Unformatted text preview: Hw1 Solutions 1. (24)
Cost
Rent
Hauling
Total Cost Site A
= $5,000
(4)(200,000)($1.50) = $1,200,000
$1,205,000 Site B
= $100,000
(3)(200,000)($1.50) = $900,000
$1,000,000 Note that the revenue of $8.00/yd3 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  0.02D2  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  0.02D2  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. 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 – 0.05D2  900,000
dπ/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.
Solving 468.5D – 0.05D2  900,000 = 0 gives us two values, D = 2,698 and
D = 6,672. D
Sign 2,698
— 6,672
+ — Therefore, domain of profitable demand is (2698, 6672).
4. (215)
(a) p = 38 + 2700/D – 5000/D2
f = 1,000
v = 40
π = Total Revenue – Total Cost
= (38 + 2700/D – 5000/D2)D – (1000 + 40D)
= 2D + 1700 – 5000/D
dπ/dD = –2 + 5000/D2
D* = 50 units per month (b) d2π/dD2 = –10000/D3
d2π/dD2(D=50) = –0.08 < 0
Therefore, at D = 50, profit is maximized. 5. (217) Breakeven point in units of production: f = $100,000/yr ; Cv = $140,000/yr (70% of capacity)
Sales = $280,000/yr (70% of capacity); p = $40/unit
Annual Sales (units) = $280,000/$40 = 7,000 units/yr (70% capacity)
v = $140,000/7,000 = $20/unit
Breakeven point: TR = TC
pD = f+ vD
40D = 100,000 + 20D
D = 5,000 units/year, or 50% utilization of capacity 6. (225)
Let x = the number of weeks to wait. x = 0, 1, 2, …
Quantity(x) = 1,000 + 1,200x – 200x
Price(x) = $3.00  0.50x
π = (1,000 + 1,200x – 200x)(3 – 0.5x) = 3,000 + 2,500x – 500x2
x* = 2.5 weeks
Profit = $6,125 7. π = x3/3 – 5 x2 + 16x + 20
where x = amount of products produced in one day. The domain of x is [0, 12].
= x2 – 10x + 16 = 0
(x – 8)(x – 2) = 0
Local optima exist at x = 2 and x = 8. The global maximum is found by
evaluating the profit function at the local optima and at the endpoints of the
domain of x.
x
π 0
20 2
29.33 8
1.33 The firm should manufacture 12 units of product each day. 12
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 Fall '08
 Vincent,G

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