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Unformatted text preview: AEM 4240 Problem Set #2 Answer Key 1. Econ review warm-up a) Hugo has decided to minimize his total costs. How much should he produce? What are his total profits? TC increases with Q at all positive values of Q. TC therefore reaches a minimum at the lowest feasible value of Q, which is 0, at which point profits = -405. b) Hugo is back in the shop. He understands the importance of amortizing the daily lease cost of the machine over a large production run. His brother-in-law, the VP of marketing firm, has convinced Hugo of the importance of dominating the market and getting as much market share as he can with his existing machine. Hugo decides to follow this friendly advice, believing that not only will he make great heaps of money but he will also achieve much more pleasant conditions at the next gathering with his in-laws. How much does he instruct you to produce in order to maximize sales? What are his total profits? We are given a market price of hugonuts that is not a function of Q. This implies that no matter how high Q is, Hugo will be able to sell all of his production at $120. So to maximize sales, Hugo simply sets Q as high as possible. Since Hugo’s capacity is limited at 100 units by his equipment, he maximizes sales by producing 100 units. At this level, profits are 120*100-(405+20*100+5*100 2 )= -$40,405. It won’t be a pleasant gathering after all. c) You notice Hugo has sprouted a few more gray hairs. Coincidentally, the marketing VP was called away to investigate a potential new client in Tierra del Fuego. Hugo has given the matter more thought, and instructs you to minimize the average cost of production. How much do you produce? What are your profits? You have a choice of two methods. At its minimum, you know AC is equal to MC. If the marginal cost of producing the next unit were below average cost, then AC would still be falling. If the marginal cost of producing the next unit were above the average cost, then AC would be increasing. So AC must reach a minimum where AC=MC. AC is found by dividing TC by Q, yielding AC=405/Q + 20 + 5Q. MC is found by taking the derivative of TC with respect to Q, yielding MC=20+10Q. Setting MC=AC and solving for Q you find that Q=9. Alternatively, AC reaches a minimum where its derivative with respect to Q is zero: AC = (405 + 20Q + 5Q 2 )/Q = 405/Q + 20 + 5Q Take the derivative of this expression: ∂ AC/ ∂ Q = -405/Q 2 + 5 Set it equal to zero and solve for Q: -405/Q 2 + 5 = 0 405/Q 2 = 5 81 = Q 2 Q = 9 So average cost is minimized at Q=9. Profits are 120*9-405-20*9-5* 9 2 =$90. d) Hugo is looking a bit better. He almost smiles now, especially when he shows everyone those spectacular post-cards of the fog and ice his brother-in-law keeps sending from Tierra del Fuego. You spot your opportunity and recommend that the time has come to maximize profits. Hugo is feeling so good, that he follows your advice. How much do you produce? What are your profits?...
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This note was uploaded on 03/03/2009 for the course AEM 4240 taught by Professor Blalock,g. during the Fall '07 term at Cornell.
- Fall '07