Microeconomia Intermedia Y Sus Aplicaciones ( Walter Nicholson) ( 1). ( Walter Nicholson) ( 1) Topics Nicholson Collection opensource Language Spanish. Microeconomia I. Internet Archive HTML5 Uploader 1.4.2. Plus-circle Add Review. Reviews There are no reviews yet. Be the first one to write a review. 16,585 Views. MICROECONOMIC THEORY BASIC PRINCIPLES AND EXTENSIONS TENTH EDITION This page intentionally left blank MICROECONOMIC THEORY BASIC PRINCIPLES AND EXTENSIONS TENTH EDITION Walter Nicholson Amherst Col lege Christopher Snyder Dartmouth College VP/Editorial Director: Jack W.
1CHAPTER 2THE MATHEMATICS OF OPTIMIZATIONThe problems in this chapter are primarily mathematical. They are intended to give students some practicewith taking derivatives and using the Lagrangian techniques, but the problems in themselves offer feweconomic insights. Consequently, no commentary is provided. All of the problems are relatively simpleand instructors might choose from among them on the basis of how they wish to approach the teaching ofthe optimization methods in class.Solutions2.12 2(, ) 4 3uf03d uf02bU x y x ya.8 6uf0b6 uf0b6uf0b6 uf0b6U U= x, = yx yb. 8, 12c.8 6uf0b6 uf0b6uf03duf0b6 uf0b6U UdU dx + dy = x dx + y dyx yd.for 0 8 6 0uf03d uf02b uf03ddydU x dx y dydx8 46 3uf02d uf02ddy x x= =dx y ye.1, 2 4 1 3 4 16uf03d uf03d uf03d uf0d7 uf02b uf0d7 uf03dx y Uf.4(1)2 / 33(2)uf02duf03d uf03d uf02ddydxg.
U = 16 contour line is an ellipse centered at the origin. With equation2 24 3 16uf02b uf03dx y, slope of the line at (x, y) is43uf03d uf02ddy xdx y.2.2 a. Profits are given by22 40 100uf070 uf03d uf02d uf03d uf02d uf02b uf02dR C q q.4 40 10uf070uf03d uf02d uf02b uf03ddq qdq2. 2(10 40(10) 100 100)uf070 uf03d uf02d uf02b uf02d uf03db. 224uf070uf03d uf02dddqso profits are maximizedc.70 2uf03d uf03d uf02ddRMR qdq2 uf076 Solutions Manual2 30uf03d uf03d uf02bdCMC qdqso q. = 10 obeys MR = MC = 50.2.3 Substitution:21 so uf03d uf02d uf03d uf03d uf02dy x f xy x x1 2 0uf0b6uf03d uf02d uf03duf0b6fxx0.5 0.5, 0.25x =, y = f =Note:2 0uf0a2uf0a2 uf03d uf02d uf03cf.
This is a local and global maximum.Lagrangian Method:? 1 )uf06cuf03d uf02b uf02d uf02dxy x y£uf06cuf0b6uf02duf0b6= y = 0x£uf06cuf0b6uf02duf0b6= x = 0yso, x = y.using the constraint gives0.5, 0.25uf03d uf03d uf03dx y xy2.4 Setting up the Lagrangian:?
0.25 )uf06cuf03d uf02b uf02b uf02dx y xy.£1£1uf06cuf06cuf0b6uf03d uf02duf0b6uf0b6uf03d uf02duf0b6yxxySo, x = y. Using the constraint gives2 0.25, 0.5uf03d uf03d uf03d uf03dxy x x y.2.5 a.2( ) 0.5 40uf03d uf02d uf02bf t gt t. 4040 0,uf03d uf02d uf02b uf03d uf03ddfg t tdt g.b. Substituting for t.,. 2( ) 0.5 (40 ) 40(40 ) 800uf03d uf02d uf02b uf03df t g g g g.2( ) 800uf0b6uf03d uf02duf0b6f tgg.c.2.1 ( )2uf0b6uf03d uf02duf0b6ftgdepends on g because t.depends on g.so. 2 2240 8000.5( ) 0.5( )uf0b6 uf02duf03d uf02d uf03d uf02d uf03duf0b6ftg g g.Chapter 2/The Mathematics of Optimization uf076 3d.800 32 25, 800 32.1 24.92uf03d uf03d, a reduction of.08. Notice that2 2800 800 32 0.8uf02d uf03d uf0bb uf02dgso a 0.1 increase in g could be predicted to reduceheight 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 3xwith the defined corner squares removed.b.2 23 16 12 0uf0b6uf03d uf02d uf02b uf03duf0b6Vx xt tt. Applying the quadratic formula to this expression yields2 216 256 144 16 10.60.225, 1.1124 24uf0b1 uf02d uf0b1uf03d uf03d uf03dx x x x xt x x. To determine truemaximum must look at second derivative - 2216 24uf0b6uf03d uf02d uf02buf0b6Vx ttwhich is negative onlyfor the first solution.c. If3 3 3 30.225, 0.67.04.05 0.68uf03d uf0bb uf02d uf02b uf0bbt x V x x x xso V increases without limit.d. This would require a solution using the Lagrangian method. The optimal solutionrequires solving three non-linear simultaneous equationsu2014a task not undertaken here.But it seems clear that the solution would involve a different relationship between t andx than in parts a-c.2.7 a.
Set up Lagrangian1 2 1 2? Ln ( )uf06cuf03d uf02b uf02b uf02d uf02dx x k x xyields the first order conditions:12 21 2£1 0?0£0uf06cuf06cuf06cuf0b6uf03d uf02d uf03duf0b6uf0b6uf03d uf02d uf03duf0b6uf0b6uf03d uf02d uf02d uf03duf0b6xx xk x xHence,2 21 5 or 5uf06c uf03d uf03d uf03dx x. With k = 10, optimal solution is1 2 5.uf03d uf03dx xb. With k = 4, solving the first order conditions yields2 15, 1.uf03d uf03d uf02dx xc.
Optimal solution is1 20, 4, 5ln 4.uf03d uf03d uf03dx x yAny positive value for x1 reduces y.d. If k = 20, optimal solution is1 215, 5.uf03d uf03dx xBecause x2 provides a diminishingmarginal increment to y whereas x1 does not, all optimal solutions require that, once x2reaches 5, any extra amounts be devoted entirely to x1.2.8 The proof is most easily accomplished through the use of the matrix algebra of quadraticforms. See, for example, Mas Colell et al., pp. Intuitively, because concavefunctions lie below any tangent plane, their level curves must also be convex. But theconverse is not true.
Quasi-concave functions may exhibit u2015increasing returns to scaleu2016;even though their level curves are convex, they may rise above the tangent plane when allvariables are increased together.4 uf076 Solutions Manual2.9 a.11 21 0.uf062uf061uf061 uf02duf03d uf03e f x x11 22 0uf062uf061uf062 uf02duf03d uf03e f.x x21 111 ( 1) 0.uf062uf061uf061 uf061 uf02duf03d uf02d uf03c f x x21 222 ( 1) 0.uf062uf061uf062 uf062 uf02duf03d uf02d uf03c f x x111 212 21 0.uf062uf061uf061 uf062 uf02duf02duf03d uf03d uf03e f f x xClearly, all the terms in Equation 2.114 are negative.b. If1 2uf062uf061uf03d uf03dy c x x/1/2 1uf061 uf062uf062 uf02duf03d x c xsince u3b1, u3b2 0, x2 is a convex function of x1.c. Using equation 2.98,2 22 2 2 22 2 2 2 21 12 211 1222( 1) ( ) ( 1) uf062 uf062uf061 uf061uf061 uf061 uf062 uf062 uf062uf061uf02d uf02duf02d uf02duf02d uf03d uf02d uf02d uf02d f f f x x x x=2 22 21 2(1 )uf062uf061uf061 uf062 uf062 uf061 uf02duf02duf02d uf02d x x which is negative for u3b1 + u3b2 1.2.10 a. Since0, 0uf0a2 uf0a2uf0a2uf03e uf03cy y, the function is concave.b. Because11 22, 0uf03cf f, and12 21 0uf03d uf03df f, Equation 2.98 is satisfied and the functionis concave.c. Y is quasi-concave as isuf067y. Butuf067yis not concave for u3b3 1.
All of these resultscan be shown by applying the various definitions to the partial derivatives of y.5CHAPTER 3PREFERENCES AND UTILITYThese problems provide some practice in examining utility functions by looking at indifferencecurve maps. The primary focus is on illustrating the notion of a diminishing MRS in variouscontexts. The concepts of the budget constraint and utility maximization are not used until the nextchapter.Comments on Problems3.1 This problem requires students to graph indifference curves for a variety of functions,some of which do not exhibit a diminishing MRS.3.2 Introduces the formal definition of quasi-concavity (from Chapter 2) to be applied to thefunctions in Problem 3.1.3.3 This problem shows that diminishing marginal utility is not required to obtain adiminishing MRS. All of the functions are monotonic transformations of one another, sothis 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 alsoshows how such problems can be treated as a composite commodity.3.6 In this problem students are asked to provide a formal, utility-based explanation for avariety of advertising slogans. The purpose is to get students to think mathematicallyabout everyday expressions.3.7 This problem shows how initial endowments can be incorporated into utility theory.3.8 This problem offers a further exploration of the Cobb-Douglas function. Part c providesan introduction to the linear expenditure system. This application is treated in more detailin the Extensions to Chapter 4.3.9 This problem shows that independent marginal utilities illustrate one situation in whichdiminishing marginal utility ensures a diminishing MRS.3.10 This problem explores various features of the CES function with weighting on the twogoods.6 uf076 Solutions ManualSolutions3.1 Here we calculate the MRS for each of these functions:a.3 1uf03d uf03dx yMRS f fu2014 MRS is constant.b. 0.50.50.5( )0.5( )uf02duf03d uf03d uf03dx yy xMRS f f y xy xu2014 MRS is diminishing.c.0.50.5 1uf02duf03d uf03dx yMRS f f xu2014 MRS is diminishingd.2 2 0.5 2 2 0.50.5( ) 2 0.5(.
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