EXERCISE 11-4 11-61
78. f(x) =
32 2
2
7 18 ( 2)( 9) ( 2) 2
() , 9.
(1)(9) 1 1
89
xx xxx x xx xx
fx x
xx x x
xx
Step 1. Analyze f(x):
(A) Domain: All real numbers except x = –9, x = 1.
(B) Intercepts: y-intercept f(0) =0
x-intercepts: x = 0, x = 2.
(C) Asymptotes:
Horizontal asymptote
y
Step 2. Analyze f ꞌ(x):
22
x
x
Step 3. Analyze f“(x):
f“(x) = 3
2
(1)x
for x ≠ –9, x ≠ 1.
Step 4. Sketch the graph of f:
80. (A) P(x) = R(x) – C(x) = 1,296x – 0.12x3 – (830 + 396x) = 900x – 0.12x3 – 830, 0 ≤ x ≤ 80
Step 2. Analyze P‘(x):
11-62 CHAPTER 11: GRAPHING AND OPTIMIZATION
Sign chart for P‘:
Increasing
x
P(x)
50
Decreasing
0
1 899.64( )
51 36.36( )
x
Step 3. Analyze P“(x):
Step 4. Sketch the graph of P:
82. N(t) = 100
9
t
t, t ≥ 0
(A) N‘(t) = 2
100( 9) 100
(9)
tt
t
= 2
900
(9)t
(B) From (A), N‘(t) = 900(t + 9)–2. Thus,
(C) lim
t N(t) = 100, thus y = 100 is a horizontal asymptote.
(D) y-intercept: N(0) = 0
(E) The graph is:
84. L(x) = 2x + 20,000
x
, x > 0
(A) Step 1. Analyze L(x):
(a) Domain: x > 0 or (0, ∞)
EXERCISE 11-4 11-63
(c) Asymptotes: For very large x, L(x) ≈ 2x. Thus, y = 2x is an oblique asymptote. As x → 0, L(x)
Step 2. Analyze L‘(x):
L‘(X) = 2 – 2
20,000
x =
2
2
220,000x
x
Critical value: x = 100
Sign chart for L‘(x):
x
Step 3. Analyze L“(x):
L“(x) = 3
40,000
Step 4. Sketch the graph of L:
(B)100′ by 200′, with the longest side parallel to the building.
86. C(x) = 500 + 2x + 0.2x2, 0 < x < ∞; ‘( ) 2 0.4 .Cx x
(A) The average cost function is: C(x) = 500
x
+ 2 + 0.2x.
x
x
– – – 0 + + +
x
C‘(x)
Test Numbers
‘( )
x
Cx
11-64 CHAPTER 11: GRAPHING AND OPTIMIZATION
minimum at x = 50.
88. (B) 0.5
0.9
() 0.9 t
t
Dt te
e
0.5(12) 6
0.9(12) 10.8
90. S(w) = 26 0.06w
w
, w ≥ 5
Step 1. Analyze S(w):
(A) Domain: w ≥ 5, i.e., [5, ∞)
(B) Intercepts: None on [5, ∞)
Step 2. Analyze S‘(w):
S(w) = 26
Step 3. Analyze S“(w):
52
Step 4. Sketch the graph of S:
EXERCISE 11-5 11-65
EXERCISE 11–5
2. Max () (4) 4;Min () (0) 0gx g gx g .
6.Max ( ) (216) 6; Min ( ) ( 125) 5.px p px p
10. Interval [2, 8]; absolute minimum: f(7) = 5;
12. Interval [2, 10]; absolute minimum: f(7) = 5;
14. Interval [0, 9]; absolute minimum: f(0) = 0;
16. Interval [0, 2]; absolute minimum: f(0) = 0;
18. Interval [5, 8]; absolute minimum: f(7) = 5
20. () 8 .
f
xx
(A) On [0,1]: Max ( ) (0) 8; Min ( ) (1) 7.fx f fx f
22. 2
( ) 100 .
f
xx
(A) On [ 10,10], Max ( ) (0) 100; Min ( ) ( 10) (10) 0.fx f fx f f
24.
ln , 1, 2fx xI
1
fx
, so there are no critical values on I.
26.
267, 0,10fx x x I
11-66 CHAPTER 11: GRAPHING AND OPTIMIZATION
28. f(x) = x2 + 2x + 1, I = (–∞, ∞)
30. f(x) = –x2 + 6x + 1, I = (–∞, ∞)
32. f(x) = 6 – x3 , I = (–∞, ∞)
34. f(x) = 8 – x4, I = (–∞, ∞)
f ꞌ(x) = – 4x3
36. f(x) = x + 9
x
, I = (–∞, 0) (0, ∞)
2
99(3)(3)
38. f(x) = 2
2
3x, I = (–∞, ∞)
x
x = 0 is the only critical value on I and f(0) = 2/3.
(3) (3) (3)
xxx
EXERCISE 11-5 11-67
40. f(x) = 2
111
,1,(,1)(1,1)(1,).
(1)(1) 1
1
xx xI
xx x
x
,
42. f(x) =
2
21
x
x, I = (–∞, ∞)
22
x
x
44. f(x) = 6x – x2 + 4, I = [0, ∞)
46. f(x) = x3 – 6x2, I = [0, ∞)
48. f(x) = (2 – x)(x + 1)2, I = [0, ∞)
50. f(x) = 4x3 – 8x4, I = (0, ∞)
f ꞌ(x) = 12x2 – 32x3 = 4x2(3 – 8x) [Note x = 0 is not a critical value since dom (f) = (0,∞).]
52. f(x) = 4 + x + 9
x
, x > 0
x
11-68 CHAPTER 11: GRAPHING AND OPTIMIZATION
54. f(x) = 20 – 4x –2
250
x
, x > 0
500
x
3
4500x
=
3
4( 125)x
x
56. f(x) = 2x – 5
x
+ 3
4
x
, I = (0, ∞)
x
x
x
f“(x) = –3
10
x
+ 5
48
x
58. f(x) =
4
x
x
e
f ꞌ(x) =
44
2
xx
x
dd
x
eex
dx dx
=
34
2
4
x
x
x
x
exe
=
34
4
x
x
x
=
3(4 )
x
x
x
x
x
x
60. f(x) =
x
e
x
x
x
dd
x
x
x
x
EXERCISE 11-5 11-69
62. f(x) = 4x ln x – 7x
f ꞌ(x) = 4 ln x + 4xd
x
–7 = 4 ln x + 4 – 7 = 4 ln x – 3, x > 0
x
64. f(x) = x3(ln x – 2), x > 0
f ꞌ(x) = 3
()
d
x
dx
(ln x – 2) + x3(ln 2)
dx
dx
= 3x2(ln x – 2) + x31
x
= 3x2(ln x – 2) + x2
x
x
11-70 CHAPTER 11: GRAPHING AND OPTIMIZATION
66. f(x) = ln(x2e-x)
2
()
x
dxe
dx
2
2
x
x
x
exe
= 2
x
x
= 2
x
x
68. f(x) = 2x3 – 3x2 – 12x + 24
f ꞌ(x) = 6x2 – 6x – 12 = 6(x2 – x – 2) = 6(x – 2)(x + 1)
critical values: x = –1, 2
(A) On the interval [–3, 4]: f(–3) = –21
(B)On the interval [–2, 3]: f(–2) = 20
(C)On the interval [–2, 1]: f(–2) = 20
70. f(x) = x4 – 8x2 + 16
f ꞌ(x) = 4x3 – 16x = 4x(x2 – 4) = 4x(x – 2)(x + 2)
critical values: x = –2, 0, 2
(A) On the interval [–1, 3]: f(–1) = 9
(B) On the interval [0, 2]:f(0) = 16
(C)On the interval [–3, 4]: f(–3) = 25
EXERCISE 11-6 11-71
Absolute maximum of f: f(4) = 144
Absolute minimum of f: f(–2) = f(2) = 0
72. f(x) = x4 – 18x2 + 32
f ꞌ(x) = 4x3 – 36x = 4x(x2 – 9) = 4x(x – 3)(x + 3)
Critical values: x = –3, x = 0, x = 3
(A) On the interval [–4, 4]: f(–4) = 0
(B) On the interval [–1, 1]: f(–1) = 15
(C) On the interval [1, 3]: f(1) = 15
f(3) = –49
76. f has a local maximum at x = –1.
80. f has a local minimum at x = 1.
EXERCISE 11–6
2. 36 36
where 36. Since , ( ) .fxy xy y fx x
x
x
x
6. Volume of a right circular cylinder of radius r and height h: 2.Vrh
We have 2.hx Therefore,
23
8. 200 400
2 2 where 200. Since , ( ) 2 .fxy xy y fx x
x
x
10. Let one number = x. Then the other number = 19 – x
Set f(x) = x(19 – x) = 19x – x2
11-72 CHAPTER 11: GRAPHING AND OPTIMIZATION
12. Let one number = x. Then the other number = x +19.
14. Let x be the first number and y be the second. Then xy = 19, y = 19
x
, and the sum of the numbers is 19
x
x
.
Set f(x) = 19
x
x
and minimize f.
2
x
16. Let x be the length and y be the width. Then A = xy = 108 and y = 108
x
,
and the perimeter is 2(x + y) = 2 108
x
x
.
x
x
18. Let x be the length and y be the width. Then P = 2x + 2y = 76 so y = 38 – x,
and the area is xy = x(38 – x).
20. (A) Revenue R(x) = x · p(x) = x(400 – 0.5x) = 400x – 0.5x2
R‘(x) = 400 – x = 0 implies x = 400.
R(400) = 400(400) – 0.5(400)2 = $80,000
when 400 cameras are produced and sold for $200 each.
22. (A) Revenue R(x) = x·p(x) = x
2
400 400
22
x
x
x
(B) Profit P(x) = R(x) – C(x) = 400x –
2
2
x
– (2,000 + 200x) =
2
200 2,000.
2
x
x
(C) The revenue at the demand level x is:
26. (A) Let x be the number of price reductions, then the price of a cup of coffee will be p = 2.40 – 0.05x
and the number of cups sold will be 1,600 + 50x.
11-74 CHAPTER 11: GRAPHING AND OPTIMIZATION
R(8) = 3,840 + 40(8) – 2.5(82) = $4,000, when 2,000 cups of coffee sold at the price of p = 2.40 – 0.05(8)
= $2.00 a cup.
(B) In this case
R(x) = (1,600 + 60x)(2.40 – 0.10x)
28. Let x = number of dollar increases in the rate per night.
Then 300 – 3 x = total number of rooms rented and 80 + x = rate/night.
Total income = (total number of rooms rented)(rate – 10)
y(x) = (300 – 3x)(80 + x – 10), 0 ≤ x ≤ 100
Thus, x = 15 is the only critical value and
30. Let x = number of additional weeks the grower waits to pick the pears. Then 60 + 6x = the yield of a pear
since every week the price is dropped 2¢ per pound.
Now we want to maximize f(x) over [0, 4].
f‘(x) = 6(0.30 – 0.02x) – 0.02(60 + 6x)
32. (A) Let x = length of the side of the square.
Then L + 4x = 108 or L = 108 – 4x.
The volume of the box = Lx2 = x2(108 – 4x).
Would like to maximize y(x) on (0, 27).
y‘(x) = 2x(108 – 4x) – 4x2
Critical value: x = 18 [Note: x = 0 is not in the domain of y(x)]
y“(x) = 216 – 24x
EXERCISE 11-6 11-75
(B) Let x be the radius and L the height of the cylindrical container. Then its volume = πx2L. Since
L + 2πx = 108, then L = 108 – 2πx and y(x) = πx2L = πx2(108 – 2πx), 0 < x < 54
.
We want to maximize y(x) on 54
0,
.
y‘(x) = 2πx(108 – 2πx) – 2π2x2
critical value: x = 36
y“(x) = 216π – 12π2x
Thus, y(x) has an absolute maximum at x = 36
Therefore, volume is maximum at
34. Cost = 840 = 18x + 6(2y + x). Solving for y, we have 24 12 840 and 70 2 .yxy y x
Area .Axy Therefore
2
( ) (70 2 ) 70 2 , 0 35.Ax x x x x x
36. Let x and y be the width and the length of the rectangle, respectively. Then the amount of fencing needed
is
2 100 2 2 100.Pxyy xy
Since the area 12,100 24,200
12,100 sq ft, and ( ) 2 100, 0.Axy x Py y y
yy
We want to minimize P(y).
11-76 CHAPTER 11: GRAPHING AND OPTIMIZATION
38. Let x be the number of times the pharmacy places orders and let y be the number of demands. Then the cost
will be
C = 40x + 10 2
y
= 40x + 5y
We also have xy = 200. From this equation we find y = 200
x
and substitute for y in the cost
function to obtain:
C(x) = 40x + 1000
x
x
40. Let x = number of saws produced in each production run. Then the number of runs = 9,000
x
.
Cost = C(x) = cost of storage + cost of the production runs
= 9,000 5 22,500,000
(5) (2,500) , 0.
22
xx
x
xx
x
42. Let x = number of hours it takes the truck to travel 800 miles at v miles per hour.
Then 800 = xv or x = 800
v.
EXERCISE 11-6 11-77
2
89,600 24 89,600
v
44. C(t) = 2
0.16
44
t
tt
= 2
0.16
(2)
t
t
2
0.16( 2) 2( 2)(0.16 )
ttt
0.16( 2) 0.32
tt
0.16 0.32 0.32
tt
0.32 0.16
t
46. (A) Let the energy to fly over land be 1 unit; then the energy to fly over the water is 1.4 units.
E(x) = total energy = (1.4) 225x + (1)(10 – x)
E(x) = 1.4(x2 + 25)1/2 + 10 – x, 0 ≤ x ≤ 10
11-78 CHAPTER 11: GRAPHING AND OPTIMIZATION
Note that: E(0) = (1.4) 25 + 10 = 17
(B) E(x) = 1.1 225x + (1)(10 – x), 0 ≤ x ≤ 10
48. C(x) =
2
8k
x
+ 2
(10 )
k
x
, 0.5 ≤ x ≤ 9.5, k > 0
C‘(x) = –3
16k
+ 3
2
k
x
33
16 (10 ) 2
kxkx
critical value: x = 20
3.
C“(x) = 4
48k
+ 4
6
k
x
50. P(x) = 96x – 24x2, 0 ≤ x ≤ 3
P‘(x) = 96 – 48x
P‘(x) = 0 implies 96 – 48x = 0 or x = 2
critical value: x = 2
CHAPTER 11 REVIEW
2. f ‘ (x) < 0 on (c1, c3), (c6, b). (11-1, 11-2)
4. A local minimum occurs at x = c3. (11-1)
6. f ‘ (x) appears to be zero at x = c1, c3, c5. (11-1)
8. x = c2, c4, c5, c7 are inflection points. (11-2)
11-80 CHAPTER 11: GRAPHING AND OPTIMIZATION
10. Domain: all real numbers
Intercepts: y-intercept: f(0) = 0
x-intercepts: x = 0
Asymptotes: Horizontal asymptote: y = 2
no vertical asymptotes
Critical values: x = 0
11. f(x) = x4 + 5x312. y = 3x + 4
x
x