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# ch05_solutions_solved edit - Solutions to Chapter 5...

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Solutions to Chapter 5 Exercises SOLVED EXERCISES S1. (a) R’s best-response rule is given by y = 10√x – x. L spends \$16 million, so x = 16. Then R’s best response is y = 10√16 – 16 = 10(4) – 16 = 40 – 16 = 24, or \$24 million. (b) R’s best response is y = 10√x – x, and L’s best response is x = 10√y – y. Solve these simultaneously: x = 10(10√x – x) 1/2 – 10√x + x √x = (10√x – x) 1/2 x = 10√x – x 2x = 10√x √x = 5 x = 25 y = 10√25 – 25 = 25 S2. (a) Xavier’s costs have not changed, nor have the demand functions, so Xavier’s best- response rule is still the same as in Figure 5.1: P x = 15 + 0.25P y . Yvonne’s new profit function is B y = (P y – 2)Q y = (P y – 2)(44 – 2P y + P x ) = – 2(44 + P x ) + (4 + 44 + P x )P y – 2(P y ) 2 . Rearranging or differentiating with respect to P y leads to Yvonne’s new best-response rule: P y = 12 + 0.25P x . Solving the two response rules simultaneously yields P x = 19.2 and P y = 16.8. (b) See the graph below. Yvonne’s best-response curve has shifted down; it has the same slope but a new, lower intercept (12 rather than 15). Yvonne is able to charge lower prices due to lower costs. The new intersection point occurs at (19.2, 16.8) as calculated above. Solutions to Chapter 5 Solved Exercises 1 of 11

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S3. (a) La Boulangerie’s profit is: Y 1 = P 1 Q 1 – Q 1 = P 1 (14 – P 1 – 0.5P 2 ) – (14 – P 1 – 0.5P 2 ) = – P 1 2 + 15P 1 – 0.5P 1 P 2 + 0.5P 2 – 14. To find the optimal P 1 without using calculus, we refer to the result in the Appendix to Chapter 5, remembering that P 2 is a constant in this situation. Using the notation of the Appendix, we have A = 0.5P 2 – 14, B = 15 – 0.5P 2 , and C = 1, so the solution is: P 1 = B/(2C) = 15 – 0.5P 2 /(2), or P 1 = 7.5 – 0.25P 2 . This is La Boulangerie’s best-response function. You get the same answer by setting 1 1 / P Y = –2P 1 + 15 – 0.5P 2 = 0 and solving for P 1 . Similarly, La Fromagerie’s profit is: Solutions to Chapter 5 Solved Exercises 2 of 11
Y 2 = P 2 Q 2 – 2Q 2 = P 2 (19 – 0.5P 1 – P 2 ) – 2(19 – 0.5P 1 – P 2 ) = – P 2 2 + 21P 2 – 0.5P 1 P 2 + P 1 – 38. Again, using the notation in the Appendix, A = P 1 – 38, B = 21 – 0.5P 1 , and C = 1, which yields: P 2 = B/(2C) = 21 – 0.5P 1 /(2), or P 2 = 10.5 – 0.25P 1 . This is La Fromagerie’s best-response function. You get the same answer by setting 2 2 / P Y = – 2P 2 + 21 – 0.5P 1 = 0 and solving for P 1 . To find the solution for the equilibrium prices analytically, substitute La Fromagerie’s best- response function for P 2 into La Boulangerie’s best-response function. This yields P 1 = 7.5 – 0.25(10.5 – 0.25P 1 ), or P 1 = 5.2. Given this value for P1, you can find P2 = 10.5 – 0.25(5.2) = 9.2. The best-response curves are shown in the diagram below. Solutions to Chapter 5 Solved Exercises 3 of 11

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(b) Colluding to set prices to maximize the sum of profits means that the firms maximize the joint-profit function: Y = Y 1 + Y 2 = 16P 1 + 21.5P 2 – P 1 2 – P 2 2 – P 1 P 2 – 52.
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ch05_solutions_solved edit - Solutions to Chapter 5...

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