Upload
vuongnhi
View
215
Download
1
Embed Size (px)
Citation preview
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]
Find critical numbers:
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]
Find critical numbers:
f 0(x) = x2 + 3x¡ 4 = 0(x+ 4)(x¡ 1) = 0x = ¡4 or x = 1
x f(x)
¡4 59
3¼ 19:67 Absolute maximum
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
4¼ 0:35 Absolute maximum
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.
Section 6.1 Absolute Extrema 379
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:
380 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
Section 6.1 Absolute Extrema 381
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
382 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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.
Section 6.1 Absolute Extrema 383
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.
384 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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:
(c) Since y = 90¡x and x and y are nonnegativenumbers, the domain of P is [0; 90]:
(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:
(c) Since y = 105¡x and x and y are nonnegativenumbers, the domain of P is [0; 105]:
(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.
386 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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:
Section 6.2 Applications of Extrema 387
(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.
388 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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.
Section 6.2 Applications of Extrema 389
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
390 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
Section 6.2 Applications of Extrema 391
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
392 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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:
Section 6.2 Applications of Extrema 393
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
from Example 4.If r = 5:454;
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
394 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
from Example 4.If r = 5:242;
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%
Section 6.2 Applications of Extrema 395
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)
396 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
Section 6.2 Applications of Extrema 397
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
398 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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.
Section 6.2 Applications of Extrema 399
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.
400 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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.
402 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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.
404 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
406 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
Section 6.4 Implicit Di¤erentiation 407
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
408 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
Section 6.4 Implicit Di¤erentiation 409
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
410 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
Section 6.4 Implicit Di¤erentiation 411
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
412 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
Section 6.4 Implicit Di¤erentiation 413
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
The point is (2; 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
The point is (1; 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
414 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
The point is (8; 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
Section 6.4 Implicit Di¤erentiation 415
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
416 Chapter 6 APPLICATIONS OF THE DERIVATIVE
(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
Section 6.4 Implicit Di¤erentiation 417
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
418 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
Section 6.4 Implicit Di¤erentiation 419
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
420 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
422 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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.
Section 6.5 Related Rates 423
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:
424 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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.
Section 6.5 Related Rates 425
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.
426 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
428 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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:
The di¤erence is j4:8¡ 4:7958j = 0:0042:
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:
The di¤erence is j4:1275¡ 4:1255j = 0:0020:
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:
Section 6.6 Di¤erentials: Linear Approximation 429
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.
430 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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.
Section 6.6 Di¤erentials: Linear Approximation 431
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
432 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
434 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
Chapter 6 Review Exercises 435
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.
436 Chapter 6 APPLICATIONS OF THE DERIVATIVE
(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.
Chapter 6 Review Exercises 437
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
=
á 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.
438 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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.
Chapter 6 Review Exercises 439
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.
440 Chapter 6 APPLICATIONS OF THE DERIVATIVE
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
(Discard the negative solution.)
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
(Discard the negative solution.)
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
442 Chapter 6 APPLICATIONS OF THE DERIVATIVE
6. Since Z(3) < Z(4); the optimal time interval is 3months.
N = mD
= 3 ¢ 3= 9
There should be 9 trainees per batch.