solutions_to_the_practice_problems_for_the_final_examination

solutions_to_the_practice_problems_for_the_final_examination...

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SOLUTIONS TO THE PRACTICE PROBLEMS FOR THE FINAL EXAMINATION (1) MR = 100 – 2Q. MC = 5. MR = MC implies 100 – 2Q = 5. So, Q M = 47.50. Thus P M = 100 – (47.5), and so P M = 52.5. Profit = (P M × Q M ) – TC = (52.5)(47.5) – [ 10 + 5(47.5)] = 2246.25. (2) Since ε 1 1 + = MC P , we have ( 29 5 1 1 10 $ - + = J P and ( 29 2 1 1 10 $ - + = US P . So, P = $12.5, and P = $20. (3) Profit-Maximization Condition for Third-Degree Price Discrimination: MR 1 = MC and MR 2 = MC . So, 100 – 2Q = 30 and 120 – 4Q = 30 which imply Q 1 * = 35 and Q 1 * = 22.5. Thus, P 1 * = 100 – (35) = 65 and P 2 * = 120 – 2(22.5) = 75. (4) Given : A B B B A A q q MR q q MR - - = - - = 2 150 ; 2 150 . (a) Cournot Duopoly Profit-Maximization Conditions: MR A = MC A and MR B = MC B . So, B A B A q q q q 5 . 0 60 30 2 150 - = = - - , the “best-response” function of Firm A. Similarly, A B A B q q q q 5 . 0 60 30 2 150 - = = - - , the “best-response” function of Firm B. By substitution we have ( 29 40 5 . 0 60 5 . 0 60 * = - - = A A A q q q . So, ( 29 40 40 5 . 0 60 * = - = B q . Thus, 80 * * * = + = B A q q Q and ( 29 70 80 150 * = - = P . Finally, 1600 $ * * * * = × - × = B A A q AC q P π and similarly . 1600 $ * = B (b) Cartel (Joint-Monopoly) Profit-Maximization Condition:
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