25

16x2 = x2 + 36

9

16x2 = 36

x2 =36 ¢ 169

x =6 ¢ 43= 8:

(Discard the negative solution.)x = 8 is impossible since Point A is only 7 milesfrom point C:

Check the endpoints.

x C(x)

0 5800

7 4610

The cost is minimized when x = 7:

7 ¡ x = 7 ¡ 7 = 0, so the company should anglethe cable at Point A.

28. Let x = the number of additionaltables.

Then 160¡ 0:50x = the cost per tableand 250 + x = the number of tables

ordered.

R = (160¡ 0:50x)(250 + x)= 40,000 + 35x¡ 0:50x2

R0 =35¡ x = 0x = 35

R00 = ¡1 < 0; so when 250 + 35 = 285 tables areordered, revenue is maximum.

Thus, the maximum revenue is

R(35) = 40,000+ 35(35)¡ 0:50(35)2= 40,612.5

The maximum revenue is $40,612.50.

Minimum revenue is found by letting R = 0:

(160¡ 0:50x)(250 + x) = 0160¡ 0:50x = 0 or 250 + x = 0

x = 320 or x = ¡250(impossible)

So when 250 + 320 = 570 tables are ordered, rev-enue is 0, that is, each table is free.I would …re the assistant.

29. From Example 4, we know that the surface areaof the can is given by

S = 2¼r2 +2000

r:

Aluminum costs 3/c/cm2; so the cost of the alu-minum to make the can is

0:03

μ2¼r2 +

2000

r

¶= 0:06¼r2 +

60

r:

The perimeter (or circumference) of the circulartop is 2¼r: Since there is a 2/c/cm charge to sealthe top and bottom, the sealing cost is

0:02(2)(2¼r) = 0:08¼r:


Thus, the total cost is given by the function

C(r) = 0:06¼r2 +60

r+ 0:08¼r

= 0:06¼r2 + 60r¡1 + 0:08¼r:

Then

C0(r) = 0:12¼r ¡ 60r¡2 + 0:08¼

= 0:12¼r ¡ 60r2+ 0:08¼:

Graph

y = 0:12¼x¡ 60x2+ 0:08¼

on a graphing calculator. Since r must be positivein this application, our window should not includenegative values of x:A suitable choice for the view-ing window is [0; 10] by [¡10; 10]: From the graph,we …nd that C 0(x) = 0 when x ¼ 5:206:Thus, the cost is minimized when the radius isabout 5.206 cm.

We can …nd the corresponding height by using theequation

h =1000

¼r2

from Example 4.If r = 5:206;

h =1000

¼(5:206)2¼ 11:75:

To minimize cost, the can should have radius 5.206cm and height 11.75 cm.

30. In Exercise 29, we found that the cost of the alu-minum to make the can is

0:03

μ2¼r2 +

2000

r

¶= 0:06¼r2 +

60

r:

The cost for the vertical seam is 0:01h: From Ex-ample 4, we see that h and r are related by theequation

h =1000

¼r2;

so the sealing cost is

0:01h = 0:01

μ1000

¼r2

¶

=10

¼r2:

Thus, the total cost is given by the function

C(r) = 0:06¼r2 +60

r+10

¼r2

or 0:06¼r2 + 60r¡1 +10

¼r¡2:

Then

C 0(x) = 0:12¼r ¡ 60r¡2 ¡ 20¼r¡3

or 0:12¼r ¡ 60r2¡ 20

¼r3:

Graph

y = 0:12¼r ¡ 60r2¡ 20

¼r3

on a graphing calculator. Since r must be posi-tive, our window should not include negative val-ues of x: A suitable choice for the viewing windowis [0; 10] by [¡10; 10]: From the graph, we …nd thatC0(x) = 0 when x ¼ 5:454:Thus, the cost is minimized when the radius isabout 5.454 cm.We can …nd the corresponding height by using theequation

h =1000

¼r2


h =1000

¼(5:454)2¼ 10:70:

Tominimize cost, the can should have radius 5.454cm and height 10.70 cm.

31. In Exercises 29 and 30, we found that the cost ofthe aluminum to make the can is 0:06¼r2+ 60

r ; thecost to seal the top and bottom is 0:08¼r; and thecost to seal the vertical seam is 10

¼r2 :

Thus, the total cost is now given by the function

C(r) = 0:06¼r2 +60

r+ 0:08¼r +

10

¼r2

or 0:06¼r2 + 60r¡1 + 0:08¼r +10

¼r¡2:

Then

C0(r) = 0:12¼r ¡ 60r¡2 + 0:08¼ ¡ 20¼r¡3

or 0:12¼r ¡ 60r2+ 0:08¼ ¡ 20

¼r3:

Graph

y = 0:12¼r ¡ 60r2+ 0:08¼ ¡ 20

¼r3


on a graphing calculator. A suitable choice for theviewing window is [0; 10] by [¡10; 10]: From thegraph, we …nd that C0(x) = 0 when x ¼ 5:242:Thus, the cost is minimized when the radius isabout 5.242 cm.To …nd the corresponding height, use the equation

h =1000

¼r2


h =1000

¼(5:242)2¼ 11:58:

To minimize cost, the can should have radius 5.242cm and height 11.58 cm.

32. p(t) =20t3 ¡ t41000

; [0; 20]

(a) p0(t) =3

50t2 ¡ 1

250t3

=1

50t2·3¡ 1

5t

¸Critical numbers:1

50t2 = 0 or 3¡ 1

5t = 0

t = 0 or t = 15

t p(t)

0 0

15 16:875

20 0

The number of people infected reaches a maximumin 15 days.

(b) P (15) = 16:875%

33. N(t) = 20μt

12¡ ln

μt

12

¶¶+ 30;

1 · t · 15

N 0(t) = 20·1

12¡ 12t

μ1

12

¶¸

= 20

μ1

12¡ 1t

¶

=20(t¡ 12)12t

N 0(t) = 0 when

t¡ 12 = 0t = 12:

N 00(t) does not exist at t = 0; but 0 is not in thedomain of N:Thus, 12 is the only critical number.

To …nd the absolute extrema on [1; 15]; evaluateN at the critical number and at the endpoints.

t N(t)

1 81:365

12 50

15 50:537

Use this table to answer the questions in (a)-(d).

(a) The number of bacteria will be a minimum att = 12; which represents 12 days.

(b) The minimum number of bacteria is given byN(12) = 50; which represents 50 bacteria per ml.

(c) The number of bacteria will be a maximum att = 1; which represents 1 day.

(d) The maximum number of bacteria is given byN(1) = 81:365; which represents 81.365 bacteriaper ml.

34. (a) p(t) = 10te¡t=8; [0; 40]

p0(t) = 10te¡t=8μ¡18

¶+ e¡t=8(10)

= 10e¡t=8μ¡ t8+ 1

¶

Critical numbers:

p0(t) = 0 when

¡ t8+ 1 = 0

t = 8:

t p(t)

0 0

8 29:43

40 2:6952

The percent of the population infected reaches amaximum in 8 days.

(b) P (8) = 29:43%


35. H(S) = f(S)¡ Sf(S) = 12S0:25

H(S) = 12S0:25 ¡ SH0(S) = 3S¡0:75 ¡ 1H 0(S) = 0 when

3S¡0:75 ¡ 1 = 0

S¡0:75 =1

31

S0:75=1

3

S0:75 = 3

S3=4 = 3

S = 34=3

S = 4:327:

The number of creatures needed to sustain thepopulation is S0 = 4:327 thousand.

H 00(S) = ¡2:25S1:75 < 0 when S = 4:327; so H(S) is

maximized.

H(4:327) = 12(4:327)0:25 ¡ 4:327¼ 12:98

The maximum sustainable harvest is 12.98 thou-sand.

36. H(S) = f(S)¡ S

f(S) =25S

S + 2

H0(S) =(S + 2)(25)¡ 25S

(S + 2)2¡ 1

=25S + 50¡ 25S ¡ (S + 2)2

(S + 2)2

=50¡ (S2 + 4S + 4)

(S + 2)2

=¡S2 ¡ 4S + 46(S + 2)2

H0(S) = 0 when

S2 + 4S ¡ 46 = 0S =

¡4§p16 + 1842

= 5:071:

(Discard the negative solution.)

The number of creatures needed to sustain thepopulation is S0 = 5:071 thousand.

H 00=(S+2)2(¡2S¡4)¡(S2¡4S+46)(2S+4)

(S + 2)4< 0;

so H is a maximum at S0 = 5:071:

H(S0) =25(5:071)

7:071¡ 5:071

¼ 12:86The maximum sustainable harvest is 12.86 thou-sand.

37. (a) H(S) = f(S)¡ S= Ser(1¡S=P ) ¡ S

H 0(S) = Ser(1¡S=P )¡¡rP

¢+ er(1¡S=P ) ¡ 1

Note that

f(S) = Ser(1¡S=P )

f 0(S) = Ser(1¡S=P )³¡ rP

´+ er(1¡S=P ):

H 0(S) =Ser(1¡S=P )³¡ rP

´+ er(1¡S=p) ¡ 1 = 0

Ser(1¡S=P )³¡ rP

´+ er(1¡S=P ) = 1

f 0(S) = 1

(b) f(S) = Ser(1¡S=P )

f 0(S) =³¡ rP

´Ser(1¡S=P ) + er(1¡S=P )

Set f 0(S0) = 1

er(1¡S0=P )·¡rS0P

+ 1

¸= 1

er(1¡S0=P ) =1

¡rS0P + 1

Using H(S) from part (a), we get

H(S0) = S0er(1¡S0=P ) ¡ S0= S0(e

r(1¡S0=P ) ¡ 1)

= S0

Ã1

1¡ rS0P

¡ 1!:

38. r = 0:1; P = 100

f(S) = Ser(1¡S=P )

f 0(S) = ¡ 1

1000¢ Se0:1(1¡S=100) + e0:1(1¡S=100)

f 0(S0) = ¡0:001S0e0:1(1¡S0=100) + e0:1(1¡S0=100)Graph

Y1 = ¡0:001xe0:1(1¡x=100) + e0:1(1¡x=100)


and

Y2 = 1

on the same screen. A suitable choice for the view-ing window is [0; 60] by [0:5; 1:5] with Xscl = 10and Yscl = 0.5. By zooming or using the “inter-sect” option, we …nd the graphs intersect whenx ¼ 49:37:The maximum sustainable harvest is 49.37.

39. r = 0:4; P = 500

f(S) = Ser(1¡S=P )

f 0(S) = ¡ 0:4500

Se0:4(1¡S=500) + e0:4(1¡S=500)

f 0(S0) = ¡0:0008S0e0:4(1¡S0=500) + e0:4(1¡S0=500)

Graph

Y1 = ¡0:0008x0e0:4(1¡x=500) + e0:4(1¡x=500)

and

Y2 = 1

on the same screen. A suitable choice for the view-ing window is [0; 300] by [0:5; 1:5] with Xscl = 50,Yscl = 0.5. By zooming or using the “intersect”option, we …nd that the graphs intersect whenx ¼ 237:10:The maximum sustainable harvest is 237.10.

40. Let x = distance from P to A:

Energy used over land: 1 unit per mile

Energy used over water: 43 units per mile

Distance over land: (2¡ x) miDistance over water:

p1 + x2 mi

Find the location of P to minimize energy used.

E(x) = 1(2¡ x) + 43

p1 + x2; where 0 · x · 2:

E0(x) = ¡1 + 43

μ1

2

¶(1 + x2)¡1=2(2x)

If E0(x) = 0;

4

3x(1 + x2)¡1=2 = 1

4x

3(1 + x2)1=2= 1

4

3x = (1 + x2)1=2

16

9x2 = 1 + x2

7

9x2 = 1

x2 =9

7

x =3p7

=3p7

7:

x E(x)

0 3:3333

1:134 2:8819

2 2:9814

The absolute minimum occurs at x ¼ 1:134:Point P is 3

p77 ¼ 1:134 mi from Point A:

41. Let x = distance from P to A:

Energy used over land: 1 unit per mileEnergy used over water: 109 units per mileDistance over land: (2¡ x) miDistance over water:

p1 + x2 mi

Find the location of P to minimize energy used.

E(x) = 1(2¡ x) + 109

p1 + x2; where 0 · x · 2:

E0(x) = ¡1 + 109

μ1

2

¶(1 + x2)¡1=2(2x)

If E0(x) = 0;

10

9x(1 + x2)¡1=2 = 1

10x

9(1 + x2)1=2= 1


10

9x = (1 + x2)1=2

100

81x2 = 1 + x2

19

81x2 = 1

x2 =81

19

x =9p19

=9p19

19

¼ 2:06:

This value cannot give the absolute maximum sincethe total distance from A to L is just 2 miles. Testthe endpoints of the domain.

x E(x)

0 319¼ 3:1111

2 2:4845

Point P must be at Point L:

42. (a) f(S) = aSe¡bS f(S) = Ser(1¡S=P )

= Ser¡rS=P

= Sere¡rS=P

= erSe¡(r=P )S

Comparing the two terms, replace a with er andb with r=P:

(b) Shepherd:

f(S) =aS

1 + (S=b)c

f 0(S) =[1 + (S=b)c](a)¡ (aS)[c(S=b)c¡1(1=b)]

[1 + (S=b)c]2

=a+ a(S=b)c ¡ (acS=b)(S=b)c¡1

[1 + (S=b)c]2

=a+ a(S=b)c ¡ ac(S=b)c

[1 + (S=b)c]2

=a[1 + (1¡ c)(S=b)c][1 + (S=b)c]2

Ricker:

f(S) = aSe¡bS

f 0(S) = ae¡bS + aSe¡bS(¡b)= ae¡bS(1¡ bS)

Berverton-Holt:

f(S) =aS

1 + (S=b)

f 0(S) =[1 + (S=b)](a)¡ aS(1=b)

[1 + (S=b)]2

=a+ a(S=b)¡ a(S=b)

[1 + (S=b)]2

=a

[1 + (S=b)]2

(c) Shepherd:

f 0(0) =a[1 + (1¡ c)(0=b)c][1 + (0=b)c]2

= a

Ricker:

f 0(0) = ae¡b(0)[1¡ b(0)] = aBeverton-Holt:

f 0(0) =a

[1 + (0=b)]2= a

The constant a represents the slope of the graphof f(S) at S = 0:

(d) First …nd the critical numbers by solvingf 0(S) = 0:

Shepherd:

f 0(S) = 0a[1 + (1¡ c)(S=b)c] = 0

(1¡ c)(S=b)c = ¡1(c¡ 1)(S=b)c = 1

Substitute b = 248:72 and c = 3:24 and solve forS:

(3:24¡ 1)(S=248:72)3:24 = 1μS

248:72

¶3:24=

1

2:24

S

248:72=

μ1

2:24

¶1=3:24

S = 248:72

μ1

2:24

¶1=3:24S ¼ 193:914


Using the Shepherd model, next year’s populationis maximized when this year’s population is about194,000 tons. This can be veri…ed by examing thegraph of f(S):

(e) First …nd the critical numbers by solvingf 0(S) = 0:

Ricker:

f 0(S) = 0ae¡bS(1¡ bS) = 0

1¡ bS = 0bS = 1

S =1

b

Substitute b = 0:0039 and solve for S:

S =1

0:0039

S ¼ 256:410

Using the Ricker model, next year’s population ismaximized when this year’s population is about256,000 tons. This can be veri…ed by examiningthe graph of f(S):

43. (a) Solve the given equation for e¤ective powerfor T; time.

kE

T= aSv3 + I

kE

aSv3 + I= T

Since distance is velocity, v; times time, T; we have

D(v) = vkE

aSv3 + I

=kEv

aSv3 + I:

(b) D0(v) =(aSv3 + I)kE ¡ kEv(3aSv2)

(aSv3 + I)2

=kE(aSv3 + I ¡ 3aSv3)

(aSv3 + I)2

=kE(I ¡ 2aSv3)(aSv3 + I)2

Find the critical numbers by solving D0(v) = 0 forv:

I ¡ 2aSv3 = 02aSv3 = I

v3 =I

2aS

v =

μI

2aS

¶1=3

44. Let x = width.Then x = heightand 108¡ 4x = length.

(since length plus girth = 108)

V (x) = l ¢ w ¢ h= (108¡ 4x)x ¢ x= 108x2 ¡ 4x3

V 0(x) = 216x¡ 12x2

Set V 0(x) = 0; and solve for x:

216x¡ 12x2 = 012x(18¡ x) = 0x = 0 or x = 18

0 is not in the domain, so the only critical numberis 18.

Width = 18Height = 18Length = 108¡ 4(18) = 36The dimensions of the box with maximum volumeare 36 inches by 18 inches by 18 inches.

45. Let 8¡ x = the distance the hunterwill travel on the river.

Thenp9 + x2 = the distance he will travel

on land.


The rate on the river is 5 mph, the rate on land is2 mph. Using t = d

r ;

8¡x5 = the time on the river,

p9+x2

2 = the time on land.

The total time is

T (x) =8¡ x5

+

p9 + x2

2

=8

5¡ 15x+

1

2(9 + x2)1=2:

T 0 = ¡15+1

4¢ 2x(9 + x2)¡1=2

¡15+

x

2(9 + x2)1=2= 0

1

5=

x

2(9 + x2)1=2

2(9 + x2)1=2 = 5x

4(9 + x2) = 25x2

36 + 4x2 = 25x2

36 = 21x2

6p21= x

6p21

21=2p21

7= x

x T (x)

0 3:1

2p217 2:98

8 4:27

Since the minimum time is 2.98 hr, the huntershould travel 8 ¡ 2

p217 = 56¡2p21

7 or about 6.7miles along the river.

46. Let 8¡ x = the distance the hunter willtravel on the river.

Thenp192 + x2 = the distance he will

travel on land.

Since the rate on the river is 5 mph, the rate onland is 2 mph, and t = d

r ;

8¡ x5

= the time on the riverp361 + x2

2= the time on the land.

The total time is

T (x) =8¡ x5

+

p361 + x2

2

=8

5¡ 15x+

1

2(361 + x2)1=2:

T 0(x) = ¡15+1

4¢ 2x(361 + x2)¡1=2

¡15+

x

2(361 + x2)1=2= 0

1

5=

x

2(361 + x2)1=2

2(361 + x2)1=2 = 5x

4(361 + x2) = 25x2

1444 + 4x2 = 25x2

1444 = 21x2

38p21= x

8:29 = x

8.29 is not possible, since the cabin is only 8 mileswest. Check the endpoints.

x T (x)

0 11:1

8 10:3

T (x) is minimized when x = 8:

The distance along the river is given by 8¡ x; sothe hunter should travel 8¡ 8 = 0 miles along theriver. He should complete the entire trip on land.


6.3 Further Business Applications:Economic Lot Size; EconomicOrder Quantity; Elasticity ofDemand

1. When q <q

2fMk , T 0(q) < ¡k

2 +k2 = 0; and

when q >q

2fMk , T 0(q) > ¡k

2 +k2 = 0: Since the

function T (q) is decreasing before q =q

2fMk and

increasing after q =q

2fMk , there must be a rela-

tive minimum at q =q

2fMk . By the critical point

theorem, there is an absolute minimum there.

3. The economic order quantity formula assumes thatM , the total units needed per year, is known.Thus, c is the correct answer.

4. Use equation (3) with k = 1;M = 100,000, andf = 500.

q =

r2fM

k=

r2(500)(100,000)

1

=p100,000,000 = 10,000

10,000 lamps should be made in each batch to min-imize production costs.

5. Use equation (3) with k = 9;M = 13,950, andf = 31:

q =

r2fM

k

=

r2(31)(13,950)

9

=p96,100 = 310

310 cases should be made in each batch to mini-mize production costs.

6. From Exercise 4, M = 100,000, and q =10,000.The number of batches per year is Mq =

100,00010,000 =

10:

7. From Exercise 5, M = 13,950 and q = 310. Thenumber of batches per year is

M

q=13,950310

= 45

45 cases should be made in each batch to minimizeproduction costs.

8. Here k = 0:50,M =100,000, and f = 60. We have

q =

r2fM

k=

r2(60)(100,000)

0:50

=p24,000,000 ¼ 4898:98

T (4898) = 2449:489792 and T (4899) =2449:489743, so ordering 4899 copies per orderminimizes the annual costs.

9. Here k = 1;M = 900, and f = 5. We have

q =

r2fM

k

=

r2(5)(900)

1

=p9000 ¼ 94:9

T (94) ¼ 94:872 and T (95) ¼ 94:868, so ordering95 bottles per order minimizes the annual costs.

10. Using maximum inventory size,

T (q) =fM

q+ gM + kq; (0;1)

T 0(q) =¡fMq2

+ k

Set the derivative equal to 0.

¡fMq2

+ k = 0

k =fM

q2

q2k = fM

q2 =fM

k

q =

rfM

k

Since limq!0 T (q) =1; limq!1 T (q) =1, and

q =q

fMk is the only critical value in (0;1),

q =q

fMk is the number of units that should be

ordered or manufactured to minimize total costs.

Section 6.3 Further Business Applications: Economic Lot Size; Economic Order Quantity; Elasticity of Demand 401

11. Use q =q

fMk from Exercise 10 with k = 6,

M = 5000, and f = 1000.

q =

rfM

k=

r(1000)(5000)

6¼ 912:9

with T (q) = fMq + kq (assume g = 0 since the

subsequent cost per book is so low that it canbe ignored), T (912) ¼ 10,954.456 and T (913) ¼10,954.451. So, 913 books should be printed ineach print run.

12. Assuming an annual cost, k1, for storing a singleunit, plus an annual cost per unit, k2, that mustbe paid for each unit up to the maximum numberof units stored, we have

T (q) =fM

q+ gM +

k1q

2+ k2q; (0;1)

T 0(q) =¡fMq2

+k12+ k2

Set this derivative equal to 0.

¡fMq2

+k12+ k2 = 0

k12+ k2 =

fM

q2

k1 + 2k22

=fM

q2

q2(k1 + 2k2)

2= fM

q2 =2fM

k1 + 2k2

q =

r2fM

k1 + 2k2

Since limq!0

T (q) =1, limq!1 T (q) =1, and

q =q

2fMk1+2k2

is the only critical value in (0;1),

q =q

2fMk1+2k2

is the number of units that shouldbe ordered or manufactured to minimize the totalcost in this case.

13. Use q =q

2fMk1+2k2

from Exercise 12 with k1 = 1;k2 = 2;M = 30,000, and f = 750. Also, note thatg = 8.

q =

r2fM

k1 + 2k2

=

s2(750)(30,000)1 + 2(2)

=p9,000,000 = 3000

The number of production runs each year to min-imize her total costs is

M

q=30,0003000

= 10.

14. q = 25,000¡ 50p

(a)dq

dp= ¡50

E = ¡pq¢ dqdp

= ¡ p

25,000 ¡ 50p(¡50)

=p

500¡ p

(b) R = pq

dR

dp= q(1¡E)

When R is maximum, q(1¡E) = 0:Since q = 0 means no revenue, set 1¡E = 0:

E = 1

From part (a),

p

500¡ p = 1

p = 500¡ pp = 250:

q = 25,000 ¡ 50p= 25,000 ¡ 50(250)= 12,500

Total revenue is maximized if q = 12,500.


15. q = 50¡ p4

(a)dq

dp= ¡1

4

E = ¡pq¢ dqdp

= ¡ p

50¡ p4

μ¡14

¶

= ¡ p200¡p4

μ¡14

¶

=p

200¡ p(b) R = pq

dR

dp= q(1¡E)

When R is maximum, q(1¡E) = 0:Since q = 0 means no revenue, set 1¡E = 0:

E = 1

From (a),

p

200¡ p = 1

p = 200¡ pp = 100:

q = 50¡ p4

= 50¡ 1004

= 25

Total revenue is maximized if q = 25:

16. q = 48,000¡ 10p2

(a)dq

dp= ¡20p

E = ¡pq¢ dqdp

=¡p

48,000 ¡ 10p2 (¡20p)

=20p2

48,000 ¡ 10p2

=2p2

4800¡ p2(b) R = pq

dR

dp= q(1¡E)

When R is maximum, q(1 ¡ E) = 0: Since q = 0means no revenue, set 1¡E = 0:

E = 1

From part (a),

2p2

4800¡ p2 = 1

2p2 = 4800¡ p23p2 = 4800

p2 = 1600

p = §40:

Since p must be positive, p = 40:

q = 48,000 ¡ 10p2= 48,000 ¡ 10(402)= 48,000 ¡ 10(1600)= 48,000 ¡ 16,000= 32,000

17. (a) q = 37,500 ¡ 5p2dq

dp= ¡10p

E =¡pq¢ dqdp

=¡p

37,500¡ 5p2 (¡10p)

=10p2

37,500¡ 5p2

=2p2

7500¡ p2(b) R = pq

dR

dp= q(1¡E)

When R is maximum, q(1 ¡ E) = 0: Since q = 0means no revenue, set 1¡E = 0:

E = 1

From (a),

2p2

7500¡ p2 = 1

2p2 = 7500¡ p23p2 = 7500

p2 = 2500

p = §50:

Section 6.3 Further Business Applications: Economic Lot Size; Economic Order Quantity; Elasticity of Demand 403

Since p must be positive, p = 50:

q = 37,500¡ 5p2= 37,500¡ 5(50)2= 37,500¡ 5(2500)= 37,500¡ 12,500= 25,000:

18. q = 10¡ ln p

(a)dq

dp= ¡1

p

E = ¡pq¢ dqdp

=¡p

10¡ ln pμ¡1p

¶

=1

10¡ ln p(b) R = pq

dR

dp= q(1¡E)

When R is maximum, q(1¡ E) = 0. Since q = 0means no revenue, set 1¡E = 0.

E = 1

From part (a),

1

10¡ ln p = 1

1 = 10¡ ln pln p = 9

p = e9

q = 10¡ ln p= 10¡ ln e9= 10¡ 9= 1

Note that E = 1q , thus we would expect E to be

maximum when q = 1.

19. p = 400e¡0:2q

In order to …nd the derivative dqdp ; we …rst need to

solve for q in the equation p = 400e¡0:2q:

(a)p

400= e¡0:2q

ln³ p

400

´= ln

¡e¡0:2q

¢= ¡0:2q

q =ln p

400

¡0:2 = ¡5 ln³ p

400

´

Now

dq

dp= ¡5 1p

400

¢ 1400

=¡5p; and

E = ¡pq¢ dqdp= ¡p

q¢ ¡5p=5

q:

(b) R = pq

dR

dp= q(1¡E)

When R is maximum, q(1 ¡ E) = 0. Since q = 0means no revenue, set 1¡E = 0.

E = 1

From part (a),5

q= 1

5 = q

20. q = 300¡ 2pdq

dp= ¡2

E = ¡pq¢ dqdp

E = ¡ p(¡2)300¡ 2p

=2p

300¡ 2p(a) When p = $100;

E =200

300¡ 200= 2:

Since E > 1; demand is elastic. This indicatesthat a percentage increase in price will result in agreater percentage decrease in demand.

(b) When p = $50;

E =100

300¡ 100=1

2< 1:

Since E < 1; supply is inelastic. This indicatesthat a percentage change in price will result in asmaller percentage change in demand.


21. q = 400¡ 0:2p2dq

dp= 0¡ 0:4p

E = ¡pq¢ dqdp

E = ¡ p

400¡ 0:2p2 (¡0:4p)

=0:4p2

400¡ 0:2p2

(a) If p = $20;

E =(0:4)(20)2

400¡ 0:2(20)2= 0:5:

Since E < 1; demand is inelastic. This indicatesthat total revenue increases as price increases.

(b) If p = $40;

E =(0:4)(40)2

400¡ 0:2(40)2= 8:

Since E > 1; demand is elastic. This indicatesthat total revenue decreases as price increases.

22. q = 342:5p¡0:5314

dq

dp= 342:5(¡0:5314)p¡0:5314¡1

=¡182p1:5314

E = ¡pq¢ dqdp

=¡p

342:5p¡0:5314¢ ¡182p1:5314

=182

342:5

¼ 0:5314

Since E < 1; the demand is inelastic.

23. (a) q = 55:2¡ 0:022pdq

dp= ¡0:022

E = ¡pq¢ dqdp

=¡p

55:2¡ 0:022p ¢ (¡0:022)

=0:022p

55:2¡ 0:022pWhen p = $166:10;

E =3:6542

55:2¡ 3:6542¼ 0:071:

(b) SinceE < 1, the demand for airfare is inelasticat this price.

(c) R = pq

dR

dp= q(1¡E)

When R is a maximum, q(1¡E) = 0:Since q = 0 means no revenue, set 1¡E = 0:

E = 1

From (a),

0:022p

55:2¡ 0:022p = 1

0:022p = 55:2¡ 0:022p0:044p = 55:2

p ¼ 1255Total revenue is maximized if p ¼ $1255:

24. p = 0:604q2 ¡ 20:16q + 263:067dp

dq= 1:208q ¡ 20:16

E = ¡pq¢ dqdp=

p

¡q³dpdq

´

=0:604q2 ¡ 20:16q + 263:067

¡q(1:208q ¡ 20:16)

(a) Since q = 11;

E =114:391

¡11(¡6:872) ¼ 1:51

(b) Since E > 1; demand is elastic.

(c) As q approaches 16.6887, the denominator ofE approaches zero, so E approaches in…nity.

Section 6.4 Implicit Di¤erentiation 405

25. q = m¡ np for 0 · p · m

n

dq

dp= ¡n

E = ¡pq¢ dqdp

E = ¡ p

m¡ np(¡n)

E =pn

m¡ np = 1

pn =m¡ np2np =m

p =m

2n

Thus, E = 1 when p = m2n ; or at the midpoint of

the demand curve on the interval 0 · p · mn :

26. E = ¡pq¢ dqdp

Since p 6= 0; E = 0 when dqdp = 0: The derivative is

zero, which implies that the demand function hasa horizontal tangent line at the value of p whereE = 0:

27. (a) q = Cp¡k

dq

dp= ¡Ckp¡k¡1

E =¡pq¢ dqdp

=¡pCp¡k

(¡Ckp¡k¡1)

=kp¡k

p¡k= k

28.

In the …gure, the slope of the tangent line is

dq

dp= ¡BR

RP=OB ¡OR¡RP =

OB ¡ q0¡p0

or

¡p0 dqdp= OB ¡ q0

¡p0q0¢ dqdp=OB

q0¡ 1

=OB

OR¡ 1

Because triangles AOB and PRB are similar,

¡p0q0¢ dqdp=AB

AP¡ 1

=AB ¡APAP

=PB

PA

But E = ¡p0q0¢ dqdp so the ratio PB

PA equals theelasticity E:

29. The demand function q(p) is positive and increas-ing, so dq

dp is positive. Since p0 and q0 are also

positive, the elasticity E = ¡p0q0¢ dqdp is negative.

6.4 Implicit Di¤erentiation1. 6x2 + 5y2 = 36

d

dx(6x2 + 5y2) =

d

dx(36)

d

dx(6x2) +

d

dx(5y2) =

d

dx(36)

12x+ 5 ¢ 2y dydx= 0

10ydy

dx= ¡12x

dy

dx= ¡6x

5y

2. 7x2 ¡ 4y2 = 24d

dx(7x2 ¡ 4y2) = d

dx(24)

d

dx(7x2)¡ d

dx(4y2) =

d

dx(24)

14x¡ 8y dydx= 0

8ydy

dx= 14x

dy

dx=7x

4y


3. 8x2 ¡ 10xy + 3y2 = 26

d

dx(8x2 ¡ 10xy + 3y2) = d

dx(26)

16x¡ d

dx(10xy) +

d

dx(3y2) = 0

16x¡ 10x dydx¡ y d

dx(10x) + 6y

dy

dx= 0

16x¡ 10x dydx¡ 10y + 6y dy

dx= 0

(¡10x+ 6y) dydx= ¡16x+ 10y

dy

dx=¡16x+ 10y¡10x+ 6y

dy

dx=8x¡ 5y5x¡ 3y

4. 7x2 = 5y2 + 4xy + 1

d

dx(7x2) =

d

dx(5y2 + 4xy + 1)

14x =d

dx(5y2 + 4xy + 1)

14x = 10yd

dxy + 4x

d

dx(y) + y

d

dx(4x)

14x = 10ydy

dx+ 4x

dy

dx(1) + 4y

14x¡ 4y = dy

dx(10y + 4x)

14x¡ 4y10y + 4x

=dy

dx

7x¡ 2y2x+ 5y

=dy

dx

5. 5x3 = 3y2 + 4y

d

dx(5x3) =

d

dx(3y2 + 4y)

15x2 =d

dx(3y2) +

d

dx(4y)

15x2 = 6ydy

dx+ 4

dy

dx

15x2

6y + 4=dy

dx

6. 3x3 ¡ 8y2 = 10y

d

dx(3x3 ¡ 8y2) = d

dx(10y)

d

dx(3x3)¡ d

dx(8y2) = 10

dy

dx

9x2 ¡ 16y dydx= 10

dy

dx

9x2 = (16y + 10)dy

dx

9x2

16y + 10=dy

dx

7. 3x2 =2¡ y2 + y

d

dx(3x2) =

d

dx

μ2¡ y2 + y

¶

6x =(2 + y) d

dx (2¡ y)¡ (2¡ y) ddx(2 + y)

(2 + y)2

6x =(2 + y)

³¡dydx

´¡ (2¡ y) dydx

(2 + y)2

6x =¡4 dydx(2 + y)2

6x(2 + y)2 = ¡4 dydx

¡3x(2 + y)2

2=dy

dx

8. 2y2 =5 + x

5¡ x

d

dx(2y2) =

d

dx

μ5 + x

5¡ x¶

4ydy

dx=(1)(5¡ x)¡ (¡1)(5 + x)

(5¡ x)2

4ydy

dx=5¡ x+ 5+ x(5¡ x)2

=10

(5¡ x)2dy

dx=

10

4y(5¡ x)2

=5

2y(5¡ x)2


9. 2px+ 4

py = 5y

d

dx(2x1=2 + 4y1=2) =

d

dx(5y)

x¡1=2 + 2y¡1=2dy

dx= 5

dy

dx

(2y¡1=2 ¡ 5)dydx= ¡x¡1=2

dy

dx=

x¡1=2

5¡ 2y¡1=2μx1=2y1=2

x1=2y1=2

¶

=y1=2

x1=2(5y1=2 ¡ 2)

=

pyp

x(5py ¡ 2)

10. 4px¡ 8py = 6y3=2

d

dx(4x1=2 ¡ 8y1=2) = 6 d

dx(y3=2)

2x¡1=2 ¡ 4y¡1=2 dydx= 6 ¢ 3

2y1=2

dy

dx

2x¡1=2 = (9y1=2 + 4y¡1=2)dy

dx

2x¡1=2

9y1=2 + 4y¡1=2=dy

dx

2pyp

x(9y + 4)=dy

dx

11. x4y3 + 4x3=2 = 6y3=2 + 5

d

dx(x4y3 + 4x3=2) =

d

dx(6y3=2 + 5)

d

dx(x4y3) +

d

dx(4x3=2) =

d

dx(6y3=2) +

d

dx(5)

4x3y3 + x4 ¢ 3y2 dydx+ 6x1=2 = 9y1=2

dy

dx+ 0

4x3y3 + 6x1=2 = 9y1=2dy

dx¡ 3x4y2 dy

dx

4x3y3 + 6x1=2 = (9y1=2 ¡ 3x4y2) dydx

4x3y3 + 6x1=2

9y1=2 ¡ 3x4y2 =dy

dx

12. (xy)4=3 + x1=3 = y6 + 1

d

dx[(xy)4=3 + x1=3] =

d

dx(y6 + 1)

d

dx(x4=3y4=3) +

d

dx(x1=3) =

d

dx(y6) +

d

dx(1)

x4=3 ¢ 43y1=3

dy

dx+4

3x1=3y4=3 +

1

3x¡2=3

= 6y5dy

dx+ 0

4

3x1=3y4=3 +

1

3x¡2=3 = 6y5

dy

dx¡ 43x4=3y1=3

dy

dx

4x1=3y4=3 + x¡2=3 = 18y5dy

dx¡ 4x4=3y1=3 dy

dx

4x1=3y4=3 + x¡2=3 = (18y5 ¡ 4x4=3y1=3) ¢ dydx

4x1=3y4=3 + x¡2=3

18y5 ¡ 4x4=3y1=3 =dy

dx

x2=3

x2=3¢ 4x

1=3y4=3 + x¡2=3

18y5 ¡ 4x4=3y1=3 =dy

dx

4xy4=3 + 1

18x2=3y5 ¡ 4x2y1=3 =dy

dx

13. ex2y = 5x+ 4y + 2

d

dx(ex

2y) =d

dx(5x+ 4y + 2)

ex2y d

dx(x2y) =

d

dx(5x) +

d

dx(4y) +

d

dx(2)

ex2y

μ2xy + x2

dy

dx

¶= 5 + 4

dy

dx+ 0

2xyex2y + x2ex

2y dy

dx= 5 + 4

dy

dx

x2ex2y dy

dx¡ 4 dy

dx= 5¡ 2xyex2y

(x2ex2y ¡ 4)dy

dx= 5¡ 2xyex2y

dy

dx=5¡ 2xyex2yx2ex2y ¡ 4


14. x2ey + y = x3

d

dx(x2ey + y) =

d

dx(x3)

d

dx(x2ey) +

d

dx(y) = 3x2

2xey + x2eydy

dx+dy

dx= 3x2

x2eydy

dx+dy

dx= 3x2 ¡ 2xey

(x2ey + 1)dy

dx= 3x2 ¡ 2xey

dy

dx=3x2 ¡ 2xeyx2ey + 1

15. x+ ln y = x2y3

d

dx(x+ ln y) =

d

dx(x2y3)

1 +1

y

dy

dx= 2xy3 + 3x2y2

dy

dx

1

y

dy

dx¡ 3x2y2 dy

dx= 2xy3 ¡ 1μ

1

y¡ 3x2y2

¶dy

dx= 2xy3 ¡ 1

dy

dx=2xy3 ¡ 11y ¡ 3x2y2

=y(2xy3 ¡ 1)1¡ 3x2y3

16. y lnx+ 2 = x3=2y5=2

d

dx(y lnx+ 2) =

d

dx(x3=2y5=2)

lnxdy

dx+y

x+ 0 =

3

2x1=2y5=2 +

5

2x3=2y3=2

dy

dx

lnxdy

dx¡ 52x3=2y3=2

dy

dx=3

2x1=2y5=2 ¡ y

x

dy

dx

μlnx¡ 5

2x3=2y3=2

¶=3

2x1=2y5=2 ¡ y

x

dy

dx=

32x

1=2y5=2 ¡ yx

lnx¡ 52x

3=2y3=2

=3x3=2y5=2 ¡ 2y

x(2 lnx¡ 5x3=2y3=2)

17. x2 + y2 = 25; tangent at (¡3; 4)d

dx(x2 + y2) =

d

dx(25)

2x+ 2ydy

dx= 0

2ydy

dx= ¡2x

dy

dx= ¡x

y

m = ¡xy= ¡¡3

4=3

4

y ¡ y1 = m(x¡ x1)

y ¡ 4 = 3

4[x¡ (¡3)]

4y ¡ 16 = 3x+ 94y = 3x+ 25

y =3

4x+

25

4

18. x2 + y2 = 100; tangent at (8;¡6)d

dx(x2 + y2) =

d

dx(100)

2x+ 2ydy

dx= 0

dy

dx= ¡x

y

m = ¡xy= ¡ 8

¡6 =4

3

y ¡ y1 = m(x¡ x1)

y + 6 =4

3(x¡ 8)

3y + 18 = 4x¡ 323y = 4x¡ 50

19. x2y2 = 1; tangent at (¡1; 1)d

dx(x2y2) =

d

dx(1)

x2d

dx(y2) + y2

d

dx(x2) = 0

x2(2y)dy

dx+ y2(2x) = 0

2x2ydy

dx= ¡2xy2

dy

dx=¡2xy22x2y

= ¡yx


m = ¡yx= ¡ 1

¡1 = 1

y ¡ 1 = 1[x¡ (¡1)]y = x+ 1+ 1

y = x+ 2

20. x2y3 = 8; tangent at (¡1; 2)d

dx(x2y3) =

d

dx(8)

2xy3 + 3x2y2dy

dx= 0

dy

dx= ¡ 2xy

3

3x2y2

m = ¡ 2xy3

3x2y2= ¡ 2(¡1)(2)

3

3(¡1)2(2)2

=16

12=4

3

y ¡ 2 = 4

3(x+ 1)

3y ¡ 6 = 4x+ 43y = 4x+ 10

21. 2y2 ¡px = 4; tangent at (16; 2)d

dx(2y2 ¡px) = d

dx(4)

4ydy

dx¡ 12x¡1=2 = 0

4ydy

dx=

1

2x1=2

dy

dx=

1

8yx1=2

m =1

8yx1=2=

1

8(2)(16)1=2

=1

8(2)(4)=1

64

y ¡ 2 = 1

64(x¡ 16)

64y ¡ 128 = x¡ 1664y = x+ 112

y =x

64+7

4


22. y +px

y= 3; tangent at (4; 2)

d

dx

μy +

px

y

¶=d

dx(3)

dy

dx+d

dx

μpx

y

¶= 0

dy

dx+y¡12

¢x¡1=2 ¡px dy

dxy2

= 0

dy

dx=¡12yx

¡1=2 +px dydx

y2

y2dy

dx= ¡1

2yx¡1=2 +

pxdy

dx

(y2 ¡px)dydx= ¡1

2yx¡1=2

dy

dx=

¡y2x1=2(y2 ¡px)

m =¡y

2x1=2(y2 ¡px)=

¡22(2)(4¡ 2)

= ¡14

y ¡ 2 = ¡14(x¡ 4)

y = ¡14x+ 3

x+ 4y = 12

23. ex2+y2 = xe5y ¡ y2e5x=2; tangent at (2; 1)

d

dx(ex

2+y2) =d

dx(xe5y ¡ y2e5x=2)

ex2+y2 ¢ d

dx(x2 + y2) = e5y + x

d

dx(e5y)¡

·2ydy

dxe5x=2 + y2e5x=2

d

dx

μ5x

2

¶¸

ex2+y2

μ2x+ 2y

dy

dx

¶= e5y + x ¢ 5e5y dy

dx¡ 2ye5x=2 dy

dx¡ 52y2e5x=2

(2yex2+y2 ¡ 5xe5y + 2ye5x=2)dy

dx= ¡2xex2+y2 + e5y ¡ 5

2y2e5x=2

dy

dx=¡2xex2+y2 + e5y ¡ 5

2y2e5x=2

2yex2+y2 ¡ 5xe5y + 2ye5x=2

m =¡4e5 + e5 ¡ 5

2e5

2e5 ¡ 10e5 + 2e5 =¡11

2 e5

¡6e5 =11

12

y ¡ 1 = 11

12(x¡ 2)

y =11

12x¡ 5

6


24. 2xexy = ex3

+ yex2

; tangent at (1; 1)

d

dx(2xexy) =

d

dx(ex

3

+ yex2

)

2exy + 2x ¢ exy ddx(xy) = ex

3 d

dx(x3) +

dy

dx¢ ex2 + yex2 ¢ d

dx(x2)

2exy + 2xexyμy + x

dy

dx

¶= ex

3 ¢ 3x2 + dy

dxex

2

+ yex2 ¢ 2x

(2x2exy ¡ ex2)dydx= ¡2exy ¡ 2xyexy + 2xyex2 + 3x2ex3

dy

dx=¡2exy ¡ 2xyexy + 2xyex2 + 3x2ex3

2x2exy ¡ ex2

m =¡2e¡ 2e+ 2e+ 3e

2e¡ e =e

e= 1

y ¡ 1 = 1(x¡ 1)y = x

25. ln(x+ y) = x3y2 + ln(x2 + 2)¡ 4; tangent at (1; 2)d

dx[ln(x+ y)] =

d

dx[x3y2 + ln(x2 + 2)¡ 4]

1

x+ y¢ ddx(x+ y) = 3x2y2 + x3 ¢ 2y dy

dx+

1

x2 + 2¢ ddx(x2 + 2)¡ d

dx(4)μ

1

x+ y¡ 2x3y

¶dy

dx= 3x2y2 +

2x

x2 + 2¡ 1

x+ y

dy

dx=3x2y2 + 2x

x2+2 ¡ 1x+y

1x+y ¡ 2x3y

m =3 ¢ 1 ¢ 4 + 2¢1

3 ¡ 13

13 ¡ 2 ¢ 1 ¢ 2

=373¡113

= ¡3711

y ¡ 2 = ¡3711(x¡ 1)

y = ¡3711x+

59

11

26. ln(x2 + y2) = ln(5x) +y

x¡ 2; tangent at (1; 2)

d

dx[ln(x2 + y2)] =

d

dx

hln(5x) +

y

x¡ 2i

1

x2 + y2¢ ddx(x2 + y2) =

1

5x¢ ddx(5x) +

x dydx ¡ yx2

¡ d

dx(2)

1

x2 + y2

μ2x+ 2y

dy

dx

¶=1

x+1

x¢ dydx¡ y

x2μ2y

x2 + y2¡ 1

x

¶dy

dx=1

x¡ y

x2¡ 2x

x2 + y2

dy

dx=

1x ¡ y

x2 ¡ 2xx2+y2

2yx2+y2 ¡ 1

x


m =1¡ 2¡ 2

545 ¡ 1

=¡75¡15

= 7

y ¡ 2 = 7(x¡ 1)y = 7x¡ 5

27. y3 + xy ¡ y = 8x4; x = 1First, …nd the y-value of the point.

y3 + (1)y ¡ y = 8(1)4y3 = 8

y = 2

The point is (1; 2):

Find dydx :

3y2dy

dx+ x

dy

dx+ y ¡ dy

dx= 32x3

(3y2 + x¡ 1) dydx= 32x3 ¡ y

dy

dx=

32x3 ¡ y3y2 + x¡ 1

At (1; 2);dy

dx=

32(1)3 ¡ 23(2)2 + 1¡ 1 =

30

12=5

2:

y ¡ 2 = 5

2(x¡ 1)

y ¡ 2 = 5

2x¡ 5

2

y =5

2x¡ 1

2

28. y3 + 2x2y ¡ 8y = x3 + 19; x = 2

3y2dy

dx+ 2x2

dy

dx+ 4xy ¡ 8 dy

dx= 3x2

(3y2 + 2x2 ¡ 8) dydx= 3x2 ¡ 4xy

dy

dx=

3x2 ¡ 4xy3y2 + 2x2 ¡ 8

Find y when x = 2:y3 + 8y ¡ 8y = 8 + 19

y3 = 27

y = 3

dy

dx=

12¡ 2427 + 8¡ 8 =

¡1227

= ¡49

y ¡ 3 = ¡49(x¡ 2)

y ¡ 3 = ¡49x+

8

9

y = ¡49x+

35

9


29. y3 + xy2 + 1 = x+ 2y2; x = 2

Find the y-value of the point.y3 + 2y2 + 1 = 2 + 2y2

y3 + 1 = 2

y3 = 1

y = 1


Find dydx :

3y2dy

dx+ x2y

dy

dx+ y2 = 1 + 4y

dy

dx

3y2dy

dx+ 2xy

dy

dx¡ 4y dy

dx= 1¡ y2

(3y2 + 2xy ¡ 4y) dydx= 1¡ y2

dy

dx=

1¡ y23y2 + 2xy ¡ 4y

At (2; 1);dy

dx=

1¡ 123(1)2 + 2(2)(1)¡ 4(1) = 0:

y ¡ 0 = 0(x¡ 2)y = 1

30. y4(1¡ x) + xy = 2; x = 1Find the y-value.

y4(1¡ 1) + y = 2y = 2


Find dydx :

y4(¡1) + 4y3(1¡ x) dydx+ x

dy

dx+ y = 0

¡y4 + 4y3 dydx¡ 4xy3 dy

dx+ x

dy

dx+ y = 0

(4y3 ¡ 4xy3 + x) dydx= y4 ¡ y

dy

dx=

y4 ¡ y4y3 ¡ 4xy3 + x

dy

dxat (1; 2) is

24 ¡ 24 ¢ 23 ¡ 4(1)23 + 1 =

16¡ 232¡ 32 + 1 = 14:

y ¡ 2 = 14(x¡ 1)y ¡ 2 = 14x¡ 14

y = 14x¡ 12


31. 2y3(x¡ 3) + xpy = 3; x = 3

Find the y-value of the point.

2y3(3¡ 3) + 3py = 33py = 3py = 1

y = 1

The point is (3; 1)

Find dydx :

2y3(1) + 6y2(x¡ 3) dydx

+ x

μ1

2

¶y¡1=2

dy

dx+py = 0

6y2(x¡ 3) dydx+

x

2py

dy

dx= ¡2y3 ¡ p

y

·6y2(x¡ 3) + x

2py

¸dy

dx= ¡2y3 ¡ p

y

dy

dx=

¡2y3 ¡py6y2(x¡ 3) + x

2py

=¡4y7=2 ¡ 2y

12y5=2(x¡ 3) + x

At (3; 1);

dy

dx=

¡4(1)¡ 212(1)(3¡ 3) + 3 =

¡63= ¡2:

y ¡ 1 = ¡2(x¡ 3)y ¡ 1 = ¡2x+ 6

y = ¡2x+ 7

32.y

18(x2 ¡ 64) + x2=3y1=3 = 12; x = 8

Find the y-value of the point.y

18(64¡ 64) + 82=3y1=3 = 12

4y1=3 = 12

y1=3 = 3

y = 27


Finddy

dx:

y

18(2x) +

1

18(x2 ¡ 64) dy

dx+1

3x2=3y¡2=3

dy

dx

+2

3x¡1=3y1=3 = 0


μx2 ¡ 6418

+x2=3y¡2=3

3

¶dy

dx=¡2xy18

¡ 2x¡1=3y1=3

3

dy

dx=

¡2xy18 ¡ 2x¡1=3y1=3

3

x2¡6418 + x2=3y¡2=3

3

=¡2xy ¡ 12x¡1=3y1=3x2 ¡ 64 + 6x2=3y¡2=3

dy

dxat (8; 27) is

¡2(8)(27)¡ 12(8)¡1=3(27)1=364¡ 64 + 6(8)2=3(27)¡2=3 =

¡432¡ 18249

= (¡450)μ9

24

¶

=¡6754

y ¡ 27 = ¡6754(x¡ 8)

y ¡ 27 = ¡6754x+ 1350

y = ¡6754x+ 1377

33. x2 + y2 = 100

(a) Lines are tangent at points where x = 6: By substituting x = 6 in the equation, we …nd that the points are(6; 8) and (6;¡8):

d

dx(x2 + y2) =

d

dx(100)

2x+ 2ydy

dx= 0

2ydy

dx= ¡2x

dy = ¡xy

m1 = ¡xy= ¡6

8= ¡3

4

m2 = ¡xy= ¡ 6

¡8 =3

4

First tangent:

y ¡ 8 = ¡34(x¡ 6)

y = ¡34x+

25

2

Second tangent:

y ¡ (¡8) = 3

4(x¡ 6)

y + 8 =3

4x¡ 18

4

y =3

4x¡ 25

2


(b)

34. x2=3 + y2=3 = 2; (1; 1)

Finddy

dx.

2

3x¡1=3 +

2

3y¡1=3

dy

dx= 0

2

3y¡1=3

dy

dx= ¡2

3x¡1=3

dy

dx=¡23x

¡1=323y¡1=3

= ¡y1=3

x1=3

At (1; 1)

dy

dx= ¡1

1=3

11=3= ¡1

y ¡ 1 = ¡1(x¡ 1)y ¡ 1 = ¡x+ 1

y = ¡x+ 2

35. 3(x2 + y2)2 = 25(x2 ¡ y2); (2; 1)

Finddy

dx.

6(x2 + y2)d

dx(x2 + y2) = 25

d

dx(x2 ¡ y2)

6(x2 + y2)

μ2x+ 2y

dy

dx

¶= 25

μ2x¡ 2y dy

dx

¶

12x3 + 12x2ydy

dx+ 12xy2 + 12y3

dy

dx= 50x¡ 50y dy

dx

12x2ydy

dx+ 12y3

dy

dx+ 50y

dy

dx= ¡12x3 ¡ 12xy2 + 50x

(12x2y + 12y3 + 50y)dy

dx= ¡12x3 ¡ 12xy2 + 50x

dy

dx=¡12x3 ¡ 12xy2 + 50x12x2y + 12y3 + 50y


At (2; 1),

dy

dx=¡12(2)3 ¡ 12(2)(1)2 + 50(2)12(2)2 + 12(1)3 + 50(1)

=¡20110

= ¡ 2

11

y ¡ 1 = ¡ 2

11(x¡ 2)

y ¡ 1 = ¡ 2

11x+

4

11

y = ¡ 2

11x+

15

11

36. y2(x2 + y2) = 20x2; (1; 2)

Finddy

dx.

2y(x2 + y2)dy

dx+ y2

μ2x+ 2y

dy

dx

¶= 40x

2x2ydy

dx+2y3

dy

dx+2xy2+2y3

dy

dx= 40x

2x2ydy

dx+ 4y3

dy

dx= ¡2xy2 + 40x

(2x2y + 4y3)

μdy

dx

¶= ¡2xy2 + 40x

dy

dx=¡2xy2 + 40x2x2y + 4y3

At (1; 2),

dy

dx=¡2(1)(2)2 + 40(1)2(1)2(2) + 4(2)3

=32

36=8

9

y ¡ 2 = 8

9(x¡ 1)

y ¡ 2 = 8

9x¡ 8

9

y =8

9x+

10

9


37. 2(x2 + y2)2 = 25xy2; (2; 1)

Finddy

dx.

4(x2 + y2)d

dx(x2 + y2) = 25

d

dx(xy2)

4(x2 + y2)

μ2x+ 2y

dy

dx

¶= 25

μy2 + 2xy

dy

dx

¶

8x3 + 8x2ydy

dx+ 8xy2 + 8y3

dy

dx= 25y2 + 50xy

dy

dx

8x2ydy

dx+ 8y3

dy

dx¡ 50xy dy

dx= ¡8x3 ¡ 8xy2 + 25y2

(8x2y + 8y3 ¡ 50xy)μdy

dx

¶= ¡8x3 ¡ 8xy2 + 25y2

dy

dx=¡8x3 ¡ 8xy2 + 25y28x2y + 8y3 ¡ 50xy

At (2; 1),

dy

dx=¡8(2)3 ¡ 8(2)(1)2 + 25(1)28(2)2(1) + 8(1)3 ¡ 50(2)(1)

=¡55¡60 =

11

12

y ¡ 1 = 11

12(x¡ 2)

y ¡ 1 = 11

12x¡ 11

6

y =11

12x¡ 5

6

38. x2 + y2 + 1 = 0

d

dx(x2 + y2) =

d

dx(¡1)

2x+ 2ydy

dx= 0

dy

dx=¡2x2y

= ¡xy

If x and y are real numbers, x2 and y2 are nonnegative. 1 plus a nonnegative number cannot equal zero, sothere is no function y = f(x) that satis…es x2 + y2 + 1 = 0:

39. y2 = x3 + ax+ bd

dx(y2) =

d

dx(x3 + ax+ b)

2ydy

dx= 3x2 + a

dy

dx=3x2 + a

2y


40.pu+

p2v + 1 = 5

du

dv(pu+

p2v + 1) =

du

dv(5)

1

2u¡1=2

du

dv+1

2(2v + 1)¡1=2(2) = 0

1

2u¡1=2

du

dv= ¡ 1

(2v + 1)1=2

du

dv= ¡ 2u1=2

(2v + 1)1=2

41.pu+

p2v + 1 = 5

dv

du(pu+

p2v + 1) =

dv

du(5)

1

2u¡1=2 +

1

2(2v + 1)¡1=2(2)

dv

du= 0

(2v + 1)¡1=2dv

du= ¡1

2u¡1=2

dv

du= ¡(2v + 1)

1=2

2u1=2

42. 2p2 + q2 = 1600

(a) 4p+ 2qdq

dp= 0

4p = ¡2q dqdp

¡2pq=dq

dp

This is the rate of change of demand with respectto price.

(b) 4pdp

dq+ 2q = 0

dp

dq= ¡ q

2p

This is the rate of change of price with respect todemand.

43. C2 = x2 + 100px+ 50

(a) 2CdC

dx= 2x+

1

2(100)x¡1=2

dC

dx=2x+ 50x¡1=2

2C

dC

dx=x+ 25x¡1=2

C¢ x

1=2

x1=2

dC

dx=x3=2 + 25

Cx1=2

When x = 5; the approximate increase in cost ofan additional unit is

(5)3=2 + 25

(52 + 100p5 + 50)1=2(5)1=2

=36:18

(17:28)p5

¼ 0:94:

(b) 900(x¡ 5)2 + 25R2 = 22; 500

R2 = 900¡ 36(x¡ 5)2

2RdR

dx= ¡72(x¡ 5)

dR

dx=¡36(x¡ 5)

R=180¡ 36x

R

When x = 5; the approximate change in revenuefor a unit increase in sales is

180¡ 36(5)R

=0

R= 0:

44. First note that

if logR(w) = 1:83¡ 0:43 log(w)then R(w) = 101:83¡0:43 log(w)

= 101:8310¡0:43 log(w)

= 101:83[10log(w)]¡0:43

= 101:83w¡0:43

(a)d

dw[logR(w)] =

d

dw[1:83¡ 0:43 log(w)]

1

ln 10

1

R(w)

dR

dw= 0¡ 0:43 1

ln 10

1

w

dR

dw= ¡0:43R(w)

w

= ¡0:43101:83w¡0:43

w

¼ ¡29:0716w¡1:43

(b) R(w) = 101:83w¡0:43

d

dw[R(w)] =

d

dw[101:83w¡0:43]

dR

dw= 101:83(¡0:43)w¡1:43¼ ¡29:0716w¡1:43


45. b¡ a = (b+ a)3

d

db(b¡ a) = d

db[(b+ a)3]

1¡ dadb= 3(b+ a)2

d

db(b+ a)

1¡ dadb= 3(b+ a)2

μ1 +

da

db

¶

1¡ dadb= 3(b+ a)2 + 3(b+ a)2

da

db

¡dadb¡ 3(b+ a)2 da

db= 3(b+ a)2 ¡ 1

[¡1¡ 3(b+ a)2]dadb= 3(b+ a)2 ¡ 1

da

db=

3(b+ a)2 ¡ 1¡1¡ 3(b+ a)2

da

db= 0

3(b+ a)2 ¡ 1 = 0

(b+ a)2 =1

3

b+ a =1p3

Since b¡ a = (b+ a)3 =μ1p3

¶3=

1

3p3:

b+ a =1p3

¡(b¡ a) = ¡ 1

3p3

2a =2

3p3

a =1

3p3

46. xya = k

d

dx(xya) =

d

dx(k)

xd

dx(ya) + ya(1) = 0

x

μaya¡1

dy

dx

¶+ ya = 0

axya¡1dy

dx= ¡ya

dy

dx= ¡ ya

axya¡1

dy

dx= ¡ y

ax

47. s3 ¡ 4st+ 2t3 ¡ 5t = 0

3s2ds

dt¡μ4tds

dt+ 4s

¶+ 6t2 ¡ 5 = 0

3s2ds

dt¡ 4t ds

dt¡ 4s+ 6t2 ¡ 5 = 0

ds

dt(3s2 ¡ 4t) = 4s¡ 6t2 + 5

ds

dt=4s¡ 6t2 + 53s2 ¡ 4t

48. 2s2 +pst¡ 4 = 3t

4sds

dt+1

2(st)¡1=2

μs+ t

ds

dt

¶= 3

4sds

dt+s+ t dsdt2pst

= 3

8s(pst) dsdt + s+ t

dsdt

2pst

= 3

(8spst+ t) dsdt + s

2pst

= 3

(8spst+ t)

ds

dt= 6

pst¡ s

ds

dt=¡s+ 6pst8spst+ t

6.5 Related Rates

1. y2 ¡ 8x3 = ¡55; dxdt= ¡4; x = 2; y = 3

2ydy

dt¡ 24x2 dx

dt= 0

ydy

dt= 12x2

dx

dt

3dy

dt= 48(¡4)

dy

dt= ¡64

2. 8y3 + x2 = 1;dx

dt= 2; x = 3; y = ¡1

24y2dy

dt+ 2x

dx

dt= 0

dy

dt=¡2x dxdt24y2

= ¡ xdxdt

12y2

= ¡ (3)(2)

12(¡1)2

= ¡12

Section 6.5 Related Rates 421

3. 2xy ¡ 5x+ 3y3 = ¡51; dxdt= ¡6; x = 3; y = ¡2

2xdy

dt+2y

dx

dt¡5dx

dt+9y2

dy

dt= 0

(2x+9y2)dy

dt+(2y¡5)dx

dt= 0

(2x+9y2)dy

dt= (5¡2y)dx

dt

dy

dt=

5¡2y2x+9y2

¢ dxdt

=5¡2(¡2)

2(3)+9(¡2)2 ¢ (¡6)

=9

42¢ (¡6) = ¡54

42= ¡9

7

4. 4x3 ¡ 6xy2 + 3y2 = 228; dxdt= 3; x = ¡3; y = 4

12x2dx

dt¡μ6y2

dx

dt+12xy

dy

dt

¶+6y

dy

dt= 0

12x2dx

dt¡ 6y2 dx

dt¡ 12xy dy

dt+ 6y

dy

dt= 0

(6y ¡ 12xy)dydt= (6y2¡12x2)dx

dt

dy

dt=y2¡2x2y¡2xy ¢

dx

dt

=42¡2 ¢(¡3)24¡2 ¢ (¡3) ¢ 4 ¢3

= ¡ 2

28¢ 3 = ¡6

28= ¡ 3

14

5.x2 + y

x¡ y = 9;dx

dt= 2; x = 4; y = 2

(x¡y)³2x dxdt+

dydt

´¡(x2 + y)

³dxdt¡dy

dt

´(x¡ y)2 = 0

2x(x¡y) dxdt + (x¡y) dydt¡(x2+y) dxdt+(x2+y) dydt(x¡ y)2 = 0

[2x(x¡y)¡(x2+y)] dxdt+[(x¡y)+(x2+y)] dy

dt= 0

dy

dt=[(x2 + y)¡ 2x(x¡ y)] dxdt(x¡ y) + (x2 + y)

dy

dt=(¡x2 + y + 2xy) dxdt

x+ x2

=[¡(4)2 + 2+ 2(4)(2)](2)

4 + 42

=4

20=1

5

6.y3 ¡ 4x2x3 + 2y

=44

31;dx

dt= 5; x = ¡3; y = ¡2

31(y3 ¡ 4x2) = 44(x3 + 2y)31y3 ¡ 124x2 = 44x3 + 88y

93y2dy

dt¡ 248x dx

dt= 132x2

dx

dt+ 88

dy

dt

(93y2 ¡ 88) dydt= (132x2 + 248x)

dx

dt

dy

dt=132x2 + 248x

93y2 ¡ 88 ¢ dxdt

=132(¡3)2 + 248(¡3)

93(¡2)2 ¡ 88 ¢ 5

=444

284¢ 5 = 2220

284=555

71

7. xey = 3+ lnx;dx

dt= 6; x = 2; y = 0

eydx

dt+ xey

dy

dt= 0 +

1

x

dx

dt

xeydy

dt=

μ1

x¡ ey

¶dx

dt

dy

dt=

¡1x ¡ ey

¢dxdt

xey

=(1¡ xey)dxdt

x2ey

=[1¡ (2)e0](6)

22e0

=¡64= ¡3

2

8. y lnx+ xey = 1;dx

dt= 5; x = 1; y = 0

d

dt(y lnx) +

d

dt(xey) =

d

dt(1)

lnxdy

dt+y

x

dx

dt+ ey

dx

dt+ x ¢ ey dy

dt= 0

(lnx+ xey)dy

dt+³yx+ ey

´ dxdt= 0

(lnx+ xey)dy

dt= ¡

³yx+ ey

´ dxdt

dy

dt= ¡(

yx + e

y)dxdtlnx+ xey

= ¡ (y + xey)dxdt

x lnx+ x2ey

= ¡ [0 + (1)e0](5)

(1) ln 1 + 12e0= ¡5


9. C = 0:2x2 + 10,000; x = 80;dx

dt= 12

dC

dt= 0:2(2x)

dx

dt= 0:2(160)(12) = 384

The cost is changing at a rate of $384 per month.

10. C =R2

450,000+ 12,000;

dC

dx= 15

R = 25,000

dC

dx=

R

225,000¢ dRdx

15 =25,000225,000

¢ dRdx

15 =1

9¢ dRdx

135 =dR

dx

Revenue is changing at a rate of $135 per unit.

11. R = 50x¡ 0:4x2; C = 5x+ 15; x = 40; dxdt = 10

(a)dR

dt= 50

dx

dt¡ 0:8x dx

dt

= 50(10)¡ 0:8(40)(10)= 500¡ 320= 180

Revenue is increasing at a rate of $180 per day.

(b)dC

dt= 5

dx

dt= 5(10) = 50

Cost is increasing at a rate of $50 per day.

(c) Pro…t = Revenue¡CostP = R¡CdP

dt=dR

dt¡ dCdt= 180¡ 50 = 130

Pro…t is increasing at a rate of $130 per day.

12. R = 50x¡ 0:4x2; C = 5x+ 15; x = 80; dxdt= 12

(a)dR

dt= 50

dx

dt¡ 0:8x dx

dt

= 50(12)¡ 0:8(80)(12)= 600¡ 768= ¡168

Revenue is decreasing at a rate of $168 per day.

(b)dC

dt= (5)

dx

dt= (5)(12) = 60

Cost is increasing at a rate of $60 per day.

(c) P = R¡CdP

dt=dR

dt¡ dCdt

= ¡168¡ 60= ¡228

Pro…t is decreasing at a rate of $228 per day.

13. pq = 8000; p = 3:50; dpdt = 0:15

pq = 8000

pdq

dt+ q

dp

dt= 0

dq

dt=¡q dpdtp

=¡ ¡80003:50

¢(0:15)

3:50

¼ ¡98

Demand is decreasing at a rate of approximately98 units per unit time.

14. R = pq;dq

dt= 25

Find the relationship between p and q by …ndingthe equation of the line through (0; 70); and(100; 60):

m =70¡ 606¡ 100 =

10

¡100 = ¡1

10

p¡ 70 = ¡ 1

10(q ¡ 0)

p¡ 70 = ¡ 1

10q

p = ¡ 1

10q + 70

R =

μ¡ 1

10q + 70

¶q

= ¡ 1

10q2 + 70q

dR

dt= ¡1

5qdq

dt+ 70

dq

dt

= ¡15(20)(25) + 70(25)

= ¡100 + 1750= $1650

Revenue is increasing at a rate of $1650 per day.


15. V = k(R2 ¡ r2); k = 555:6; R = 0:02 mm,dRdt = 0:003 mm per minute; r is constant.

V = k(R2 ¡ r2)V = 555:6(R2 ¡ r2)dV

dt= 555:6

μ2R

dR

dt¡ 0¶

= 555:6(2)(0:02)(0:003)

= 0:067 mm/min

16. y = nxm

Note that n is a constant.

ln y = ln (nxm)

ln y = ln n+ ln xm

ln y = ln n+m ln x

Now take the derivative of both sides with respectto t:

1

y

dy

dt= 0 +m

1

x

dx

dt

1

y

dy

dt=m

1

x

dx

dt

17. b = 0:22m0:87

db

dt= 0:22(0:87)m¡0:13 dm

dt

= 0:1914m¡0:13 dmdt

dm

dt=m0:13

0:1914

db

dt

=250:13

0:1914(0:25)

¼ 1:9849

The rate of change of the total weight is about1.9849 g/day.

18. E = 429m¡0:35

dE

dt= 429(¡0:35)m¡1:35 dm

dt

= ¡150:15m¡1:35 dmdt

= ¡150:15(10)¡1:35(0:001)¼ ¡0:0067

The rate of change of the energy expenditure isabout ¡0:0067 cal/g/hr2:

19. r = 140:2m0:75

(a)dr

dt= 140:2(0:75)m¡0:25 dm

dt

= 105:15m¡0:25 dmdt

(b)dr

dt= 105:15(250)¡0:25(2)

¼ 52:89

The rate of change of the average daily metabolicrate is about 52.89 kcal/day2:

20. E = 26:5w¡0:34

dE

dt= 26:5(¡0:34)w¡1:34 dw

dt

= ¡9:01w¡1:34 dwdt

= ¡9:01(5)¡1:34(0:05)¼ ¡0:0521

The rate of change of the energy expenditure isabout ¡0:0521 kcal/kg/km/day.

21. C =1

10(T ¡ 60)2 + 100

dC

dt=1

5(T ¡ 60) dT

dt

If T = 76± anddT

dt= 8;

dC

dt=1

5(76¡ 60)(8) = 1

5(16)(8)

= 25:6:

The crime rate is rising at the rate of 25.6crimes/month.

22.W (t) =¡0:02t2 + tt+ 1

dW

dt=(¡0:04t+ 1)(t+ 1)¡ (1)(¡0:02t2 + t)

(t+ 1)2

If t = 5;

dW

dt=(¡0:2 + 1)(6)¡ (¡0:5 + 5)

62

=4:8¡ 4:536

= 0:008:


23. Let x = the distance of the base ofthe ladder from the base ofthe building;

y = the distance up the side ofthe building to the top ofthe ladder.

Find dydt when x = 8 ft and

dxdt = 9 ft/min:

Since y =p172 ¡ x2; when x = 8;

y = 15:

By the Pythagorean theorem,

x2 + y2 = 172:

d

dt(x2 + y2) =

d

dt(172)

2xdx

dt+ 2y

dy

dt= 0

2ydy

dt= ¡2x dx

dt

dy

dt=¡2x2y

¢ dxdt= ¡x

y¢ dxdt

= ¡ 8

15(9)

= ¡245

The ladder is sliding down the building at the rateof 245 ft/min.

24. (a) Let x = the distance one car travels west;y = the distance the other car travels

north;s = the distance between the two cars.

s2 = x2 + y2

2sds

dt= 2x

dx

dt+ 2y

dy

dt

sds

dt= x

dx

dt+ y

dy

dt

Use d = rt to …nd x and y:

x = (40)(2) = 80 miy = (30)(2) = 60 mis =

px2 + y2

=p(80)2 + (60)2

= 100

The distance between the cars after 2 hours is 100mi.dxdt = 40 mph and

dydt = 30 mph

(100)ds

dt= (80)(40) + (60)(30)

ds

dt=5000

100= 50

The distance between the two cars is changing atthe rate of 50 mph.

(b) From part (a), we have

sds

dt= x

dx

dt+ y

dy

dt

Use d = rt to …nd x and y. When the second carhas traveled 1 hour, the …rst car has traveled 2hours.

x = 40(1) = 40 miy = 30(2) = 60 mis =

px2 + y2

=p(40)2 + (60)2

= 72:11

The distance between the cars after the second carhas traveled 1 hour is about 72.11 mi.

72:11ds

dt= (40)(40) + (60)(30)

ds

dt=3400

72:11¼ 47:15

The distance between the two cars is changing ata rate of about 47.15 mph.


25. Let r = the radius of the circle formedby the ripple.

Find dAdt when r = 4 ft and

drdt = 2 ft/min:

A = ¼r2

dA

dt= 2¼r

dr

dt

= 2¼(4)(2)

= 16¼

The area is changing at the rate of 16¼ ft2/min.

26. V = 43¼r

3; r = 4 in, and drdt = ¡1

4 in/hr

dV

dt= 4¼r2

dr

dt

= 4¼(4)2μ¡14

¶

= ¡16¼ in3/hr

27. V = x3; x = 3 cm, and dVdt = 2 cm

3=min

dV

dt= 3x2

dx

dt

dx

dt=

1

3x2dV

dt

=1

3 ¢ 32 (2)

=2

27cm/min

28. Let r = the radius of the base of theconical pile.

Find dVdt when r = 6 in, and

drdt = 0:75 in/min.

h = 2r for all t.

V =¼

3r2h

V =¼

3r2(2r)

=2¼

3r3

dV

dt=3 ¢ 2¼r23

¢ drdt

dV

dt= 2¼(62)(0:75)

= 54¼

The volume is changing at the rate of 54¼ in3/min.

29. Let y = the length of the man’s shadow;x = the distance of the man from the

lamp post;h = the height of the lamp post.

dx

dt= 50 ft/min

Find dydt when x = 25 ft.

Now hx+y =

6y ; by similar triangles.

When x = 8; y = 10;

h

18=6

10

h = 10:8:

10:8

x+ y=6

y;

10:8y = 6x+ 6y

4:8y = 6x

y = 1:25x

dy

dt= 1:25

dx

dt

= 1:25(50)

dy

dt= 62:5

The length of the shadow is increasing at the rateof 62.5 ft/min.


30. Let x = one-half the width of thetriangular cross section;

h = the height of the water;V = the volume of the water.

dV

dt= 4 cu ft per min

Find dhdt when h = 4:

V =

0@ Area of

triangularcross section

1A ¢ (length)

Area of triangular cross section

=1

2(base)(height)

=1

2(2x)(h) = xh

By similar triangles,

6

2x=6

h;

so x =h

2:

V = (xh)(16)

=

μh

2h

¶16

= 8h2

dV

dt= 16h

dh

dt

1

16h

dV

dt=dh

dt

1

16(4)(4) =

dh

dt

dh

dt=1

16

The height of the water is increasing at a rate of116 ft/min.

31. Let x = the distance from the dockss = the length of the rope.

ds

dt= 1 ft/sec

s2 = x2 + (8)2

2sds

dt= 2x

dx

dt+ 0

sds

dt= x

dx

dt

If x = 8;

s =p(8)2 + (8)2 =

p128 = 8

p2:

Then,

8p2(1) = 8

dx

dt

dx

dt=p2 ¼ 1:41

The boat is approaching the deck atp2 ¼ 1:41

ft/sec.

32. Let x = the horizontal length;r = the rope length.

dx

dt= 50 ft/min

Section 6.6 Di¤erentials: Linear Approximation 427

By the Pythagorean theorem,

x2 + 1002 = 2002

x =p30,000 = 100

p3:

r2 = x2 + 1002

2rdr

dt= 2x

dx

dt+ 0

rdr

dt= x

dx

dt

dr

dt=x dxdtr

dr

dt=100p3(50)

200= 25

p3

¼ 43:3

She must let out the string at a rate of 25p3 ¼

43:3 ft/min.

6.6 Di¤erentials: LinearApproximation

1. y = 2x3 ¡ 5x; x = ¡2; ¢x = 0:1dy = (6x2 ¡ 5)dx¢y ¼ (6x2 ¡ 5)¢x ¼ [6(¡2)2 ¡ 5](0:1) ¼ 1:9

2. y = 4x3 ¡ 3x; x = 3; ¢x = 0:2dy = (12x2 ¡ 3)dx¢y ¼ (12x2 ¡ 3)¢x ¼ [12(3)2 ¡ 3](0:2) ¼ 21

3. y = x3 ¡ 2x2 + 3; x = 1; ¢x = ¡0:1dy = (3x2 ¡ 4x)dx¼ (3x2 ¡ 4x)¢x= [3(12)¡ 4(1)](¡0:1)= 0:1

4. y = 2x3 + x2 ¡ 4x; x = 2; ¢x = ¡0:2dy = (6x2 + 2x¡ 4)dx¼ (6x2 + 2x¡ 4)¢x= [6(2)2 + 2(2)¡ 4](¡0:2)= (24 + 4¡ 4)(¡0:2) = ¡4:8

5. y =p3x+ 2; x = 4; ¢x = 0:15

dy = 3

μ1

2(3x+ 2)¡1=2

¶dx

¢y ¼ 3

2p3x+ 2

¢x ¼ 3

2(3:74)(0:15) ¼ 0:060

6. y =p4x¡ 1; x = 5; ¢x = 0:08

dy =1

2(4x¡ 1)¡1=2(4)dx

= 2(4x¡ 1)¡1=2 dx¼ 2(4x¡ 1)¡1=2¢x= 2[4(5)¡ 1]¡1=2(0:08)= 2(19)¡1=2(0:08)

=2(0:08)

(19)1=2= 0:037

7. y =2x¡ 5x+ 1

; x = 2; ¢x = ¡0:03

dy =(x+ 1)(2)¡ (2x¡ 5)(1)

(x+ 1)2dx

=7

(x+ 1)2dx

=7

(x+ 1)2¢x

=7

(2 + 1)2(¡0:03)

= ¡0:023

8. y =6x¡ 32x+ 1

; x = 3; ¢x = ¡0:04

dy =6(2x+ 1)¡ 2(6x¡ 3)

(2x+ 1)2dx

=12

(2x+ 1)2dx

¼ 12

(2x+ 1)2¢x

=12

[2(3) + 1]2(¡0:04)

=¡0:4849

= ¡0:010

9.p145

We knowp144 = 12; so f(x) =

px; x = 144;

dx = 1:

dy

dx=1

2x¡1=2

dy =1

2pxdx

dy =1

2p144

(1) =1

24

p145 ¼ f(x) + dy = 12 + 1

24

¼ 12:0417


By calculator,p145 ¼ 12:0416:

The di¤erence is j12:0417¡ 12:0416j = 0:0001:

10.p23

We knowp25 = 5; f(x) =

px; x = 25; and

dx = ¡2:dy

dx=1

2x¡1=2

dy =1

2pxdx =

1

2p25(¡2) = ¡1

5= ¡0:2:

p23 ¼ f(x) + dy = 5¡ 0:2 = 4:8

By calculator,p23 ¼ 4:7958:


11.p0:99

We knowp1 = 1; so f(x) =

px; x = 1; dx =

¡0:01:

dy

dx=1

2x¡1=2

dy =1

2pxdx

dy =1

2p1(¡0:01) = ¡0:005

p0:99 ¼ f(x) + dy = 1¡ 0:005

= 0:995

By calculator,p0:99 ¼ 0:9950:

The di¤erence is j0:995¡ 0:9950j = 0:

12.p17:02

We knowp16 = 4; f(x) =

px; x = 16; and

dx = 1:02:

dy

dx=1

2x¡1=2

dy =1

2pxdx =

1

2p16(1:02)

=1

8(1:02) = 0:1275

p17:02 ¼ f(x) + dy = 4 + 0:1275 = 4:1275

By calculator,p17:02 ¼ 4:1255:


13. e0:01

We know e0 = 1; so f(x) = ex; x = 0; dx = 0:01:

dy

dx= ex

dy = ex dx

dy = e0(0:01) = 0:01

e0:01 ¼ f(x) + dy = 1 + 0:01 = 1:01

By calculator, e0:01 ¼ 1:0101:The di¤erence is j1:01¡ 1:0101j = 0:0001:

14. e¡0:002

We know e0 = 1; f(x) = ex; x = 0; anddx = ¡0:002:

dy

dx= ex

dy = ex dx = e0(¡0:002) = ¡0:002e¡0:002 ¼ f(x) + dy = 1¡ 0:002 = 0:998By calculator, e¡0:002 ¼ 0:9980:The di¤erence is j0:9980¡ 0:998j = 0:

15. ln 1:05

We know ln 1 = 0; so f(x) = ln x; x = 1; dx =0:05:

dy

dx=1

x

dy =1

xdx

dy =1

1(0:05) = 0:05

ln 1:05 ¼ f(x) + dy = 0 + 0:05 = 0:05

By calculator, ln 1:05 ¼ 0:0488:The di¤erence is j0:05¡ 0:0488j = 0:0012:

16. ln 0:98

We know ln 1 = 0; f(x) = ln x; x = 1; anddx = ¡0:02:

dy

dx=1

x

dy =1

xdx =

1

1(¡0:02) = ¡0:02

ln 0:98 ¼ f(x) + dy = 0¡ 0:02 = ¡0:02By calculator, ln 0:98 ¼ ¡0:0202:The di¤erence is j¡0:02¡ (¡0:0202)j = 0:0002:


17. Let D = the demand in thousands of pounds;x = the price in dollars.

D(q) = ¡3q3 ¡ 2q2 + 1500(a) q = 2; ¢q = 0:10

dD = (¡9q2 ¡ 4q)dq¢D ¼ (¡9q2 ¡ 4q)¢q

¼ [¡9(4)¡ 4(2)](0:10)¼ ¡4:4 thousand pounds

(b) q = 6; ¢q = 0:15

¢D ¼ [¡9(36)¡ 4(6)](0:15)¼ ¡52:2 thousand pounds

18. A(x) = 0:04x3 + 0:1x2 + 0:5x+ 6

(a) x = 3; ¢x = 1

dA = (0:12x2 + 0:2x+ 0:5)dx

= (0:12x2 + 0:2x+ 0:5)¢x

= [(0:12)(3)2 + (0:2)(3) + (0:5)](1)

= 2:18

(b) x = 5; ¢x = 1

dA = [(0:12)(5)2 + (0:2)(5) + (0:5)](1)

= 4:5

19. R(x) = 12,000 ln(0:01x+ 1)x = 100; ¢x = 1

dR =12,0000:01x+ 1

(0:01)dx

¢R ¼ 120

0:01x+ 1¢x

¼ 120

0:01(100) + 1(1)

¼ $60

20. P = R¡C= 12,000 ln(0:01x+ 1)¡ (150 + 75x)

x = 100;¢x = 1

dP =12,0000:01x+ 1

(0:01)dx¡ 75dx

¢P ¼ 12,0000:01x+ 1

(0:01)¢x¡ 75¢x

¼ 12,0000:01(100) + 1

(0:01)(1)¡ 75(1)

¼ 60¡ 75¼ ¡15

The change in pro…t is a loss of about $15:

21. If a cube is given a coating 0.1 in. thick, eachedge increases in length by twice that amount, or0.2 in. because there is a face at both ends of theedge.

V = x3; x = 4; ¢x = 0:2

dV = 3x2 dx

¢V ¼ 3x2¢x= 3(42)(0:2)

= 9:6

For 1000 cubes 9.6(1000) = 9600 in.3 of coatingshould be ordered.

22. Let x = the number of beach balls;V = the volume of x beach balls.

ThendV

dr¼ the volume of material inbeach balls since they arehollow.

V =4

3¼r3x

r = 6 in., x = 5000; ¢r = 0:03 in.

dV =4

3¼(3r2x+ r3)¢r

=4

3¼(3 ¢ 36 ¢ 5000 + 216)(0:03)

= 21; 608:64¼

21,608¼ in.3 of material would be needed.

23. (a) A(x) = y = 0:003631x3 ¡ 0:03746x2+ 0:1012x+ 0:009

Let x = 1; dx = 0:2:

dy

dx= 0:010893x2 ¡ 0:07492x+ 0:1012

dy = (0:010893x2 ¡ 0:07492x+ 0:1012)dx¢y ¼ (0:010893x2 ¡ 0:07492x+ 0:1012)¢x

¼ (0:010893 ¢ 12 ¡ 0:07492 ¢ 1 + 0:1012) ¢ 0:2¼ 0:007435

The alcohol concentration increases by about 0.74percent.

(b)¢y ¼ (0:010893 ¢ 32 ¡ 0:07492 ¢ 3 + 0:1012) ¢ 0:2¼ ¡0:005105

The alcohol concentration decreases by about 0.51percent.


24. C =5x

9 + x2

dC =5(9 + x2)¡ 2x(5x)

(9 + x2)2dx

=45 + 5x2 ¡ 10x2(9 + x2)2

dx

=45¡ 5x2(9 + x2)2

dx

¼ 45¡ 5x2(9 + x2)2

¢x

(a) x = 1; ¢x = 0:5

dC ¼ 45¡ 5(1)2(9 + 1)2

(0:5)

=40

100(0:5)

= 0:2

(b) x = 2; ¢x = 0:25

dC ¼ 45¡ 5(2)2(9 + 4)2

(0:25)

= 0:037

25. P (x) =25x

8 + x2

dP =(8 + x2)(25)¡ 25x(2x)

(8 + x2)2dx

=(8 + x2)(25)¡ 25x(2x)

(8 + x2)2¢x

(a) x = 2; ¢x = 0:5

dP =[(8 + 4)(25)¡ (25)(2)(4)](0:5)

(8 + 4)2

= 0:347 million

(b) x = 3; ¢x = 0:25

dP =[(8 + 9)(25)¡ 25(3)(6)]0:25

(8 + 9)2

¼ ¡0:022 million

26. A = ¼r2; r = 1:7 mm; ¢r = ¡0:1 mmdA = 2¼r dr

¢A ¼ 2¼r¢r= 2¼(1:7)(¡0:1)= ¡0:34¼ mm2

27. r changes from 14 mm to 16 mm, so ¢r = 2:

V =4

3¼r3

dV =4

3(3)¼r2 dr

¢V ¼ 4¼r2¢r= 4¼(14)2(2)

= 1568¼ mm3

28. A = ¼r2; r = 1:2 mi; ¢r = 0:2 mi

dA = 2¼r dr

¢A ¼ 2¼r¢r= 2¼(1:2)(0:2)

= 0:48¼ mi2

29. r increases from 20 mm to 22 mm, so ¢r = 2:

A = ¼r2

dA = 2¼r dr

¢A ¼ 2¼r¢r= 2¼(20)(2)

= 80¼ mm2

30. A(p) =1:181p

94:359¡ p(a) Since values for p must be non-negative andthe denominator can’t be zero, a sensible domainwould be from 0 to about 94.

(b) dA =(94:359¡ p)(1:181)¡ 1:181p(¡1)

(94:359¡ p)2 dp

=111:437979¡ 1:181p+ 1:181p

(94:359¡ p)2 dp

=111:437979

(94:359¡ p)2 dp

We are given p = 60 and dp = 65¡ 60 = 5:

dA ¼ 111:437979

(94:359¡ 60)2 (5) ¼ 0:472

It will take about 0.47 years.The actual value is

A(65)¡A(60) ¼ 2:615¡ 2:062 = 0:553

or about 0.55 years.


31. W (t) = ¡3:5 + 197:5e¡e¡0:01394(t¡108:4)

(a) dW = 197:5e¡e¡0:01394(t¡108:4)

(¡1)e¡0:01394(t¡108:4)(¡0:01394)dt= 2:75315e¡e

¡0:01394(t¡108:4)e¡0:01394(t¡108:4)dt

We are given t = 80 and dt = 90¡ 80 = 10:dW ¼ 9:258

The pig will gain about 9.3 kg.

(b) The actual weight gain is calculated as

W(90)¡W(80) ¼ 50:736¡ 41:202 = 9:534or about 9.5 kg.

32. V =4

3¼r3; r = 4 cm; ¢r = 0:2 cm

dV = 4¼r2 dr

¢V ¼ 4¼r2¢r= 4¼(4)2(0:2)

= 12:8¼ cm3

33. r = 3 cm, ¢r = ¡0:2 cmV =

4

3¼r3

dV = 4¼r2 dr

¢V ¼ 4¼r2¢r= 4¼(9)(¡0:2)= ¡7:2¼ cm3

34. V = x3; V = 27; x = 3;¢V = 0:1dV = 3x2dx

¢V ¼ 3x2¢x

¢x ¼ ¢V

3x2

¼ 0:1

3 ¢ 32¼ 0:0037 mm

35. V =1

3¼r2h; h = 13; dh = 0:2

V =1

3¼

μh

15

¶2h

=¼

775h3

dV =¼

775¢ 3h2dh

=¼

225h2dh

¢V ¼ ¼

225h2¢h

¼ ¼

225(132)(0:2)

¼ 0:472 cm3


36. A = x2; x = 3:45; ¢x = §0:002dA = 2xdx

¢A ¼ 2x¢x = 2(3:45)(§0:002) = §0:0138 in.2

37. A = x2; x = 4; dA = 0:01

dA = 2xdx

¢A ¼ 2x¢x

¢x ¼ ¢A

2x¼ 0:01

2(4)¼ 0:00125 cm

38. r = 4:87 in., ¢r = §0:040

A = ¼r2

dA = 2¼r dr

¢A ¼ 2¼r¢r = 2¼(4:87)(§0:040) = §1:224 in.2

39. V =4

3¼r3; r = 5:81; ¢r = §0:003

dV =4

3¼(3r2)dr

¢V ¼ 4

3¼(3r2)¢r

= 4¼(5:81)2(§0:003)= §0:405¼ ¼ §1:273 in.3

40. V = x3; x = 5; dV = 0:3

dV = 3x2dx

¢V ¼ 3x2¢x

¢x ¼ ¢V

3x2

¼ 0:3

3(52)

¼ 0:004 ft

41. h = 7:284 in., r = 1:09§ 0:007 in.

V =1

3¼r2h

dV =2

3¼rhdr

¢V ¼ 2

3¼rh¢r

=2

3¼(1:09)(7:284)(0:007)

= §0:116 in.3

Chapter 6 Review Exercises1. f(x) = ¡x3 + 6x2 + 1; [¡1; 6]

f 0(x) = ¡3x2 + 12x = 0 when x = 0; 4:f(¡1) = 8f(0) = 1

f(4) = 33

f(6) = 1

Absolute maximum of 33 at 4; absolute minimumof 1 at 0 and 6.

2. f(x) = 4x3 ¡ 9x2 ¡ 3; [¡1; 2]f 0(x) = 12x2 ¡ 18x = 0 when x = 0; 32 :

f(¡1) = ¡16f(0) = ¡3

f

μ3

2

¶= ¡9:75

f(2) = ¡7Absolute maximum of ¡3 at 0; absolute minimumof ¡16 at ¡1:

3. f(x) = x3 + 2x2 ¡ 15x+ 3; [¡4; 2]f 0(x) = 3x2 + 4x¡ 15 = 0 when

(3x¡5)(x+3) = 0

x =5

3or x = ¡3:f(¡4) = 31f(¡3) = 39

f

μ5

3

¶= ¡319

27

f(2) = ¡11Absolute maximum of 39 at ¡3; absolute mini-mum of ¡319

27 at53

4. f(x) = ¡2x3 ¡ 2x2 + 2x¡ 1; [¡3; 1]f 0(x) = ¡6x2 ¡ 4x+ 2f 0(x) = 0 when

3x2 + 2x¡ 1 = 0(3x¡ 1)(x+ 1) = 0

x =1

3or x = ¡1:

f(¡3) = 29f(¡1) = ¡3

f

μ1

3

¶= ¡17

27

f(1) = ¡3

Chapter 6 Review Exercises 433

Absolute maximum of 29 at ¡3; absolute mini-mum of ¡3 at ¡1 and 1.

7. (a) f(x) =2 lnx

x2; [1; 4]

f 0(x) =x2¡2x

¢¡ (2 lnx)(2x)x4

=2x¡ 4x lnx

x4

=2¡ 4 lnxx3

f 0(x) = 0 when

2¡ 4 lnx = 02 = 4 lnx

0:5 = lnx

e0:5 = x

x ¼ 1:6487:x f(x)

1 0

e0:5 0:367884 0:17329

Maximum is 0.37; minimum is 0.

(b) [2; 5]

Note that the critical number of f is not in thedomain, so we only test the endpoints.

x f(x)

2 0:346575 0:12876

Maximum is 0.35, minimum is 0.13.

10. x2y3 + 4xy = 2 d

dx(x2y3 + 4xy) =

d

dx(2)

2xy3 + 3y2μdy

dx

¶x2 + 4y + 4x

dy

dx= 0

(3x2y2 + 4x)dy

dx= ¡2xy3 ¡ 4y

dy

dx=¡2xy2 ¡ 4y3x2y2 + 4x

11. x2 ¡ 4y2 = 3x3y4d

dx(x2 ¡ 4y2) = d

dx(3x3y4)

2x¡ 8y dydx= 9x2y4 + 3x3 ¢ 4y3 dy

dx

(¡8y ¡ 3x3 ¢ 4y3) dydx= 9x2y4 ¡ 2x

dy

dx=2x¡ 9x2y48y + 12x3y3

12. 9px+ 4y3 = 2

py

d

dx

¡9px+ 4y3

¢= 2 ¢ d

dx

¡y1=2

¢9

2x¡1=2 + 12y2

dy

dx= 2 ¢ 1

2y¡1=2 ¢ dy

dx

9

2x1=2=

μ1

y1=2¡ 12y2

¶dy

dx

9

2x1=2=1¡ 12y5=2y1=2

dy

dx

dy

dx=

9y1=2

2x1=2(1¡ 12y5=2)

=9py

2px(1¡ 12y5=2)

13. 2py ¡ 1 = 9x2=3 + y

d

dx[2(y ¡ 1)1=2] = d

dx(9x2=3 + y)

2 ¢ 12¢ (y ¡ 1)¡1=2 dy

dx= 6x¡1=3 +

dy

dx

[(y ¡ 1)¡1=2 ¡ 1]dydx= 6x¡1=3

1¡py ¡ 1py ¡ 1 ¢ dy

dx=

6

x1=3

dy

dx=

6py ¡ 1

x1=3(1¡py ¡ 1)

14.x+ 2y

x¡ 3y = y1=2

x+ 2y = y1=2(x¡ 3y)d

dx(x+ 2y) =

d

dx[y1=2(x¡ 3y)]

1 + 2dy

dx= y1=2

μ1¡ 3 dy

dx

¶

+1

2(x¡ 3y)y¡1=2 dy

dx

1 + 2dy

dx= y1=2 ¡ 3y1=2 dy

dx+1

2xy¡1=2

dy

dx

¡ 32y1=2

dy

dx

(2 + 3y1=2 ¡ 12xy¡1=2 +

3

2y1=2)

dy

dx= y1=2 ¡ 1

2y1=2¡2 + 9

2y1=2 ¡ 1

2xy¡1=2¢

2y1=2dy

dx= y1=2 ¡ 1

μ4y1=2 + 9y ¡ x

2y1=2

¶dy

dx= y1=2 ¡ 1

dy

dx=

2y ¡ 2y1=24y1=2 + 9y ¡ x


15.6 + 5x

2¡ 3y =1

5x

5x(6 + 5x) = 2¡ 3y30x+ 25x2 = 2¡ 3y

d

dx(30x+ 25x2) =

d

dx(2¡ 3y)

30 + 50x = ¡3 dydx

¡30 + 50x3

=dy

dx

16. ln(x+ y) = 1 + x2 + y3

d

dx[ln(x+ y)] =

d

dx(1 + x2 + y3)

1

x+ y¢ ddx(x+ y) = 2x+ 3y2 ¢ dy

dx

1

x+ y

μ1 +

dy

dx

¶= 2x+ 3y2 ¢ dy

dxμ1

x+ y¡ 3y2

¶dy

dx= 2x¡ 1

x+ y

dy

dx=2x¡ 1

x+y1

x+y ¡ 3y2

=2x(x+ y)¡ 11¡ 3y2(x+ y)

=2x2 + 2xy ¡ 11¡ 3xy2 ¡ 3y3

=1¡ 2x2 ¡ 2xy3xy2 + 3y3 ¡ 1

17. ln(xy + 1) = 2xy3 + 4

d

dx[ln(xy + 1)] =

d

dx(2xy3 + 4)

1

xy + 1¢ ddx(xy + 1) = 2y3+2x ¢ 3y2 dy

dx+d

dx(4)

1

xy + 1

μy + x

dy

dx+d

dx(1)

¶= 2y3 + 6xy2

dy

dx

y

xy + 1+

x

xy + 1¢ dydx= 2y3 + 6xy2

dy

dxμx

xy + 1¡ 6xy2

¶dy

dx= 2y3 ¡ y

xy + 1

dy

dx=2y3 ¡ y

xy+1x

xy+1 ¡ 6xy2

=2y3(xy + 1)¡ yx¡ 6xy2(xy + 1)

=2xy4 + 2y3 ¡ yx¡ 6x2y3 ¡ 6xy2

18.p2x¡ 4yx = ¡22; tangent line at (2; 3)

d

dx(p2x¡ 4yx) = d

dx(¡22)

1

2(2x)¡1=2(2)¡ 4y ¡ 4x dy

dx= 0

(2x)¡1=2 ¡ 4y = 4x dydx

dy

dx=(2x)¡1=2 ¡ 4y

4x=

1p2x¡ 4y4x

At (2; 3);

m =

1p2¢2 ¡ 4 ¢ 34 ¢ 2 =

12 ¡ 128

= ¡2316:

The equation of the tangent line is

y ¡ y1 = m(x¡ x1)

y ¡ 3 = ¡2316(x¡ 2)

23x+ 16y = 94:

21. y = 8x3 ¡ 7x2; dxdt = 4; x = 2dy

dt=d

dt(8x3 ¡ 7x2)

= 24x2dx

dt¡ 14x dx

dt

= 24(2)2(4)¡ 14(2)(4)= 272

22. y =9¡ 4x3 + 2x

;dx

dt= ¡1; x = ¡3

dy

dt=(¡4)(3 + 2x)¡ (2)(9¡ 4x)

(3 + 2x)2dx

dt

=¡30

(3 + 2x)2dx

dt

=¡30

[3 + 2(¡3)]2 (¡1) =30

9=10

3

23. y =1 +

px

1¡px;dx

dt= ¡4; x = 4

dy

dt=d

dt

·1 +

px

1¡px¸

=(1¡px) ¡12x¡1=2dxdt¢¡(1+px) ¡¡1

2

¢ ¡x¡1=2dxdt

¢(1¡px)2

=(1¡ 2) ¡ 1

2¢2¢(¡4)¡ (1 + 2) ¡¡12¢2¢ (¡4)(1¡ 2)2

=1¡ 31

= ¡2


24.x2 + 5y

x¡ 2y = 2;dx

dt= 1; x = 2; y = 0

x2 + 5y = 2(x¡ 2y)x2 + 5y = 2x¡ 4y

9y = ¡x2 + 2x

y =1

9(¡x2 + 2x)

= ¡19x2 +

2

9x

dy

dt=

μ¡29x+

2

9

¶dx

dt

=

·μ¡29

¶(2) +

2

9

¸(1)

= ¡49+2

9= ¡2

9

25. y = xe3x;dx

dt= ¡2; x = 1

dy

dt=d

dt(xe3x)

=dx

dt¢ e3x + x ¢ d

dt(e3x)

=dx

dt¢ e3x + xe3x ¢ 3dx

dt

= (1 + 3x)e3xdx

dt

= (1 + 3 ¢ 1)e3(1)(¡2) = ¡8e3

26. y =1

ex2 + 1;dx

dt= 3; x = 1

dy

dt=d

dt

μ1

ex2 + 1

¶

=¡1

(ex2 + 1)2¢ ddt(ex

2

+ 1)

=¡1

(ex2 + 1)2

·ex

2 ¢ ddt(x2) +

d

dt(1)

¸

=¡1

(ex2 + 1)2

·ex

2 ¢ 2x dxdt

¸

=¡1

(e+ 1)2[e ¢ 2 ¢ 1 ¢ 3]

=¡6e

(e+ 1)2

28. y = 8¡ x2 + x3; x = ¡1; ¢x = 0:02dy = (¡2x+ 3x2)dx¼ (¡2x+ 3x2)¢x= [¡2(¡1) + 3(¡1)2](0:02)= 0:1

29. y =3x¡ 72x+ 1

; x = 2; ¢x = 0:003

dy =(3)(2x+ 1)¡ (2)(3x¡ 7)

(2x+ 1)2dx

dy =17

(2x+ 1)2dx

¼ 17

(2x+ 1)2¢x

=17

(2[2] + 1)2(0:003)

= 0:00204

30. ¡12x+ x3 + y + y2 = 4dy

dx(¡12x+ x3 + y + y2) = d

dx(4)

¡12 + 3x2 + dy

dx+ 2y

dy

dx= 0

(1 + 2y)dy

dx= 12¡ 3x2

dy

dx=12¡ 3x21 + 2y

(a) Ifdy

dx= 0;

12¡ 3x2 = 012 = 3x2

§2 = x:

x = 2:

¡ 24 + 8 + y + y2 = 4y + y2 = 20

y2 + y ¡ 20 = 0(y + 5)(y ¡ 4) = 0y = ¡5 or y = 4

(2;¡5) and (2; 4) are critical points.x = ¡2 :

24¡ 8 + y + y2 = 4y + y2 = ¡12

y2 + y + 12 = 0

y =¡1§p12 ¡ 48

2

This leads to imaginary roots.x = ¡2 does not produce critical points.


(b) x y1 y21:9 ¡4:99 3:99

2 ¡5 4

2:1 ¡4:99 3:99

The point (2;¡5) is a relative minimum.The point (2; 4) is a relative maximum.

(c) There is no absolute maximum or minimumfor x or y:

32. (a) P (x) = ¡x3 + 10x2 ¡ 12xP 0(x) = ¡3x2 + 20x¡ 12 = 0

3x2 ¡ 20x+ 12 = 0(3x¡ 2)(x¡ 6) = 0

3x¡ 2 = 0 or x¡ 6 = 0x =

2

3or x = 6

P 00(x) = ¡6x+ 20

P 00μ2

3

¶= 16;

which implies that x = 23 is location of minimum.

P 00(6) = ¡16;

which implies that x = 6 is location of maximum.Thus, 600 boxes will produce a maximum pro…t.

(b) Maximum pro…t = P (6)= ¡(6)3 + 10(6)2 ¡ 12(6)= 72

The maximum pro…t is $720.

33. Let x = the length and width of a sideof the base;

h = the height.

The volume is 32 m3; the base is square and thereis no top. Find the height, length, and width forminimum surface area.

Volume = x2hx2h = 32

h =32

x2

Surface area = x2 + 4xh

A = x2 + 4x

μ32

x2

¶

= x2 + 128x¡1

A0 = 2x¡ 128x¡2

If A0 = 0;

2x3 ¡ 128x2

= 0

x3 = 64

x = 4:

A00(x) = 2 + 2(128)x¡3

A00(4) = 6 > 0

The minimum is at x = 4; where

h =32

42= 2:

The dimensions are 2 m by 4 m by 4 m.

34. V = ¼r2h = 40; so h =40

¼r2:

A = 2¼r2 + 2¼rh

= 2¼r2 + 2¼r

μ40

¼r2

¶

= 2¼r2 +80

r

Cost = C(r) = 4(2¼r2) + 3μ80

r

¶

= 8¼r2 +240

r

C0(r) = 16¼r ¡ 240r2

16¼r ¡ 240r2

= 0

16¼r3 = 240

r3 =15

¼

r ¼ 1:684

C00(r) = 16¼ +480

r3> 0; so r = 1:684 minimizes

cost.

h =40

¼r2=

40

¼(1:684)2= 4:490

The radius should be 1.684 in. and the height shouldbe 4.490 in.


35. Volume of cylinder = ¼r2hSurface area of cylinder openat one end = 2¼rh+ ¼r2:

V = ¼r2h = 27¼

h =27¼

¼r2=27

r2

A = 2¼r

μ27

r2

¶+ ¼r2

= 54¼r¡1 + ¼r2

A0 = ¡54¼r¡2 + 2¼r

If A0 = 0;

2¼r =54¼

r2

r3 = 27

r = 3:

If r = 3;

A00 = 108¼r¡3 + 2¼ > 0;

so the value at r = 3 is a minimum.

For the minimum cost, the radius of the bottomshould be 3 inches.

36. M = 180,000 cases sold per yeark = $12, cost to store 1 case for 1 yrf = $20, …xed cost for orderx = the number of cases per order

x =

r2fM

k

=

r2(20)(180,000)

12

=p600,000

= 100p60

¼ 775

The store should order 775 cases each time.

37. Here k = 0:15;M = 20,000, and f = 12. We have

q =

r2fM

k=

r2(12)20,000

0:15

=p3,200,000 ¼ 1789

Ordering 1789 rolls each time minimizes annualcost.

38. Use equation (3) from Section 6.3 with k = 2,M =240,000, and f = 15:

q =

r2fM

k=

r2(15)(240,000)

2

=p3,600,000 ¼ 1897:4

T (1897) ¼ 3794:7333 and T (1898) ¼ 3794:7334.Since T (1897) < T (1898), then the number ofbatches per year should be

M

q=240,0001897

¼ 127.

39. Use equation (3) from Section 6.3 with k = 1,M = 128,000, and f = 10.

q =

r2fM

k=

r2(10)(128,000)

1

=p2,560,000 = 1600

The number of lots that should be produced an-nually is

M

q=128,0001600

= 80:

40. q =A

pk

dq

dp= ¡k A

pk+1

E = ¡pq¢ dqdp

=

Ã¡ p

Apk

!(¡k)

μA

pk+1

¶

=

μ¡p

k+1

A

¶μ¡k A

pk+1

¶

= k

The demand is elastic when k > 1 and inelasticwhen k < 1:

41. A = ¼r2; drdt = 4 ft/min; r = 7 ft

dA

dt= 2¼r

dr

dt

dA

dt= 2¼(7)(4)

dA

dt= 56¼

The rate of change of the area is 56¼ ft2/min.


42.dx

dt= rx(N ¡ x)

= rxN ¡ rx2d2x

dt2= rN

dx

dt¡ 2rx dx

dt

= rdx

dt(N ¡ 2x)

= r[rx(N ¡ x)](N ¡ 2x)= r2x(N ¡ x)(N ¡ 2x)

d2x

dt2= 0 when x = 0; x = N; or x = N

2 :

On¡0; N2

¢; d2xdt2 > 0; therefore, the curve is con-

cave upward.On

¡N2 ;N

¢; d2xdt2 < 0; therefore, the curve is con-

cave downward.Hence x = N

2 is a point of in‡ection.

43. (a)

(b) We use a graphing calculator to graph

M 0(t) = ¡0:4321173 + 0:1129024t¡ 0:0061518t2+ 0:0001260t3 ¡ 0:0000008925t4

on [3; 51] by [0; 0:3]: We …nd the maximum valueofM 0(t) on this graph at about 15.41, or on aboutthe 15th day.

44. (a)

(b) To …nd where the maximum and minimumnumbers occur, use a graphing calculator to locateany extreme points on the graph. One criticalnumber is formed at about 87.78.

t P (t)

0 237.0987.78 43.5695 48.66

The maximum number of polygons is about 237at birth. The minimum number is about 44.

45. Let x = the distance from the base ofthe ladder to the building;

y = the height on the building atthe top of the ladder.

dy

dt= ¡2

502 = x2 + y2

0 = 2xdx

dt+ 2y

dy

dt

dx

dt= ¡y

x

dy

dt

When x = 30; y =p2500¡ (30)2 = 40:

Sodx

dt=¡4030(¡2) = 80

30=8

3

The base of the ladder is slipping away from thebuilding at a rate of 83 ft/min:

46.dV

dt= 0:9 ft3=min

Find drdt when r = 1:7 ft

V =4

3¼r3

dV

dt=4

3¼(3)r2

dr

dt

= 4¼r2dr

dt

0:9 = 4¼(1:72)dr

dt

0:9

4¼(1:72)=dr

dt

dr

dt=

0:9

11:56¼¼ 0:0248

The radius is changing at the rate of 0:911:56¼ ¼

0:0248 ft/min.


47. Let x = one-half the width of thetriangular cross section;

h = the height of the water;V = the volume of the water.dV

dt= 3.5 ft3/min

Find dVdt when h =

1

3:

V =

0@ Area oftriangularside

1A (length)

Area of triangular cross section

=1

2(base)(altitude)

=1

2(2x)(h) = xh

By similar triangles, 2xh =21 ; so x = h:

V = (xh)(4)

= h2 ¢ 4= 4h2

dV

dt= 8h

dh

dt

1

8h¢ dVdt=dh

dt

1

8¡13

¢(3:5) = dh

dt

dh

dt=21

16= 1:3125

The depth of water is changing at the rate of1.3125 ft/min.

48. V =4

3¼r3; r = 4 in., ¢r = 0:02 in.

dV = 4¼r2 dr

¢V ¼ 4¼r2¢r= 4¼(4)2(0:02)

= 1:28¼ or about 4.021

The volume of the coating is 1:28¼ in.3 or about4.021 in.3:

49. A = s2; s = 9:2; ¢s = §0:04ds = 2sds

¢A ¼ 2s¢s= 2(9:2)(§0:04)= §0:736 in.2

50. V = l ¢w ¢ hw = 4 + h

l + g = 130; g = 2(w + h)

l + 2(w + h) = 130

l + 2w + 2h = 130

l = 130¡ 2w ¡ 2h= 130¡ 2(4 + h)¡ 2h= 122¡ 4h

V = l ¢w ¢ h= (122¡ 4h)(4 + h)h= 488h+ 106h2 ¡ 4h3

dV

dh= 488 + 212h¡ 12h2

SetdV

dh= 0:

488 + 212h¡ 12h2 = 012h2 ¡ 212h¡ 488 = 03h2 ¡ 53h¡ 122 = 0

h =53§p2809 + 1464

6

¼ ¡2:06 or h = 19:73

h can’t be negative, so

h ¼ 19:73:

Thus,

l = 122¡ 4h¼ 43:1:

The length that produces the maximum volume isabout 43.1 inches.


51. We need to minimize y. Note that x > 0:

dy

dx=x

8¡ 2

x

Set the derivative equal to 0.

x

8¡ 2

x= 0

x

8=2

x

x2 = 16

x = 4

Since limx!0

y = 1, limx!1 y = 1, and x = 4 is

the only critical value in (0;1), x = 4 produces aminimum value.

y =42

16¡ 2 ln 4 + 1

4+ 2 ln 6

= 1:25 + 2(ln 6¡ ln 4)= 1:25 + 2 ln 1:5

The y coordinate of the Southern most point ofthe second boat’s path is 1:25 + 2 ln 1:5.

52. Let x = width of play area;y = length of play area.

An equation describing the amount of fencing is

900 = 2x+ y

y = 900¡ 2x:

Then,A = xy

A(x) = x(900¡ 2x)= 900x¡ 2x2:

If A0(x) = 900¡ 4x = 0;x = 225:

Then y = 900¡ 2(225) = 450:A00(x) = ¡4 < 0; so the area is maximized if thedimensions are 225 m by 450 m.

53. Distance on shore: 40¡ x feetSpeed on shore: 5 feet per secondDistance in water:

px2 + 402 feet

Speed in water: 3 feet per second

The total travel time t is t = t1 + t2 =d1v1+d2v2:

t(x) =40¡ x5

+

px2 + 402

3

= 8¡ x5+

px2 + 1600

3

t0(x) = ¡15+1

3¢ 12(x2 + 1600)¡1=2(2x)

= ¡15+

x

3px2 + 1600

Minimize the travel time t(x). If t0(x) = 0:

x

3px2 + 1600

=1

5

5x = 3px2 + 1600

5x

3=px2 + 1600

25

9x2 = x2 + 1600

16

9x2 = 1600

x2 =1600 ¢ 916

x =40 ¢ 34

= 30


To minimize the time, he should walk 40¡ x =40¡30 = 10 ft along the shore before paddling to-ward the desired destination. The minimum traveltime is

40¡ 305

+

p302 + 402

3¼ 18:67 seconds.

Extended Application: A Total Cost Model for a Training Program 441

54. Distance on shore: 25¡ x feetSpeed on shore: 5 feet per secondDistance in water:

px2 + 402 feet

Speed in water: 3 feet per second

The total travel time t is t = t1 + t2 =d1v1+d2v2:

t(x) =25¡ x5

+

px2 + 402

3

= 5¡ x5+

px2 + 1600

3

t0(x) = ¡15+1

3¢ 12(x2 + 1600)¡1=2(2x)

= ¡15+

x

3px2 + 1600

Minimize the travel time t(x). If t0(x) = 0:

x

3px2 + 1600

=1

5

5x = 3px2 + 1600

5x

3=px2 + 1600

25

9x2 = x2 + 1600

16

9x2 = 1600

x2 =1600 ¢ 916

x =40 ¢ 34

= 30


x = 30 is impossible since the closest point on theshore to the desired destination is only 25 ft fromwhere he is standing.Check the endpoints.

x t(x)

0 18.3325 15.72

The time is mininized when x = 25:

25 ¡ x = 25 ¡ 25 = 0 ft, so the mathematicianshould start paddling where he is standing.

Extended Application: A Total CostModel for a Training Program

1. Z(m) =C1m+DtC2 +DC3

μm¡ 12

¶

Z0(m) = ¡C1m2

+ 0+DC32

= ¡C1m2

+DC32

2. Z0(m) = 0 when

DC32

=C1m2

m2 =2C1DC3

m =

r2C1DC3

:

3. D = 3; t = 12; C1 = 15,000; C3 = 900

m =

s2(15,000)3(900)

=

r30,0002700

=

r100

9

¼ 3:334. 3 < 3:33 < 4m+ = 4 and m¡ = 3

5. Z(m) =C1 +mDtC2

m+DC3

hm(m¡1)

2

im

=C1m+DtC2 +DC3

μm¡ 12

¶

C1 = 15,000; D = 3; t = 12;C2 = 100; C3 = 900

Z(m+) = Z(4)

=15,0004

+ 3(12)(100) + 3(900)

μ4¡ 12

¶

= 3750 + 3600 + 4050

= $11,400

Z(m¡) = Z(3)

=15,0003

+ 3600 + 3(900)(3¡ 1)

2

= 5000 + 3600 + 2700

= $11,300


6. Since Z(3) < Z(4); the optimal time interval is 3months.

N = mD

= 3 ¢ 3= 9

There should be 9 trainees per batch.

applications of the derivative - grosse pointe...

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