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Suggested problems_Chp 2 sol

# Suggested problems_Chp 2 sol - Solutions for Suggested...

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Solutions for Suggested Problems to Chp 2 1. (2-4) 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. (2-12) (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.

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3. (2-14) (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.
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Suggested problems_Chp 2 sol - Solutions for Suggested...

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