6724180-Solutions-Manual

# 6724180-Solutions-Manual - CHAPTER 2 THE MATHEMATICS OF...

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1 CHAPTER 2 THE MATHEMATICS OF OPTIMIZATION The problems in this chapter are primarily mathematical. They are intended to give students some practice with taking derivatives and using the Lagrangian techniques, but the problems in themselves offer few economic insights. Consequently, no commentary is provided. All of the problems are relatively simple and instructors might choose from among them on the basis of how they wish to approach the teaching of the optimization methods in class. Solutions 2.1 2 2 ( , ) 4 3 U x y x y a. 8 6 U U = x , = y x y b. 8, 12 c. 8 6 U U dU dx + dy = x dx + y dy x y d. for 0 8 6 0 dy dU x dx y dy dx 8 4 6 3 dy x x = = dx y y e. 1, 2 4 1 3 4 16   x y U f. 4(1) 2/3 3(2)   dy dx g. U = 16 contour line is an ellipse centered at the origin. With equation 2 2 4 3 16 x y , slope of the line at ( x, y ) is 4 3   dy x dx y . 2.2 a. Profits are given by 2 2 40 100   R C q q * 4 40 10   d q q dq 2 * 2(10 40(10) 100 100 )   b. 2 2 4   d dq so profits are maximized c. 70 2 dR MR q dq

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2 Solutions Manual 2 30 dC MC q dq so q * = 10 obeys MR = MC = 50. 2.3 Substitution: 2 1 so y x f xy x x 1 2 0 f x x 0.5 0.5, 0.25 x = , y = f = Note: 2 0     f . This is a local and global maximum. Lagrangian Method: ? 1 ) xy x y £ = y = 0 x £ = x = 0 y so, x = y. using the constraint gives 0.5, 0.25 x y xy 2.4 Setting up the Lagrangian: ? 0.25 ) x y xy . £ 1 £ 1 y x x y So, x = y . Using the constraint gives 2 0.25, 0.5 xy x x y . 2.5 a. 2 ( ) 0.5 40   f t gt t * 40 40 0,   df g t t dt g . b. Substituting for t* , * 2 ( ) 0.5 (40 ) 40(40 ) 800   f t g g g g . * 2 ( ) 800   f t g g . c. 2 * 1 ( ) 2   f t g depends on g because t * depends on g . so * 2 2 2 40 800 0.5( ) 0.5( )     f t g g g .
Chapter 2/The Mathematics of Optimization 3 d. 800 32 25, 800 32.1 24.92 , a reduction of .08. Notice that 2 2 800 800 32 0.8   g so a 0.1 increase in g could be predicted to reduce height by 0.08 from the envelope theorem. 2.6 a. This is the volume of a rectangular solid made from a piece of metal which is x by 3 x with the defined corner squares removed. b. 2 2 3 16 12 0 V x xt t t . Applying the quadratic formula to this expression yields 2 2 16 256 144 16 10.6 0.225 , 1.11 24 24 x x x x x t x x . To determine true maximum must look at second derivative -- 2 2 16 24   V x t t which is negative only for the first solution. c. If 3 3 3 3 0.225 , 0.67 .04 .05 0.68 t x V x x x x so V increases without limit.

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