Download 4. SECOND DERIVATIVE TEST Let c be a critical value for f(x). f`(c) f

4.
5.
SECOND DERIVATIVE TEST
Let c be a critical value for f(x).
f'(c)
f"(c) GRAPH OF f IS:
f(c)
0
+
Concave upward
Local minimum
EXAMPLE
0
Concave downward Local maximum
0
0
?
Test does not apply
SECOND DERIVATIVE TEST FOR ABSOLUTE EXTREMUM
Let f be continuous on an interval I with only one critical value
c on I:
If f'(c) = 0 and f"(c) > 0,
then f(c) is the absolute
minimum of f on I.
(
If f'(c) = 0 and f"(c) < 0,
then f(c) is the absolute
maximum of f on I.
)
I
(
I
)
x
x
1. Interval [0, 10]; absolute minimum: f(0) = 0;
absolute maximum: f(10) = 14
3. Interval [0, 8]; absolute minimum: f(0) = 0;
absolute maximum: f(3) = 9
5. Interval [1, 10]; absolute minimum: f(1) = f(7) = 5;
absolute maximum: f(10) = 14
7. Interval [1, 9]; absolute minimum: f(1) = f(7) = 5;
absolute maximum: f(3) = f(9) = 9
9. Interval [2, 5]; absolute minimum: f(5) = 7;
absolute maximum: f(3) = 9
11. f(x) = x2 - 2x + 3, I = (-∞, ∞)
f'(x) = 2x - 2 = 2(x - 1)
f'(x) = 0: 2(x - 1) = 0
x = 1
x = 1 is the ONLY critical value on I, and f(1) = 12 - 2(1) + 3 = 2
f"(x) = 2 and f"(1) = 2 > 0. Therefore, f(1) = 2 is the absolute
minimum. The function does not have an absolute maximum since
lim f(x) = ∞.
x ! ±"
13. f(x) = -x2 - 6x + 9, I = (-∞, ∞)
f'(x) = -2x - 6 = -2(x + 3)
f'(x) = 0: -2(x + 3) = 0
x = -3
x = -3 is the ONLY critical value on I, and
f(-3) = -(-3)2 - 6(-3) + 9 = 18
f"(x) = -2 and f"(-3) = -2 < 0. Therefore, f(-3) = 18 is the absolute
maximum. The function does not have an absolute minimum since
lim f(x) = -∞.
x ! ±"
310
CHAPTER 5
GRAPHING AND OPTIMIZATION
15. f(x) = x3 + x, I = (-∞, ∞)
f'(x) = 3x2 + 1 ≥ 1 on I; f is increasing on I and
lim f(x) = -∞,
x ! "#
lim f(x) = ∞. Therefore, f does not have any absolute extrema.
x !"
17. f(x) = 8x3 - 2x4; domain: all real numbers
f'(x) = 24x2 - 8x3 = 8x2(3 - x)
f"(x) = 48x - 24x2 = 24x(2 - x)
Critical values: x = 0, x = 3
f"(0) = 0 (second derivative test fails)
f"(3) = -72 f has a local maximum at x = 3.
+ + + 0 + + + 0 - - Sign chart for f'(x) = 8x2(3 - x)
f'(x)
x
(0 and 3 are partition numbers)
f(x)
0
3
From the sign chart, f does not
Increasing Increasing Decreasing
have a local extremum at x = 0;
Local
f has a local maximum at x = 3
maximum
which must be an absolute maximum
since f is increasing on (-∞, 3) and decreasing on (3, ∞); f(3) = 54
is the absolute maximum of f. f does not have an absolute minimum
since lim f(x) = lim f(x) = -∞.
x !"
x ! "#
16
; domain: all real numbers except x = 0.
x
16
x 2 ! 16
(x ! 4)(x + 4)
f'(x) = 1 - 2 =
=
2
x
x
x2
32
f"(x) = 3
x
Critical values: x = -4, x = 4
1
f"(-4) = - < 0; f has a local maximum at x = -4
2
1
f"(4) =
> 0; f has a local minimum at x = 4
2
16#
16#
!
!
lim f(x) = lim " x +
=
∞;
lim
f
(
x
)
=
lim
x
+
= -∞;
x$
x$
x !"
x !"
x ! "#
x ! "# "
f has no absolute extrema.
19. f(x) = x +
x2
; domain: all real numbers
x2 + 1
(x 2 + 1)2x ! x 2(2x)
2x
=
f'(x) =
2
2
2
(x + 1)
(x + 1)2
(x 2 + 1)2(2) ! 2x(2)(x 2 + 1)(2x)
2 ! 6x 2
=
f"(x) =
(x 2 + 1)4
(x 2 + 1)3
Critical value: x = 0
Since f has only one critical value and f"(0) = 2 > 0, f(0) = 0 is
x2
the absolute minimum of f. Since lim f(x) = lim 2
= 1, f has no
x !"
x !" x + 1
absolute maximum; y = 1 is a horizontal asymptote for the graph of f.
21. f(x) =
EXERCISE 5-5
311
23. f(x) =
2x
; domain: all real numbers
x2 + 1
(x 2 + 1)2 ! 2x(2x)
2 ! 2x 2
2(1 ! x 2)
=
=
f'(x) =
(x 2 + 1)2
(x 2 + 1)2
(x 2 + 1)2
(x 2 + 1)2(!4x) ! 2(1 ! x 2)(2)(x 2 + 1)(2x)
4x(x 2 ! 3)
=
(x 2 + 1)3
f"(x) =
(x 2 + 1)4
Critical values: x = -1, x = 1
f"(-1) = 8 > 0; f has a local minimum at x = -1
f"(1) = -8 < 0; f has a local maximum at x = 1
Sign chart for f'(x)
- - - 0 + + + 0 - - f'(x)
(partition numbers are -1 and 1)
lim f(x) =
x ! ±"
lim
2x
= 0
2
x ! ±" x + 1
f(x)
-1
0
x
1
Decreasing Increasing Decreasing
Local
Local
minimum
maximum
(the x-axis is a horizontal
asymptote)
We can now conclude that f(1) = 1 is the absolute maximum of f and
f(-1) = -1 is the absolute minimum of f.
x2 ! 1
; domain: all real numbers
x2 + 1
4x
(x 2 + 1)(2x) ! (x 2 ! 1)(2x)
f'(x) =
=
2
2
2
(x + 1)2
(x + 1)
(x 2 + 1)2(4) ! 4x(2)(x 2 + 1)2x
4(3 ! x 2)
f"(x) =
=
(x 2 + 1)4
(x 2 + 1)3
Critical value: x = 0
f"(0) = 12 > 0; f has a local minimum at x = 0
4x
- - - - - 0 + + + + +
Sign chart for f'(x) =
2
2
f
'(
x
)
(x + 1)
(0 is the partition number)
f(x)
0
25. f(x) =
Decreasing Increasing
x
x2 ! 1
Local
lim f(x) = lim 2
= 1
minimum
x ! ±"
x ! ±" x + 1
(y = 1 is a horizontal asymptote)
We can now conclude that f(0) = -1 is the absolute minimum and f does
not have an absolute maximum.
27. f(x) = 2x2 - 8x + 6 on I = [0, ∞)
f'(x) = 4x - 8 = 4(x - 2)
f"(x) = 4
Critical value: x = 2
f"(2) = 4 > 0; f has a local minimum at x = 2
Since x = 2 is the only critical value of f on I, f(2) = -2 is the
absolute minimum of f on I.
312
CHAPTER 5
GRAPHING AND OPTIMIZATION
29. f(x) = 3x2 - x3 on I = [0, ∞)
f'(x) = 6x - 3x2 = 3x(2 - x)
f"(x) = 6 - 6x
Critical value (in (0, ∞)): x = 2
f"(2) = -6 < 0; f has a local maximum at x = 2
Since f(0) = 0 and x = 2 is the only critical value of f in (0, ∞),
f(2) = 4 is the absolute maximum value of f on I.
31. f(x) = (x + 4)(x - 2)2 on I = [0, ∞)
f'(x) = (x + 4)(2)(x - 2) + (x - 2)2 = (x - 2)[2x + 8 + x - 2]
= (x - 2)(3x + 6)
= 3x2 - 12
f"(x) = 6x
Critical value in I: x = 2
f"(2) = 12 > 0; f has a local minimum at x = 2
Since f(0) = 16 and x = 2 is the only critical value of f in (0, ∞),
f(2) = 0 is the absolute minimum of f on I.
33. f(x) = 2x4 - 8x3 on I = (0, ∞)
Since lim f(x) = lim (2x4 - 8x3) = ∞, f does not have an absolute
x !"
x !"
maximum on I.
12
, x > 0; I = (0, ∞)
x
35. f(x) = 20 - 3x -
f'(x) = -3 +
12
x2
f'(x) = 0: -3 +
12
= 0
x2
3x2 = 12
x2 = 4
x = 2 (-2 is not in I)
x = 2 is the only critical value of f on I, and
12
f(2) = 20 - 3(2) = 8.
2
24
24
f"(x) = - 2 ; f"(2) = = -6 < 0. Therefore, f(2) = 8 is the
x
4
absolute maximum of f. The function does not have an absolute minimum
since
lim f(x) = -∞. (Also, lim + f(x) = -∞.)
x !"
x !0
EXERCISE 5-5
313
37. f(x) = 10 + 2x +
f'(x) = 2 -
128
x3
64
, x > 0; I = (0, ∞)
x2
128
= 0
x3
2x3 = 128
x3 = 64
x = 4
x = 4 is the only critical value of f on I and
64
f(4) = 10 + 2(4) + 2 = 22
4
384
384
3
f"(x) =
> 0. Therefore, f(4) = 22 is the
4 ; f"(4) =
4 =
x
4
2
absolute minimum of f. The function does not have an absolute maximum
since lim f(x) = ∞. (Also, lim + f(x) = ∞.)
f'(x) = 0: 2 -
x !"
x !0
1
30
+ 3 on I = (0, ∞)
39. f(x) = x +
x
x
1
90
x 4 ! x 2 ! 90
(x 2 ! 10)(x 2 + 9)
f'(x) = 1 - 2 ! 4 =
=
x
x
x4
x4
2
360
f"(x) = 3 + 5
x
x
Critical value (in (0, ∞)): x = 10
2
360
+
f"( 10 ) =
> 0; f has a local minimum at x =
3
2
(10)
(10)5 2
Since
!
!
value of f on I, f( 10 ) =
10 is the only critical
absolute minimum of f on I.
ex
!
, x > 0
x2
d x
d 2
x2
e ! ex
x
x 2e x ! 2xe x
xe x(x !
dx
dx
f'(x) =
=
=
x4
x4
x4
ex(x ! 2)
Critical values: f'(x) =
= 0
x3
ex(x - 2) = 0
x = 2 [Note: ex ≠
Thus, x = 2 is the only critical value of f on (0,
Sign chart for f' [Note: This approach is a little
calculating f"(x)]:
Test Numbers
- - - 0 + + +
f'(x)
x
f'(x)
x
1 !e (!)
f(x)
1
2
3
0
e 3 (+)
3
27
Decr.
Incr.
41. !f(x) =
314
CHAPTER 5
GRAPHING AND OPTIMIZATION
!
10
14
is the
10
!
2)
=
ex(x ! 2)
x3
0 for all x.]
∞).
easier than
By the first derivative test, f has a minimum value at x = 2;
e2
e2
f(2) = 2 =
≈ 1.847 is the absolute minimum value of f.
2
4
x3
ex
" d 3% x
" d x% 3
x 'e ( $
e 'x
$
# dx
&
# dx
&
f'(x) =
(ex )2
3x 2e x ! x 3e x
x 2(3 ! x)e x
x 2(3 ! x)
=
=
=
e 2x
e2x
ex
x 2(3 ! x)
!
Critical
values: f'(x) =
= 0
ex
x2(3 - x) = 0
x = 0 and x = 3
Sign chart for f' [Note: This approach is a little easier than
calculating f"(x)]:
Test Numbers
0
0
x
f '(x)
f'(x) + + + + + + + + - - 4 (+)
!1
e!1
43. f(x) =
x
f(x)
-1
Increasing
0
1
Increasing
3
4
Decreasing
2 (+)
1
e
4 ! 16
e 4 (!)
By the first derivative test, f has a maximum value at x = 3;
27
f(3) = 3 ≈ 1.344 is the absolute maximum value of f.
e
45. f(x) = 5x - 2x ln x, x > 0
d
f'(x) = 5 - 2x
(ln x) - ln x
dx
"1%
= 5 - 2x $ ' - 2 ln x = 3
#x &
Critical values: f'(x) = 3 - 2
d
(2x)
dx
- 2 ln x, x > 0
ln x = 0
3
ln x =
= 1.5; x = e1.5
2
!
1.5
Thus, x = e
is the only critical value of f on (0, ∞).
d
2
Now, f"(x) =
(3 - 2 ln x) = dx
x
2
1.5
and f"(e ) = - 1.5 < 0.
e
Therefore, f has a maximum value at x = e1.5, and
f(e1.5) = 5e1.5 - 2e1.5 ln(e1.5) = 5e1.5 - 2(1.5)e1.5 = 2e1.5 ≈ 8.963
is the absolute maximum of f.
EXERCISE 5-5
315
47. f(x) = x2(3 - ln
d
f'(x) = x2
(3 dx
= x2 ! x1 +
Critical values:
x), x > 0
d 2
x
dx
(3 - ln x)2x = -x + 6x - 2x ln x = 5x - 2x ln x
f'(x) = 5x - 2x ln x = 0
x(5 - 2 ln x) = 0
5 - 2 ln x = 0
5
ln x =
= 2.5
2
x = e2.5 [Note: x ≠ 0 on (0, ∞)]
Now
f"(x) = 5 - 2x 1x - 2 ln x
= 3 - 2 ln x
2.5
and f"(e ) = 3 - 2·ln(e2.5) = 3 - 2(2.5) = 3 - 5 = -2 < 0
Therefore, f has a maximum value at x = e2.5 and
e5
f(e2.5) = (e2.5)2(3 - ln e2.5) = e5(3 - 2.5) =
≈ 74.207
2
is the absolute maximum value of f.
ln x) + (3 - ln x)
( )
( )
49. f(x) = ln(xe-x), x > 0
1 d
1
1 ! x
f'(x) =
(xe-x) =
[e-x - xe-x] =
!
x
!
x
xe
xe
dx
x
1 ! x
Critical values: f'(x) =
= 0; x = 1
x
Sign chart for f'(x):
f'(x)
+ + + + 0 - - - -
x
f(x)
0
1
Increasing
2
Decreasing
Local
maximum
Test Numbers
x
f(x)
1
1(+)
2
2 ! 12 (!)
By the first derivative test, f has a maximum value at x = 1;
f(1) = ln(e-1) = -1 is the absolute maximum value of f.
51. f(x) = x3 - 6x2 + 9x - 6
f'(x) = 3x2 - 12x + 9 = 3(x2 - 4x + 3) = 3(x - 3)(x - 1)
Critical values: x = 1, 3
(A) On the interval [-1, 5]: f(-1) = -1 - 6 - 9 - 6 = -22
f(1) = 1 - 6 + 9 - 6 = -2
f(3) = 27 - 54 + 27 - 6 = -6
f(5) = 125 - 150 + 45 - 6 = 14
Thus, the absolute maximum of f is f(5) = 14, and the absolute
minimum of f is f(-1) = -22.
(B) On the interval [-1, 3]: f(-1) = -22
f(1) = -2
f(3) = -6
Absolute maximum of f: f(1) = -2
Absolute minimum of f: f(-1) = -22
316
CHAPTER 5
GRAPHING AND OPTIMIZATION
(C) On the interval [2, 5]: f(2) = 8 - 24 + 18 - 6 = -4
f(3) = -6
f(5) = 14
Absolute maximum of f: f(5) = 14
Absolute minimum of f: f(3) = -6
53. f(x) = (x - 1)(x - 5)3 + 1
f'(x) = (x - 1)3(x - 5)2 + (x - 5)3
= (x - 5)2(3x - 3 + x - 5)
= (x - 5)2(4x - 8)
Critical values: x = 2, 5
(A) Interval [0, 3]: f(0) = (-1)(-5)3 + 1 = 126
f(2) = (2 - 1)(2 - 5)3 + 1 = -26
f(3) = (3 - 1)(3 - 5)3 + 1 = -15
Absolute maximum of f: f(0) = 126
Absolute minimum of f: f(2) = -26
(B) Interval [1, 7]: f(1) =
f(2) =
f(5) =
f(7) =
1
-26
1
(7 - 1)(7 - 5)3 + 1 = 6·8 + 1 = 49
Absolute maximum of f: f(7) = 49
Absolute minimum of f: f(2) = -26
(C) Interval [3, 6]: f(3) = (3 - 1)(3 - 5)3 + 1 = -15
f(5) = 1
f(6) = (6 - 1)(6 - 5)3 + 1 = 6
Absolute maximum of f: f(6) = 6
Absolute minimum of f: f(3) = -15
55. f(x) = x4 - 4x3 + 5
f'(x) = 4x3 - 12x2 = 4x2(x - 3)
Critical values: x = 0, x = 3
(A) On the interval [-1, 2]: f(-1) = 10
f(0) = 5
f(2) = -11
Thus, the absolute maximum of f is f(-1) = 10; the absolute minimum
of f is f(2) = -11.
(B) On the interval [0, 4]: f(0) = 5
f(3) = -22
f(4) = 5
Absolute maximum of f: f(0) = f(4) = 5
Absolute minimum of f: f(3) = -22
(C) On the interval [-1, 1]: f(-1) = 10
f(0) = 5
f(1) = 2
Absolute maximum of f: f(-1) = 10
Absolute minimum of f: f(1) = 2
EXERCISE 5-5
317
57. f has a local minimum at x = 2.
59. Unable to determine from the given information (f'(-3) = f"(-3) = 0).
61. Neither a local maximum nor a local minimum at x = 6; x = 6 is not a
critical value of f.
63. f has a local maximum at x = 2.
EXERCISE 5-6
Things to remember:
STRATEGY FOR SOLVING OPTIMIZATION PROBLEMS
Step 1. Introduce variables, look for relationships among these
variables, and construct a mathematical model of the form:
Maximize (or minimize) f(x) on the interval I
Step 2. Find the critical values of f(x).
Step 3. Use the procedures developed in Section 5-5 to find the
absolute maximum (or minimum) value of f(x) on the interval I
and the value(s) of x where this occurs.
Step 4. Use the solution to the mathematical model to answer all the
questions asked in the problem.
1. Let one length = x and the other = 10 - x.
Since neither length can be negative, we have x ≥ 0 and 10 - x ≥ 0, or
x ≤ 10. We want the maximum value of the product x(10 - x), where
0 ≤ x ≤ 10.
f(x) = x(10 - x) = 10x - x2; domain I = [0, 10]
f'(x) = 10 - 2x; x = 5 is the only critical value
f"(x) = -2
f"(5) = -2 < 0
Thus, f(5) = 25 is the absolute maximum; divide the line in half.
Let
3. Let one number = x. Then the other number = x + 30.
f(x) = x(x + 30) = x2 + 30x; domain I = (-∞, ∞)
f'(x) = 2x + 30; x = -15 is the only critical value
f"(x) = 2
f"(-15) = 2 > 0
Thus, the absolute minimum of f occurs at x = -15. The numbers,
then, are -15 and -15 + 30 = 15.
318
CHAPTER 5
GRAPHING AND OPTIMIZATION
x
5. Let x = the length of the rectangle
and y = the width of the rectangle.
Then, 2x + 2y = 100
y
y
x + y = 50
y = 50 - x
We want to find the maximum of the area:
x
A(x) = x·y = x(50 - x) = 50x - x2.
Since x ≥ 0 and y ≥ 0, we must have 0 ≤ x ≤ 50.
[Note: A(0) = A(50) = 0.]
dA
A'(x) = dx = 50 - 2x; x = 25 is the only critical value.
Now, A" = -2 and A"(25) = -2 < 0. Thus, A(25) is the absolute maximum.
The maximum area is A(25) = 25(50 - 25) = 625 cm2, which means that the
rectangle is actually a square with sides measuring 25 cm each.
7. Let the rectangle of fixed area A have
dimensions x and y. Then A = xy and y =
The cost of the fence is
A
.
x
y
2AB
C = 2Bx + 2By = 2Bx +
, x > 0
x
Thus, we want to find the absolute minimum of
x
2AB
C(x) = 2Bx +
, x > 0
x
Since lim C(x) = lim C(x) = ∞, and C(x) > 0 for all x > 0, we can
x !0
x !"
conclude that C has an absolute minimum on (0, ∞). This agrees with
our intuition that there should be a cheapest way to build the fence.
9. Let x and y be the dimensions of the rectangle and let C be the fixed
amount which can be spent. Then
C ! 2Bx
C = 2Bx + 2By and y =
2B
The area enclosed by the fence is:
#C " 2Bx &
A = xy = x %
$ 2B ('
Thus, we want to find the absolute maximum value of
C
C
A(x) =
x - x2, 0 ≤ x ≤
2B
2B
!
" C%
Since A(x) is a continuous function on the closed interval $0, ', it
# 2B &
has an absolute maximum value. This agrees with our intuition that
there should be a largest rectangular area that can be enclosed with
a fixed amount of fencing.
!
EXERCISE 5-6
319
11. Price-demand: p(x) = 500 - 0.5x; cost: C(x) = 20,000 + 135x
(A) Revenue: R(x) = x·p(x) = 500x - 0.5x2, 0 ≤ x < ∞
R'(x) = 500 - x
R'(x) = 500 - x = 0 implies x = 500
R"(x) = -1; R"(500) = -1 < 0
R has an absolute maximum at x = 500.
p(500) = 500 - 0.5(500) = 250; R(500) = (500)2 - 0.5(500)2 = 125,000
The company should produce 500 phones each week at a price of $250
per phone to maximize their revenue. The maximum revenue is $125,000
(B) Profit: P(x) = R(x) - C(x) = 500x - 0.5x2 - (20,000 + 135x)
= 365x - 0.5x2 - 20,000
P'(x) = 365 - x
P'(x) = 365 - x = 0 implies x = 365
P"(x) = -1; P"(365) = -1 < 0
P has an absolute maximum at x = 365
p(365) = 500 - 0.5(365) = 317.50;
P(365) = (365)2 - 0.5(365)2 - 20,000 = 46,612.50
To maximize profit, the company should produce 365 phones each week
at a price of $317.50 per phone. The maximum profit is $46,612.50.
#
x2
x&
13. (A) Revenue R(x) = x·p(x) = x %200 "
, 0 ≤ x ≤ 6,000
( = 200x $
30
30'
x
2x
R'(x) = 200 = 200 15
30
x
Now R'(x) = 200 - ! = 0 implies x = 3000.
15
1
R"(x) = < 0.
15
1
Thus, R"(3000) = < 0 and we conclude that R has an absolute
15
maximum at x = 3000. The maximum revenue is
(3000)2
R(3000) = 200(3000) = $300,000
30
x2
(B) Profit P(x) = R(x) - C(x) = 200x - (72,000 + 60x)
30
x2
= 140x - 72,000
30
x
P'(x) = 140 15
x
Now 140 = 0 implies x = 2,100.
15
1
1
P"(x) = and P"(2,100) = < 0. Thus, the maximum profit
15
15
occurs when 2,100 television sets are produced. The maximum
profit is
(2,100)2
P(2,100) = 140(2,100) - 72,000 = $75,000
30
the price that the company should charge is
2,100
p(2,100) = 200 = $130 for each set.
30
320
CHAPTER 5
GRAPHING AND OPTIMIZATION
(C) If the government taxes the company $5 for each set, then the
profit P(x) is given by
x2
P(x) = 200x - (72,000 + 60x) - 5x
30
x2
= 135x - 72,000.
30
x
P'(x) = 135 .
15
x
Now 135 = 0 implies x = 2,025.
15
1
1
P"(x) = and P"(2,025) = < 0. Thus, the maximum profit in
15
15
this case occurs when 2,025 television sets are produced. The
maximum profit is
(2,025)2
P(2,025) = 135(2,025) - 72,000 = $64,687.50
30
2,025
and the company should charge p(2,025) = 200 = $132.50/set.
30
15. (A)
(B)
(C) The revenue at the demand level x is:
R(x) = xp(x)
where p(x) is the quadratic regression equation in (A).
The cost at the demand level x is C(x) given by the linear
regression equation in (B). The profit P(x) = R(x) - C(x).
The maximum profit is $118,996 at the demand level x = 1422.
The price per saw at the demand level x = 1422 is $195.
17. (A) Let x = number of 10¢ reductions in price. Then
640 + 40x = number of sandwiches sold at x 10¢ reductions
8 - 0.1x = price per sandwich 0 ≤ x ≤ 80
Revenue: R(x) = (640 + 40x)(8 - 0.1x)
= 5120 + 256x - 4x2, 0 ≤ x ≤ 80
R'(x) = 256 - 8x
R'(x) = 256 - 8x = 0 implies x = 32
R(0) = 5120, R(32) = 9216, R(80) = 0
Thus, the deli should charge 8 - 3.20 = $4.80 per sandwich to
realize a maximum revenue of $9216.
EXERCISE 5-6
321
(B) Let x = number of 20¢ reductions in price. Then
640 + 15x = number of sandwiches sold
8 - 0.2x = price per sandwich 0 ≤ x ≤ 40
Revenue: R(x) = (640 + 15x)(8 - 0.2x)
= 5120 - 8x - 3x2, 0 ≤ x ≤ 40
R'(x) = -8 - 6x
R'(x) = -8 - 6x = 0 has no solutions in (0, 40)
Now, R(0) = 640·8 = $5120
R(40) = 0
Thus, the deli should charge $8 per sandwich to maximize their
revenue under these conditions.
19. Let x = number of dollar increases in the rate per day. Then
200 - 5x = total number of cars rented and 30 + x = rate per day.
Total income = (total number of cars rented)(rate)
y(x) = (200 - 5x)(30 + x), 0 ≤ x ≤ 40
y'(x) = (200 - 5x)(1) + (30 + x)(-5)
= 200 - 5x - 150 - 5x
= 50 - 10x
= 10(5 - x)
Thus, x = 5 is the only critical value and
y(5) = (200 - 25)(30 + 5) = 6125.
y"(x) = -10
y"(5) = -10 < 0
Therefore, the absolute maximum income is y(5) = $6125 when the rate
is $35 per day.
21. Let x = number of additional trees planted per acre. Then
30 + x = total number of trees per acre and 50 - x = yield per tree.
Yield per acre = (total number of trees per acre)(yield per tree)
y(x) = (30 + x)(50 - x), 0 ≤ x ≤ 20
y'(x) = (30 + x)(-1) + (50 - x)
= 20 - 2x
= 2(10 - x)
The only critical value is x = 10, y(10) = 40(40) = 1600 pounds per
acre.
y"(x) = -2
y"(10) = -2 < 0
Therefore, the absolute maximum yield is y(10) = 1600 pounds per acre
when the number of trees per acre is 40.
V(x) = (12 - 2x)(8 - 2x)x, 0 ≤ x ≤ 4
= 96x - 40x2 + 4x3
V'(x) = 96 - 80x + 12x2
= 4(24 - 20x + 3x2)
We solve 24 - 20x + 3x2 = 0 by using the
quadratic formula:
20 ± 400 " 4 # 24 # 3
10 ± 2 7
x =
=
6
3
23. Volume =
322
!
CHAPTER 5
GRAPHING AND OPTIMIZATION
!
8
x
12
Thus, x =
10 " 2 7
≈ 1.57 is the only critical value on the interval
3
[0, 4].
V "(x) = -80 + 24x
V "(1.57) = -80 + 24(1.57) < 0
!
Therefore,
a square with a side of length x = 1.57 inches should be
cut from each corner to obtain the maximum volume.
25. Area = 800 square feet = xy
(1)
Cost = 18x + 6(2y + x)
800
y
From (1), we have y =
.
x
! 1600
Hence, cost C(x) = 18x + 6 "
+ x #$ , or
x
x
9600
C(x) = 24x +
, x > 0,
x
9600
24(x ! 20)(x + 20)
24(x 2 ! 400)
C'(x) = 24 =
=
.
2
2
x
x2
x
Therefore, x = 20 is the only critical value.
19, 200
C"(x) =
x3
19, 200
C"(20) =
> 0. Therefore, x = 20 for the
8000
minimum cost.
The dimensions of the fence are shown in the
diagram at the right.
40
20
(Expensive
side)
27. Let x = number of cans of paint produced in each production run.
Then,
16,000
number of production runs:
, 1 ≤ x ≤ 16,000
x
Cost: C(x) = cost of storage + cost of set up
x
16,000
=
(4) +
(500)
2
x
[Note: x2 is the average number of cans of paint in storage per day.]
Thus,
8,000,000
C(x) = 2x +
, 1 ≤ x ≤ 16,000
x
8,000,000
2x 2 ! 8,000,000
2(x 2 ! 4,000,000)
C'(x) = 2 =
=
x2
x2
x2
Critical value: x = 2000
16,000,000
C"(x) =
; C"(2000) > 0
x3
Thus, the minimum cost occurs when x = 2000 and the number of
16,000
production runs is
= 8.
2,000
EXERCISE 5-6
323
29. Let x = number of books produced each printing. Then, the number of
50,000
printings =
.
x
Cost = C(x) = cost of storage + cost of printing
x
50,000
=
+
(1000), x > 0
2
x
x
[Note:
is the average number in storage each day.]
2
50,000,000
(x + 10,000)(x ! 10,000)
x 2 ! 100,000,000
1
C'(x) =
=
=
2
2
x
2x 2
2x
2
Critical value: x = 10,000
100,000,000
C"(x) =
x3
100,000,000
C"(10,000) =
> 0
(10,000)3
Thus, the minimum cost occurs when x = 10,000 and the number of
50,000
printings is
= 5.
10,000
31. (A) Let the cost to lay the pipe on the land be 1 unit; then the cost
to lay the pipe in the lake is 1.4 units.
C(x) = total cost = (1.4) x 2 + 25 + (1)(10 - x), 0 ≤ x ≤ 10
= (1.4)(x2 + 25)1/2 + 10 - x
1
C'(x) = (1.4) (x 2 + 25)-1/2(2x) - 1
2
!
= (1.4)x(x2 + 25)-1/2 - 1
=
1.4x "
x 2 + 25
x 2 + 25
x 2 + 25 = 0 or 1.96x2 = x2 + 25
.96x2 = 25
25
!
x2 =
= 26.04
.96
!
x = ±5.1
Thus, the critical value is x = 5.1.
# 1&
C"(x) = (1.4)(x2 + 25)-1/2 + (1.4)x %" ((x 2 + 25)-3/22x
$ 2'
2
1.4
35
(1.4)x
=
=
2
1
2
2
2
3
2
(x + 25)
(x + 25)3 2
(x + 25)
35
C"(5.1) =
> 0!
2
[(5.1) + 25]3 2
Thus, the cost will be a minimum when x = 5.1.
C'(x) = 0 when 1.4x -
Note that: C(0) = (1.4) 25 + 10 = 17
C(5.1) = (1.4) 51.01 + (10 - 5.1) = 14.9
C(10) = (1.4) 125 = 15.65
Thus, the absolute
! minimum occurs when x = 5.1 miles.
324
CHAPTER 5
!
GRAPHING AND!OPTIMIZATION
(B) C(x) = (1.1) x 2 + 25 + (1)(10 - x), 0 ≤ x ≤ 10
(1.1)x " x 2 + 25
C'(x) =
x 2 + 25
!
C'(x) = 0 when 1.1x - x 2 + 25 = 0 or (1.21)x2 = x2 + 25
.21x2 = 25
25
!
x2 =
= 119.05
.21
!
x = ±10.91
Critical value: x = 10.91 > 10, i.e., there are no critical values
on the interval [0, 10]. Now,
C(0) = (1.1) 25 + 10 = 15.5,
C(10) = (1.1) 125 ≈ 12.30.
Therefore, the absolute minimum occurs when x = 10 miles.
!
33. C(t) = 30t2! - 240t + 500, 0 ≤ t ≤ 8
C'(t) = 60t - 240; t = 4 is the only critical value.
C"(t) = 60
C"(4) = 60 > 0
Now, C(0) = 500
C(4) = 30(4)2 - 240(4) + 500 = 20,
C(8) = 30(8)2 - 240(8) + 500 = 500.
Thus, 4 days after a treatment, the concentration will be minimum;
the minimum concentration is 20 bacteria per cm3.
35. Let x = the number of mice ordered in each order. Then the number of
500
orders =
.
x
x
500
[Note: Cost = cost of feeding + cost of order,
C(x) = Cost =
·4+
(10)
x is the average number of mice at any one
2
x
2
time.]
5000
C(x) = 2x +
, 0 < x ≤ 500
x
5000
2(x + 50)(x ! 50)
2x 2 ! 5000
2(x 2 ! 2500)
C'(x) = 2 =
=
=
2
2
2
x
x2
x
x
Critical value: x = 50 (-50 is not a critical value, since the domain
of C is x > 0.
10,000
10,000
C"(x) =
and C"(50) =
> 0
3
x
503
Therefore, the minimum cost occurs when 50 mice are ordered each
time.
500
The total number of orders is
= 10.
50
EXERCISE 5-6
325
37. H(t) = 4t1/2 - 2t, 0 ≤ t ≤ 2
H'(t) = 2t-1/2 - 2
Thus, t = 1 is the only critical value.
Now, H(0) = 4·01/2 - 2(0) = 0,
H(1) = 4·11/2 - 2(1) = 2,
H(2) = 4·21/2 - 4 ≈ 1.66.
Therefore, H(1) is the absolute maximum, and after one month the
maximum height will be 2 feet.
39. N(t) = 30 + 12t2 - t3, 0 ≤ t ≤ 8
The rate of increase = R(t) = N'(t) = 24t - 3t2, and
R'(t) = N"(t) = 24 - 6t.
Thus, t = 4 is the only critical value of R(t).
Now, R(0) = 0,
R(4) = 24·4 - 3·42 = 48,
R(8) = 24·8 - 3·82 = 0.
Therefore, the absolute maximum value of R occurs when t = 4; the
maximum rate of increase will occur four years from now.
CHAPTER 5 REVIEW
1. The function f is increasing on (a, c1), (c3, c6).
(5-1, 5-2)
2. f'(x) < 0 on (c1, c3), (c6, b).
(5-1, 5-2)
3. The graph of f is concave downward on (a, c2), (c4, c5), (c7, b).
(5-1, 5-2)
4. A local minimum occurs at x = c3.
(5-1)
5. The absolute maximum occurs at x = c1, c6.
(5-1, 5-5)
6. f'(x) appears to be zero at x = c1, c3, c5.
(5-1)
7. f'(x) does not exist at x = c4, c6.
(5-1)
8. x = c2, c4, c5, c7 are inflection points.
(5-2)
9.
f"(x)
- - - - 0 + + + 0 - - - - - ND - - - -
x
Graph
of f
-1
Concave
Downward
0
Concave
Upward
Point of
Inflection
326
CHAPTER 5
2
Concave
Downward
Concave
Downward
Point of
Inflection
GRAPHING AND OPTIMIZATION
f'(x)
+ + 0 - - - - 0 - - - - ND + +
x
f(x)
-2
Incr.
0
2
Decr.
Decr.
Local
maximum
Incr.
Local
minimum
Using this information together with the
points (-3, 0), (-2, 3), (-1, 2), (0, 0),
(2, -3), (3, 0) on the graph, we have
f(x)
3
2
1
x
-3 -2 -1
-1
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
f'(x)
1
2
3
-2
(5-2)
-3
- - - - - - - - 0 + + + + + + + +
x
0
f(x)
Decreasing
Increasing
Local
minimum
f"(x)
f(x)
2
- - 0+ + + + + + + + + +0 - x
f(x)
-2
Concave
Downward
Inflection
point
Concave
Upward
2
Concave
Downward
Inflection
point
1
–5
5
x
(5-1, 5-2)
11. f(x) = x4 + 5x3
f'(x) = 4x3 + 15x2
f"(x) = 12x2 + 30x
4
x
4
y' = 3 - 2
x
8
y" = 3
x
12. y = 3x +
(5-2)
CHAPTER 5 REVIEW
(5-2)
327
13. f(x) = x3 - 18x2 + 81x
Step 1. Analyze f(x):
(A) Domain: All real numbers, (-∞, ∞)
(B) Intercepts: y-intercept: f(0) = 03 - 18(0)2 + 81(0) = 0
x-intercepts: x3 - 18x2 + 81x = 0
x(x2 - 18x + 81) = 0
x(x - 9)2 = 0
x = 0, 9
(C) Asymptotes: No horizontal or vertical asymptotes.
Step 2. Analyze f'(x):
f'(x) = 3x2 - 36x + 81 = 3(x2 - 12x + 27) = 3(x - 3)(x - 9)
Critical values: x = 3, x = 9
Partition numbers: x = 3, x = 9
Sign chart for f':
f'(x)
Test Numbers
+ + + + 0 - - - - - 0 + + + +
x
0
3
5
9 10
f(x) Increasing Decreasing
Increasing
Local
Local
Maximum
Minimum
x
0
5
10
f'(x)
81 (+)
-24 (-)
21 (+)
Thus, f is increasing on (-∞, 3) and on (9, ∞); f is decreasing on
(3, 9). There is a local maximum at x = 3 and a local minimum at
x = 9.
Step 3. Analyze f"(x):
f"(x) = 6x - 36 = 6(x - 6)
Thus, x = 6 is a partition number for f".
Sign chart for f":
f"(x)
Test Numbers
- - - - - - 0 + + + + + +
x
Graph
of f
0
6 7
Concave
Downward
Concave
Upward
x
0
7
f"(x)
-36 (-)
6 (+)
Inflection
Point
Thus, the graph of f is concave downward on (-∞, 6) and concave
upward on (6, ∞). The point x = 6 is an inflection point.
328
CHAPTER 5
GRAPHING AND OPTIMIZATION