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尼克尔森微观经济学课后答案CHAPTER 2THE MATHEMATICS OF OPTIMIZATIONThe 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.Solutions2.1 22(,)43=+U x y x y a.86 U U = x , = y x y b.8, 12 c. 86??=?? U U dU dx + dy = x dx + y dy x y d. for 0 8 6 0=+=dy dU x dx y dy dx 8463--dy x x = = dx y ye.1,2413416===?+?=x y U f.4(1)2/33(2)-==-dy dx g. U = 16 contour line is an ellipse centered at the origin.With equation 224316+=x y , slope of the line at (x, y ) is43=-dy x dx y.2.2 a. Profits are given by 2240100π=-=-+-R C q q*44010π=-+=d q q dq2*2(1040(10)100100)π=-+-= b. 224π=-d dq so profits are maximized c. 702==-dR MR q dq 230==+dC MCq dq so q * = 10 obeys MR = MC = 50.2.3 Substitution:21 so =-==-y x f xy x x120?=-=?f x x0.50.5,0.25x =, y = f =Note: 20''=-Lagrangian Method: ?1)λ=+--xy x y￡λ?-? = y = 0x￡λ?-? = x = 0yso, x = y.using the constraint gives 0.5,0.25===x y xy2.4 Setting up the Lagrangian: ?0.25)λ=++-x y xy .￡1￡1λλ?=-??=-?y x x ySo, x = y . Using the constraint gives 20.25,0.5====xy x x y .2.5 a. 2()0.540=-+f t gt t*40400,=-+==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*222408000.5()0.5()?-=-=-=?f t g g g . d. 8003225,80032.124.92==, a reduction of .08. Notice that 22800800320.8-=≈-g so a 0.1 increase in g could bepredicted 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 ofmetal which is x by 3x with the defined corner squares removed. b. 22316120?=-+=?V x xt t t. Applying the quadratic formula to this expressionyields1610.60.225, 1.1124±===x x t x x . Todetermine true maximum must look at second derivative --221624?=-+?V x t twhich is negative only for the first solution.c. If 33330.225,0.67.04.050.68=≈-+≈t x V x x x x so V increaseswithout limit.d. This would require a solution using the Lagrangian method. Theoptimal solution requires solving three non-linear simultaneousequations —a task not undertaken here. But it seems clear thatthe solution would involve a different relationship between tand x than in parts a-c.2.7 a. Set up Lagrangian 1212?ln ()λ=++--x x k x x yields the first order conditions: 12212￡10?0￡0λλλ?=-=??=-=??=--=?x x x k x xHence, 2215 or 5λ===x x . With k = 10, optimal solution is 12 5.==x xb. With k = 4, solving the first order conditions yields215, 1.==-x xc. Optimal solution is 120,4,5ln 4.===x x y Any positive value forx 1 reduces y.d. If k = 20, optimal solution is 1215, 5.==x x Because x 2provides a diminishing marginal increment to y whereas x 1 doesnot, all optimal solutions require that, once x 2 reaches 5, anyextra amounts be devoted entirely to x 1.2.8 The proof is most easily accomplished through the use of the matrixalgebra of quadratic forms. See, for example, Mas Colell et al.,pp. 937–939. Intuitively, because concave functions lie belowany tangent plane, their level curves must also be convex. Butthe converse is not true. Quasi-concave functions may exhibit“increasing returns to scale”; even though their level curvesare convex, they may rise above the tangent plane when allvariables are increased together.2.9 a.11210.βαα-=> f x x 11220βαβ-=> f .x x21111(1)0.βααα-=-< f x x21222(1)0.βαββ-=-< f x x111212210.βααβ--==> f f x xClearly, all the terms in Equation 2.114 are negative. b.If 12βα==y c x x /1/21αββ-= x c x since α, β > 0, x 2 is a convex function of x 1 .c.Using equation 2.98, 222222222221122111222(1)()(1)ββααααβββα-----=--- f f f x x x x= 222212(1)βααββα---- x x which is negative for α + β > 1.2.10 a.Since 0,0'''>Because 1122,0y is quasi-concave as is γy . But γy is not concave for γ > 1. All of these results can be shown by applying the various definitions to the partial derivatives of y .CHAPTER 3PREFERENCES AND UTILITYThese problems provide some practice in examining utility functions by looking at indifference curve maps. The primary focus is on illustrating the notion of a diminishing MRS in various contexts. The concepts of the budget constraint and utility maximization are not used until the next chapter.Comments on Problems3.1 This problem requires students to graph indifference curves for a varietyof functions, some of which do not exhibit a diminishing MRS.3.2 Introduces the formal definition of quasi-concavity (from Chapter 2) to beapplied to the functions in Problem 3.1.3.3 This problem shows that diminishing marginal utility is not required toobtain a diminishing MRS. All of the functions are monotonic transformations of one another, so this problem illustrates that diminishing MRS is preserved by monotonic transformations, but diminishing marginal utility is not.3.4 This problem focuses on whether some simple utility functions exhibit convexindifference curves.3.5 This problem is an exploration of the fixed-proportions utility function.The problem also shows how such problems can be treated as a composite commodity.3.6 In this problem students are asked to provide a formal, utility-basedexplanation for a variety of advertising slogans. The purpose is to get students to think mathematically about everyday expressions.3.7 This problem shows how initial endowments can be incorporated into utilitytheory.3.8This problem offers a further exploration of the Cobb-Douglas function.Part c provides an introduction to the linear expenditure system. This application is treated in more detail in the Extensions to Chapter 4.。
