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Mathematical Economics Practice Problems and Solutions – Second Edition – G. Stolyarov II MatheMatical econoMics Practice ProbleMs and solutions Second Edition G. Stolyarov II, ASA, ACAS, MAAA, CPCU, ARe, ARC, API, AIS, AIE, AIAF First Edition Published in March-April 2008 Second Edition Published in July 2014 Note: Here, I will present solve problems typical of those offered in a mathematical economics or advanced microeconomics course. The problems were originally compiled by Dr. Charles N. Steele and are reprinted with his generous permission. The solutions to the problems are my own work and not necessarily the only way to solve the problems. Table of Contents Section Page Section 1: Profit Maximization in Mathematical Economics 2 Section 2: The Lagrangian Method of Constrained Optimization 4 Section 3: Intertemporal Allocation of a Depletable Resource: Optimization Using the Kuhn- 7 Tucker Conditions Section 4: Optimization with Inequality Constraints 9 Section 5: The Economics of Fisheries 13 Section 6: Additional Practice Problems Involving the Kuhn-Tucker Conditions 16 Section 7: Additional Problems on the Economics of Fisheries 18 Section 8: The Deacon Model of Forest Economics 20 Section 9: The Second-Order Conditions for Multiple Choice Variables 22 Section 10: Second-Order Conditions: Practice Problems and Solutions 24 Section 11: Expected Utility 26 Section 12: Principal-Agent Problems and Designing Contracts Under Asymmetric Information 31 About Mr. Stolyarov 35 © 2008, 2014, G. Stolyarov II. This work is distributed under a Creative Commons Attribution Share-Alike 3.0 Unported License. Permission to reprint this work, in whole or in part, is granted, as long as full credit is given to the author by identification of the author’s name, and no additional rights are claimed by the party reprinting the work, beyond the rights provided by the aforementioned Creative Commons License. In particular, no entity may claim the right to restrict other parties from obtaining copies of this work, or any derivative works created from it. Commercial use of this work is permitted, as long as the user does not claim any manner of exclusive rights arising from such use. While not mandatory, notification to the author of any commercial use or derivative works would be appreciated. Such notification may be sent electronically to gennadystolyarovii@gmail.com. 1 Mathematical Economics Practice Problems and Solutions – Second Edition – G. Stolyarov II Section 1 Profit Maximization in Mathematical Economics Problem 1. Suppose a firm faces a demand curve for its product P = a - bQ, and the firm's costs of production and marketing are C(Q) = cQ + d, where P is price, Q is quantity, and a, b, c, and d are positive constants. Find the following: a. The formula for profit Π in terms of Q. b. The first order condition (FOC) for maximum profit. c. The second order condition (SOC) for maximum profit. 2 2 Solution 1a. Π = TR - TC = PQ - C(Q) = aQ - bQ - cQ - d = Π = - bQ + (a-c)Q - d Solution 1b. FOC: dΠ/dQ = -2bQ + (a-c) ≡ 0. Thus, -2bQ = -(a-c) and Q = (a-c)/2b. 2 2 Solution 1c. SOC: d Π/dQ = -2b < 0, since it is given that b > 0. Thus, Q = (a-c)/2b is a maximum. Problem 2. Suppose the firm faces a demand curve for its product P = 32 - 2Q, and the firm's 2 costs of production and marketing are C(Q) = 2Q . Find the following. a. The formula for profit Π in terms of Q. b. The FOC and SOC for maximum total revenue. c. The price and quantity that maximize total revenue, and the corresponding value of total revenue. d. The FOC and SOC for maximum profit. e. The price and quantity that maximize profit, and the corresponding value of profit. 2 f. What would the competitive price and quantity be, assuming C(Q) = 2Q represented the industry cost function? 2 2 2 Solution 2a. Π = TR - TC = PQ - C(Q) = 32Q - 2Q - 2Q = Π = 32Q - 4Q 2 Solution 2b. TR = 32Q - 2Q 2 Mathematical Economics Practice Problems and Solutions – Second Edition – G. Stolyarov II FOC: d[TR]/dQ = 32 - 4Q ≡ 0. Thus, Q = 8. 2 2 SOC: d [TR]/dQ = -4 < 0. Thus, Q = 8 is a maximum. Solution 2c. The quantity that maximizes total revenue is Q = 8, according to the first and second-order conditions in Solution 2b. The price that maximizes total revenue is 32 - 2*8 = P = 16. Total revenue at this level is PQ = 16*8 = TR =128. We note that AVC here is 2Q = 2*8 = 16, so price is at least equal to average variable cost. Solution 2d. FOC: dΠ/dQ = 32 - 8Q = 0. Thus, Q = 4. 2 2 SOC:d Π/dQ = -8 < 0. Thus, Q = 4 is a maximum. Solution 2e. The quantity that maximizes profit is Q = 4, according to the first and second-order conditions in Solution 2d. The price that maximizes profit is 2 32 - 2*4 = P = 24. Total profit at this level is 32*4 - 4*4 = Π = 64. Here, 24 > 16, so P > AVC, and it is optimal for the firm to produce Q = 4. 2 Solution 2f. The firm will produce at P = MC, where P = 32 - 2Q. TC = 2Q , so MC = 4Q. Thus, 32 - 2Q = 4Q. Thus, 32 = 6Q and Q = 32/6 = Q = 16/3. P = 32 - 2(16/3) = P = 64/3 3 Mathematical Economics Practice Problems and Solutions – Second Edition – G. Stolyarov II Section 2 The Lagrangian Method of Constrained Optimization Note: Here, I will present solve problems typical of those offered in a mathematical economics or advanced microeconomics course. The problems were authored by Dr. Charles N. Steele and are reprinted with his generous permission. The solutions to the problems are my own work and not necessarily the only way to solve the problems. 3. Find the maximum values of the objective function F subject to the given constraint for each of the following, using the Lagrangian method. a. F(x, y) = xy, subject to 5x + 2y = 20 1/2 1/2 2 2 b. F(x, y) = 2x y subject to x + y = 8 2 2 2 c. F(x, y, z) = xyz subject to x + y + z = 12 2 2 2 d. F(x, y, z) = x + y + z subject to x + y + z = 12 Solution 3a. Lagrangian: L(x, y, λ) = xy + λ[20 - 5x - 2y] Lx = y - 5λ ≡ 0 Ly = x - 2λ ≡ 0 Lλ = 20 - 5x - 2y ≡ 0 Thus, 2λ = x and 5λ = y (from the transformed for Lx and Ly). So 20 - 5x - 2y = 20 - 5*2λ - 2*5λ = 20 - 20λ = 0, so 20 = 20λ and λ =1, whereby x = 2 and y = 5. 1/2 1/2 2 2 Solution 3b. Lagrangian: L(x, y, λ) = 2x y + λ[8 - x - y ] -1/2 1/2 Lx = x y - 2λx ≡ 0 1/2 -1/2 Ly = x y - 2λy ≡ 0 2 2 Lλ = 8 - x - y ≡ 0 4
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