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Unformatted text preview: Solutions for Suggested Problems to Chp 2 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 = 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) (c) Max Profit = Profit(D=2425) = 97(2425) 0.02(2425)2 42,000 = $75,612.50 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 = Total Revenue  Total Cost = (600  0.05D)D  (900,000 + 131.5D) = 468.5D 0.05D2  900,000 d/dD = 468.5 0.1D D* = 4,685 units per month = Profit(D=4685) = $197,461 (b) Find the range of values for D such that > 0. Solving 468.5D 0.05D2  900,000 = 0 gives us two values, D = 2,698 and D = 6,672. 2,698  + 6,672  D Sign Therefore, domain of profitable demand is (2698, 6672). 4. (215) (a) p = 38 + 2700/D 5000/D2 f = 1,000 v = 40 = Total Revenue Total Cost = (38 + 2700/D 5000/D2)D (1000 + 40D) = 2D + 1700 5000/D d/dD = 2 + 5000/D2 D* = 50 units per month (b) d2/dD2 = 10000/D3 d2/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 = (1,000 + 1,200x 200x)(3 0.5x) = 3,000 + 2,500x 500x2 x* = 2.5 weeks Profit = $6,125 7. = 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 0 20 2 29.33 8 1.33 12 68 The firm should manufacture 12 units of product each day. ...
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This note was uploaded on 02/07/2012 for the course IE 343 taught by Professor Vincent,g during the Spring '08 term at Purdue UniversityWest Lafayette.
 Spring '08
 Vincent,G

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