2009PS_3_Solutions - AEM 4150 Price Analysis Fall Semester...

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1 AEM 4150 Price Analysis Fall Semester 2009 Homework Assignment 3 Suggested Solutions Problem 1 (a) (1 point) Profit maximization for the monopolist implies that MC = MR TC = 4,000 + 20Q MC = dTC/dQ = 20 At profit max MC = MR Therefore, 20 = 300 – 0.4Q => Q* = 700 From demand curve P = 300 - 0.2Q = 300 - 0.2(700) P*= 160 Profit = TR – TC at P = 160 and Q = 700 = PQ – (4,000 + 20Q) = (160x700) – [4,000 + 20(700)] = 94,000 Graph (2 points for monopoly and perfect competition) (b) (1 point) In perfect competition the firm takes the price as given, in this case profit maximization implies that: P = MC P = 300 - 0.2Q = 20 P PC = 20 D MR P P M =160 Q M = 700 Q PC = 1400 MC=20
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2 => Q* = 1,400 and P* = 20 Profit = TR – TC at P = 20 and Q = 1,400 = PQ – (4,000 + 20Q) = (20x1,400) – [4,000 + 20(1,400)] = -4,000 = Loss See graph in part (a) Problem 2 (a) (1 point) Calculation of the MC of milk production: TC = 8 + 2Q MC = dTC/dQ = 2 (b) Fluid milk market (1 point) At profit max. MR = MC Therefore 15 – 4Q f = 2 or Q f *= 3.25 P f *= 15 – 2(3.25) = 8.5 Processing milk market (1 point) At profit max. MR = MC Therefore 10 – 1Q p = 2 or Q p *= 8 P p *= 10 – 0.5(8) = 6 They should therefore sell 3.25 units of milk at $8.5 per unit in the Fluid Milk Market and 8 units of milk at $6 per unit in the Processing Market. (c) Fluid milk buyers are paying more for their milk ($8.50 - $6.50 =
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2009PS_3_Solutions - AEM 4150 Price Analysis Fall Semester...

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