Solutions for Suggested Problems to Chp 2
1.
(24)
Cost
Site A
Site B
Rent
= $5,000
= $100,000
Hauling
(4)(200,000)($1.50) = $1,200,000
(3)(200,000)($1.50) = $900,000
Total Cost
$1,205,000
$1,000,000
Note that the revenue of $8.00/yd
3
is independent of the site selected. Thus, we
can maximize profit by minimizing total cost.
The solid waste site should be
located in Site B.
2.
(212)
(a)
p
= 150  0.02D
v =
53
f = 42,000
π
= Total Revenue  Total Cost
= (150  0.02D)D  (42,000 + 53D)
= 97D  0.02D
2
 42,000
D
*
= 2425
units
Since the second derivative is negative, profit has been maximized at D
*
.
(b)
Max Profit = Profit(D=2425) = 97(2425) – 0.02(2425)
2
– 42,000
= $75,612.50
(c)
Profit = 97D  0.02D
2
 42,000 = 0
Solving quadratically, we obtain two solutions:
D = 481, or, D = 4,369.
Our maximum output is 4,000, so the only
breakeven is at D = 481.
(d)
The range of profitable demand is (481, 4,000]. Breakeven occurs at 481
units, and maximum output is 4000. So by (c), producing in this range
ensures profit.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
3.
(214)
(a)
p
= 600  0.05D
f
= 900,000
v
= 131.50
π
= Total Revenue  Total Cost
= (600  0.05D)D  (900,000 + 131.5D)
= 468.5D – 0.05D
2
 900,000
d
π
/dD = 468.5 – 0.1D
D
*
= 4,685 units per month
π
= Profit(D=4685) = $197,461
(b)
Find the range of values for D such that
π
> 0.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Vincent,G
 Economics, Microeconomics

Click to edit the document details