dx
dp
dt
(B) From (2),
=
(5x + 100 p)
dt
Setting x = 150 in (1), we get
!(4x + 5p)
or
45,000 + 750p + 50p2 = 80,000
p2 + 15p - 700 = 0
!15 ± 55
225 + 2800
=
= -35, 20
2
2
Since p ≥ 0, p = 20.
dx
Now, for x = 150, p = 20 and
= -6, we have
dt
!
dp
[4(150) + 5(20)]("6)
4200
=
= ≈ 1.53
dt
5(150) + 100(20)
2750
and p =
"15 ±
Thus, the price is increasing at the rate of $1.53 per month.
!
31. Volume V = πR2h, where h = thickness of the circular oil slick.
1
Since h = 0.1 =
, we have:
10
! 2
V =
R
10
Differentiating with respect to t:
# " 2&
d%
R (
dV
!
dR
! dR
$ 10
'
=
=
2R
=
R
dt
10
dt
5 dt
dt
dR
Given:
= 0.32 when R = 500. Therefore,
dt
!dV = ! (500)(0.32) = 100π(0.32) ≈ 100.48 cubic feet per minute.
dt
5
EXERCISE 4-7
Things to remember:
1.
RELATIVE AND PERCENTAGE RATES OF CHANGE
The RELATIVE RATE OF CHANGE of a function f(x) is
The PERCENTAGE RATE OF CHANGE is 100 ×
2.
198
f '(x)
.
f(x)
f '(x)
.
f(x)
ELASTICITY OF DEMAND
If price and demand are related by x = f(p), then the
ELASTICITY OF DEMAND is given by
pf '(p)
E(p) = f(p)
CHAPTER 4
ADDITIONAL DERIVATIVE TOPICS
3.
INTERPRETATION OF ELASTICITY OF DEMAND
E(p)
Demand
Interpretation
0 < E(p) < 1 Inelastic
Demand is not sensitive to
changes in price. A change in
price produces a smaller
change in demand.
E(p) > 1
Elastic
Demand is sensitive to changes
in price. A change in price
produces a larger change in
demand.
E(p) = 1
Unit
A change in price produces the
same change in demand.
4.
REVENUE AND ELASTICITY OF DEMAND
If R(p) = pf(p) is the revenue function, then R'(p) and
[1 - E(p)] always have the same sign.
Demand is inelastic [E(p) < 1, R'(p) > 0]:
A price increase will increase revenue.
A price decrease will decrease revenue.
Demand is elastic [E(p) > 1, R'(p) < 0]:
A price increase will decrease revenue.
A price decrease will increase revenue.
1. f(x) = 25
f’(x) = 0
Relative rate of change of f:
f "(x)
0
=
= 0
f(x)
25
3. f(x) = 30x
f’(x) = 30
f "(x)
30
1
Relative rate of change !of f: !
=
=
f(x)
30x
x
5. f(x) = 10x + 500
f'(x) = 10
f '(x) !
10
1
Relative rate of change !of f: !
=
=
f(x)
10x + 500
x + 50
7. f(x) = 100x - 0.5x2
f'(x) = 100 - x
Relative rate of change of f:
100 ! x
f '(x)
=
100x ! 0.5x 2
f(x)
9. f(x) = 4 + 2e-2x
f'(x) = -4e-2x
Relative rate of change of f:
!4e !2x
f '(x)
=
4 + 2e !2x
f(x)
2
2e !2x
e2x
= "
= !2
x
2
x
1 + 2e2x
2+ e
e
EXERCISE 4-7
199
11. f(x) = 25x + 3x ln x
f'(x) = 25 + 3 ln x + 3 = 28 + 3 ln x
f '(x)
28 + 3 ln x
Relative rate of change of f:
=
f(x)
25x + 3x ln x
13. x = f(p) = 12,000 - 10p2
f'(p) = -20p
! pf '(p)
20 p2
=
f(p)
12,000 ! 10p 2
2000
2000
2
=
=
10: E(10) =
12,000 ! 1000
11,000
11
is inelastic.
8000
8000
=
20: E(20) =
= 1; unit elasticity.
12,000 ! 4000
8000
18,000
18,000
=
30: E(30) =
= 6
12,000 ! 9,000
3,000
is elastic.
Elasticity of demand: E(p) =
(A) At p =
Demand
(B) At p =
(C) At p =
Demand
15. x = f(p) = 950 - 2p - 0.1p2
f'(p) = -2 - 0.2p
! pf '(p)
2p + 0.2p2
=
f(p)
950 ! 2p ! 0.1 p2
60 + 180
240
3
30: E(30) =
=
=
950 ! 60 ! 90
800
10
is inelastic.
100 + 500
600
50: E(50) =
= 1; unit elasticity.
=
950 ! 100 ! 250
600
140 + 980
1120
70: E(70) =
= 3.5
=
950 ! 140 ! 490
320
is elastic.
Elasticity of demand: E(p) =
(A) At p =
Demand
(B) At p =
(C) At p =
Demand
17. p + 0.005x = 30
(A) x =
(B)
30 ! p
= 6000 - 200p, 0 ≤ p ≤ 30
0.005
f(p) = 6000 - 200p
f'(p) = -200
! pf '(p)
200p
=
f(p)
6000 ! 200 p
p
=
30 ! p
10
1
(C) At p = 10: E(10) =
= 0.5
=
30 ! 10
2
If the price increases by 10%, the demand will decrease by
approximately 0.5(10%) = 5%.
Elasticity of demand: E(p) =
200
CHAPTER 4
ADDITIONAL DERIVATIVE TOPICS
25
= 5
30 ! 25
If the price increases by 10%, the demand will decrease by
approximately 5(10%) = 50%.
15
(E) At p = 15: E(15) =
= 1
30 ! 15
If the price increases by 10%, the demand will decrease by
approximately 10%.
(D) At p = 25: E(25) =
19. 0.02x + p = 60
60 ! p
(A) x =
= 3000 - 50p, 0 ≤ p ≤ 60
0.02
(B) R(p) = p(3000 - 50p) = 3000p - 50p2
(C) f(p) = 3000 - 50p
f'(p) = -50
! pf '(p)
p
50p
Elasticity of demand: E(p) =
=
=
f(p)
60 ! p
3000 ! 50p
p
(D) Elastic: E(p) =
> 1
60 ! p
p > 60 - p
p > 30,
30 < p < 60
p
Inelastic: E(p) =
< 1
60 ! p
p < 60 - p
p < 30,
0 < p < 30
(E) R'(p) = f(p) [1 - E(p)] (equation (9))
R'(p) > 0 if E(p) < 1; R'(p) < 0 if E(p) > 1
Therefore, revenue is increasing for 0 < p < 30 and decreasing
for 30 < p < 60.
(F) If p = $10 and the price is decreased, revenue will also decrease.
(G) If p = $40 and the price is decreased, revenue will increase.
21. x = f(p) = 10(p - 30)2, 0 ≤ p ≤ 30
f'(p) = 20(p - 30)
! p[20(p ! 30)]
!2p
Elasticity of demand: E(p) =
=
2
10(p ! 30)
p ! 30
2p
Elastic: E(p) = > 1
p ! 30
-2p < p - 30 (p - 30 < 0 reverses inequality)
-3p < -30
p > 10;
10 < p < 30
2p
Inelastic: E(p) = < 1
p ! 30
-2p > p - 30 (p - 30 < 0 reverses inequality)
-3p > -30
p < 10;
0 < p < 10
EXERCISE 4-7
201
23. x = f(p) =
144 " 2p , 0 ≤ p ≤ 72
f'(p) =
1
(144 - 2p)-1/2(-2) =
2
!
Elasticity
of demand: E(p) =
Elastic: E(p) =
p
144 ! 2 p
p
> !1
144 ! 2 p
p > 144 - 2p
3p > 144
p > 48,
48 < p < 72
Inelastic: E(p) =
p
< 1
144 ! 2 p
p < 144 - 2p
3p < 144
p < 48,
0 < p < 48
2, 500 " 2p2
25. x = f(p) =
"1
144 " 2p
0 ≤ p ≤ 25 2
1
!2 p
(2,500 - 2p2)-1/2(-4p) =
2
(2, 500 ! 2p 2)1 2
!
!
2p 2
p2
Elasticity of demand: E(p) =
=
2,500 ! 2p 2
1,250 ! p 2
f'(p) =
Elastic: E(p) =
p2
> 1
1,250 ! p 2
p2 > 1,250 - p2
2p2 > 1,250
p2 > 625
p > 25,
Inelastic: E(p) =
25 < p < 25 2
p2
< 1
1,250 ! p 2
!
p2 < 1,250 - p2
2p2 < 1,250
p2 < 625
p < 25,
202
CHAPTER 4
0 < p < 25
ADDITIONAL DERIVATIVE TOPICS
27. x = f(p) = 20(10 - p)
0 ≤ p ≤ 10
R(p) = pf(p) = 20p(10 - p) = 200p - 20p2
R'(p) = 200 - 40p
Critical value: R'(p) = 200 - 40p = 0; p = 5
Sign chart for R'(p):
Test Numbers
+ + + + 0 - - - R'(p)
p
R'(p)
x
0
200(+)
R(p)
0
5
10
10 !200(!)
Increasing
Demand: Inelastic
Decreasing
Elastic
29. x = f(p) = 40(p - 15)2
0 ≤ p ≤ 15
R(p) = pf(p) = 40p(p - 15)2
R'(p) = 40(p - 15)2 + 40p(2)(p - 15)
= 40(p - 15)[p - 15 + 2p]
= 40(p - 15)(3p - 15)
= 120(p - 15)(p - 5)
Critical values [in (0, 15)]: p = 5
Sign chart for R'(p):
Test Numbers
p
R'(p)
R'(p)
+ + + 0 - - - - 0
(+)
x
10
(!)
R(p)
0
5
Increasing
Demand: Inelastic
10
15
Decreasing
Elastic
31. x = f(p) = 30 - 10 p
0 ≤ p ≤ 9
R(p) = pf(p) = 30p - 10p p
1
R'(p) = 30 - 10 p - 10p · p-1/2
2
!
5p
= 30 - 10 p! = 30 - 15 p
p
Critical !values: R'(p) = 30 - 15 p = 0
p = 2; p = 4
!
!
Sign chart for !R'(p):
Test Numbers
R'(p)
+ + + 0 - - !
p R'(p)
x!
0 30(+)
R(p)
0
Increasing
Demand: Inelastic
4
9
5
(!)
Decreasing
Elastic
EXERCISE 4-7
203
33. p = g(x) = 50 - 0.1x
g'(x) = -0.1
g(x)
50 ! 0.1x
500
E(x) = = - 1
=
xg '(x)
!0.1x
x
500
3
E(200) =
- 1 =
200
2
35. p = g(x) = 50 - 2 x
1
g'(x) = x
g(!x)
50 " 2 x
50
=
E(x) = = - 2
#% 1 &(
xg '(x)
x
x "
$
x'
!
1
50
E(400) =
- 2 =
2
20
-k
37. x = f(p) = Ap
! , A, k positive constants
f'(p) = -Akp-k-1
! pf '(p)
Akp ! k
E(p) =
=
= k
f(p)
Ap !k
39. The company's daily cost is increasing by
1.25(20) = $25 per day.
41. x + 400p = 2,000
x = f(p) = 2,000 - 400p
f'(p) = -400
Elasticity of demand: E(p) =
400p
p
=
2,000 ! 400 p
5 ! p
2
< 1
3
The demand is inelastic; a price increase will increase revenue.
E(2) =
43. x + 1,000p = 800
x = f(p) = 800 - 1,000p
f'(p) = -1,000
Elasticity of demand: E(p) =
1,000p
5p
=
800 ! 1,000 p
4 ! 5p
1.5
1.5
3
< 1
=
=
4 ! 1.5
2.5
5
The demand is inelastic; a price decrease will decrease revenue.
E(0.30) =
45. From Problem 41, R(p) = pf(p) = 2,000p - 400p2
R'(p) = 2,000 - 800p
Critical values: R'(p) = 2,000 - 800p = 0
800p = 2000
p = 2.50
R"(p) = -800
Since p = 2.50 is the only critical value and R"(2.50) = -800 < 0,
the maximum revenue occurs when the price p = $2.50.
204
CHAPTER 4
ADDITIONAL DERIVATIVE TOPICS
y
47.
f(t) = 0.34t + 14.6, 0 ≤ t ≤ 50
f'(t) = 0.34
3
Percentage rate of change:
100
f '(t)
34
=
f(t)
0.34t + 14.6
50
t
49. r(t) = 11.3 - 3.6 ln t
3.6
r'(t) = t
!3.6
f '(t)
t
Relative rate of change of f(t):
=
f(t)
11.3 ! 3.6 ln t
!3.6
=
= C(t)
11.3t ! 3.6 t ln t
!3.6
Relative rate of change in 2002: C(12) =
≈ -0.13
11.3(12) ! 3.6(12)ln(12)
The relative rate of change for robberies annually per 1,000
population age 12 and over is approximately -0.13.
CHAPTER 4 REVIEW
1. A(t) = 2000e0.09t
A(5) = 2000e0.09(5) = 2000e0.45 ≈ 3136.62 or $3136.62
A(10) = 2000e0.09(10) = 2000e0.9 ≈ 4919.21 or $4919.21
A(20) = 2000e0.09(20) = 2000e1.8 ≈ 12,099.29 or $12,099.29
d
d
d x
2
(2 ln x + 3ex) = 2
ln x + 3
e =
+ 3ex
dx
dx
dx
x
d 2x-3
d
3.
e
= e2x-3
(2x - 3)
(by the chain rule)
dx
dx
= 2e2x-3
2.
4. y = ln(2x + 7)
1
y' =
(2)
2x + 7
2
=
2x + 7
(4-1)
(4-2)
(4-4)
(by the chain rule)
(4-4)
5. y = ln u, where u = 3 + ex.
(A) y = ln[3 + ex]
1
ex
dy
dy
du
1 x
x
(e ) =
(B)
=
·
=
(e ) =
3 + ex
3 + ex
dx
du
dx
u
(4-4)
CHAPTER 4 REVIEW
205