范里安高级微观经济学(第三版)答案.doc

Answers toExercisesMicroeconomicAnalysisThird EditionHal R. VarianUniversity of California at BerkeleyW. W. Norton & Company New York LondonCopyright c1992, 1984, 1978 by W. W. Norton & Company, Inc.All rights reservedPrinted in the United States of AmericaTHIRD EDITION0-393-96282-2W. W. Norton & Company, Inc., 500 Fifth Avenue, New York, N.Y. 10110W. W. Norton Ltd., 10 Coptic Street, London WC1A 1PU2 3 4 5 6 7 8 9 0ANSWERSChapter 1. Technology1.1 False.There are many counterexamples.Consider the technologygenerated by a production functionf (x) =x2.The production set isY=(y, x) : yx2which is certainly not convex, but the input re-quirement set is V (y) = x : x y which is a convex set.1.2 It doesnt change.1.31= a and2= b.1.4 Let y(t) = f(tx). Thendydt=nf(x)xixi,so that1 dyy dt=i=11f(x)nf (x)xixi.1.5 Substitute txifor i = 1, 2 to geti=111f(tx1, tx2) = (tx1)+ (tx2)= tx1+ x2= tf (x1, x2).This implies that the CES function exhibits constant returns to scale andhence has an elasticity of scale of 1.1.6 This is half true: ifg0(x)0, then the function must be strictlyincreasing, but the converse is not true. Consider, for example, the functiong(x) = x3. This is strictly increasing, but g0(0) = 0.1.7 Letf (x) =g(h(x) and suppose thatg(h(x) =g(h(x0 ). Since gismonotonic, it follows thath(x) =h(x0 ). Now g(h(tx) =g(th(x) andg(h(tx0) = g(th(x0 ) which gives us the required result.1.8 A homothetic function can be written asg(h(x) where h(x) is ho-mogeneous of degree 1. Hence the TRS of a homothetic function has the2ANSWERSformg0(h(x) hx1g0(h(x) h=hx1h.x2x2That is, the TRS of a homothetic function is just the TRS of the un-derlying homogeneous function. But we already know that the TRS of ahomogeneous function has the required property.1.9 Note that we can write11(a1+ a2) a1a1+ a2x1+a2a1+ a2x2.1Now simply dene b = a1/(a1+ a2) and A = (a1+ a2) .1.10 To prove convexity, we must show that for allyandy0inYand0 t 1, we must have ty + (1 t)y0inY . But divisibility implies thattyand (1 t)y0are in Y , and additivity implies that their sum is in Y .To show constant returns to scale, we must show that ify is in Y , ands 0, we must have sy in Y . Given any s 0, let n be a nonnegativeinteger such that n s n 1. By additivity, nyis in Y ; since s/n 1,divisibility implies (s/n)ny = sy is in Y .1.11.a This is closed and nonempty for all y 0 (if we allow inputs to benegative). The isoquants look just like the Leontief technology except weare measuring output in units of log yrather than y. Hence, the shape ofthe isoquants will be the same. It follows that the technology is monotonicand convex.1.11.b This is nonempty but not closed. It is monotonic and convex.1.11.c This is regular. The derivatives of f (x1, x2) are both positive so thetechnology is monotonic. For the isoquant to be convex to the origin, it issucient (but not necessary) that the production function is concave. Tocheck this, form a matrix using the second derivatives of the productionfunction, and see if it is negative semidenite. The rst principal minor ofthe Hessian must have a negative determinant, and the second principalminor must have a nonnegative determinant.2f (x)= 1x3212f (x)11 1x2141x22x1x2=4x 21x222f (x)113= x22x2241x2Hessian =Ch. 2 PROFIT MAXIMIZATION3#14x13/2x12/214x11/2x21/211/21/214x11/2x23/2D1= 4x1x21x3/21/2411x2 1. It ismonotonic and weakly convex.1.11.f This is regular. To check monotonicity, write down the productionfunction f(x) = ax1 x1x2+ bx2and computef(x)x1= a 12x1/21/21x2.This is positive only if a 1qx2always monotonic.2x1, thus the input requirement set is notLooking at the Hessian of f , its determinant is zero, and the determinantof the rst principal minor is positive. Thereforefis not concave. Thisalone is not sucient to show that the input requirement sets are notconvex. But we can say even more:fis convex; therefore, all sets of theformx1, x2: ax1 x1x2+ bx2 yfor all choices of yare convex. Except for the border points this is just the complement ofthe input requirement sets we are interested in (the inequality sign goes inthe wrong direction). As complements of convex sets (such that the borderline is not a straight line) our input requirement sets can therefore not bethemselves convex.1.11.g This function is the successive application of a linear and a Leontieffunction, so it has all of the properties possessed by these two types offunctions, including being regular, monotonic, and convex.Chapter 2. Profit Maximization4ANSWERS2.1 For prot maximization, the Kuhn-Tucker theorem requires the follow-ing three inequalities to hold(f(x)pxj wjxj= 0,pf(x)xj wj 0,xj 0.Note that if xj 0, then we must have wj/p = f(x)/xj.2.2 Suppose thatx0is a prot-maximizing bundle with positive prots(x0) 0. Sincef(tx0) tf (x0),for t 1, we have(tx0) = pf(tx0) twx0 t(pf(x0) wx0) t(x0) (x0).Therefore, x0could not possibly be a prot-maximizing bundle.2.3 In the text the supply function and the factor demands were computedfor this technology. Using those results, the prot function is given by(p, w) = p(wapaa1 w(wap1a1.To prove homogeneity, note that(a(1(tp, tw) = tpwapa1 twwapa1= t(p, w),which implies that (p, w) is a homogeneous function of degree 1.Before computing the Hessian matrix, factor the prot function in thefollowing way:1aa11a(p, w) = p1a w a1a1a a 1a= p 1a w a1 (a),where (a) is strictly positive for 0 a 0and 0.Therefore, the Hessian is a positive semidenite matrix, whichimplies that (p, w) is convex in (p, w).2.4 By prot maximization, we havefNow, note that|T RS| =x1fx2=w1w2.Therefore,ln(w2x2/w1x1) = (ln(w1/w2) + ln(x1/x2).d ln(w2x2/w1x1)d ln(x1/x2)=d ln(w1/w2) 1 =dln |T RS|d ln(x2/x1)d ln(x2/x1) 1 = 1/ 1.2.5 From the previous exercise, we know thatln(w2x2/w1x1) = ln(w2/w1) + ln(x2/x1),Dierentiating, we getd ln(w2x2/w1x1)d ln(w2/w1)= 1 dln(x2/x1)d ln |T RS|= 1 .2.6 We know from the text that Y O YY I. Hence for any p, themaximum of pyover Y O must be larger than the maximum over Y , andthis in turn must be larger than the maximum over Y I.2.7.a We want to maximize 20x x2 wx. The rst-order condition is20 2x w = 0.2.7.b For the optimal x to be zero, the derivative of prot with respect tox must be nonpositive at x = 0: 20 2x w 0 when x = 0, or w 20.2.7.c The optimal x will be 10 when w = 0.2.7.d The factor demand function is x = 10 w/2, or, to be more precise,x = max10 w/2, 0.6ANSWERS2.7.e Prots as a function of output are20x x2 wx = 20 w xx.Substitute x = 10 w/2 to nd(w) =h10 w2i2.2.7.f The derivative of prot with respect to w is (10 w/2), which is, ofcourse, the negative of the factor demand.Chapter 3. Profit Function3.1.a Since the prot function is convex and a decreasing function of thefactor prices, we know that 0i(wi) 0 and 00i(wi) 0.3.1.b It is zero.3.1.c The demand for factor i is only a function of the ithprice. Thereforethe marginal product of factor i can only depend on the amount of factori. It follows that f (x1, x2) = g1(x1) + g2(x2).3.2 The rst-order conditions arep/x =w, which gives us the demandfunctionx=p/wand the supply functiony= ln(p/w).The protsfrom operating at this point arep ln(p/w) p.Since the rm can al-ways choosex= 0 and make zero prots, the prot function becomes(p, w) = maxp ln(p/w) p, 0.3.3 The rst-order conditions arepa1 w1= 0x1pa2x2 w2= 0,which can easily be solved for the factor demand functions. Substitutinginto the objective function gives us the prot function.3.4 The rst-order conditions arepa1xa111xa22 w1= 0pa2xa221xa11 w2= 0,which can easily be solved for the factor demands. Substituting into theobjective function gives us the prot function for this technology. In orderCh. 4 COST MINIMIZATION7for this to be meaningful, the technology must exhibit decreasing returnsto scale, so a1+ a2 1.3.5 If wiis strictly positive, the rm will never use more of factor i than itneeds to, which implies x1= x2. Hence the prot maximization problemcan be written asmaxpxa1 w1x1 w2x2.The rst-order condition ispaxa11 (w1+ w2) = 0.The factor demand function and the prot function are the same as if theproduction function were f (x) = xa, but the factor price is w1+ w2ratherthan w. In order for a maximum to exist, a 1.Chapter 4. Cost Minimization4.1 Letxbe a prot-maximizing input vector for prices (p, w).Thismeans that xmust satisfy pf (x) wx pf(x) wxfor all permissiblex. Assume that xdoes not minimize cost for the output f(x); i.e., thereexists a vector xsuch that f (x) f (x ) and w(x x) pf(x) wx,which contradicts the assumption that xwas prot-maximizing.4.2 The complete set of conditions turns out to be(tf(x)xj wjxj= 0,tf(x)xj wj 0,xj 0,(y f(x)t = 0,y f (x) 0,t 0.If, for instance, we have xi 0 and xj= 0, the above conditions implyf(x)xiwi.f(x)xjwj8ANSWERSThis means that it would decrease cost to substitute xifor xj, but sincethere is no xjused, this is not possible. If we have interior solutions forboth xiand xj, equality must hold.4.3 Following the logic of the previous exercise, we equate marginal coststo ndy1= 1.We also know y1+ y2= y, so we can combine these two equations to gety2= y 1. It appears that the cost function is c(y) = 1/2 +y 1 = y 1/2.However, on reection this cant be right: it is obviously better to produceeverything in plant 1 if y1 1. As it happens, we have ignored the implicitconstraint that y2 0. The actual cost function isc(y) =y2/2if y 1.4.4 According to the text, we can write the cost function for the rst plantasc1(y) = Ayand for the second plant as c2(y) = By, where Aand Bdepend on a, b, w1, and w2. It follows from the form of the cost functionsthatc(y) = minA, By.4.5 The cost of using activity a is a1w1+a2w2, and the cost of using activityb is b1w1+ b2w2. The rm will use whichever is cheaper, soc(w1, w2, y) = y mina1w1+ a2w2, b1w1+ b2w2.The demand function for factor 1, for example, is given byx1=a1yb1yif a1w1+ a2w2 b1w1+ b2w2 any amount betweena1y and b1yotherwise.The cost function will not be dierentiable whena1w1+ a2w2= b1w1+ b2w2.4.6 By the now standard argument,c(y) = min4y1+ 2y2: y1+ y2 y.It is tempting to setM C1(y1) =M C2(y2) to nd thaty1=y/5 andy2= 4y/5. However, if you think about it a minute you will see that thisCh. 5 COST FUNCTION9doesnt make senseyou are producing more output in the plant with thehigher costs!It turns out that this corresponds to a constrained maximum and not tothe desired minimum. Check the second-order conditions to verify this.Since the cost function is concave, rather than convex, the optimal solu-tion will always occur at a boundary. That is, you will produce all outputat the cheaper plant so c(y) = 2y.4.7 No, the data violate WACM. It costs 40 to produce 100 units of output,but at the same prices it would only cost 38 to produce 110 units of output.4.8 Set up the minimization problemmin x1+ x2x1x2= y.Substitute to get the unconstrained minimization problemminx1+ y/x1.The rst-order condition is1 y/x21,which impliesx1=y.By symmetry,x2=y.We are given that2y= 4, so y= 2, from which it follows that y = 4.Chapter 5. Cost Function5.1 The rm wants to minimize the cost of producing a given level of output:y22c(y) = miny1,y21+y2such that y1+ y2= y.The solution has y1= y2= y/2. Substituting into the objective functionyieldsc(y) = (y/2)2+ (y/2)2= y2/2.5.2 The rst-order conditions are 6y1= 2y2, or y2= 3y1. We also requirey1+ y2= y. Solving these two equations in two unknowns yields y1= y/4and y2= 3y/4. The cost function isc(y) = 3hyi24+3y42=3y24.10ANSWERS5.3 Consider the rst technique. If this is used, then we need to have2x1+ x2=y. Since this is linear, the rm will typically specialize andsetx2=yorx1=y/2 depending on which is cheaper. Hence the costfunction for this technique is y minw1/2, w2. Similarly, the cost functionfor the other technique is y minw3, w4/2. Since both techniques must beused to produce y units of output,c(w1, w2, y) = y minw1/2, w2 + minw3, w4/2 .5.4 The easiest way to answer this question is to sketch an isoquant. Firstdraw the line 2x1+ x2= yand then the line x1+ 2x2= y. The isoquantis the upper northeast boundary of this “cross.” The slope is2 to theleft of the diagonal and 1/2 to the right of the diagonal. This means thatwhen w1/w2 1/2, we have x1= 0 and x2= y. When w1/w2 w1/w2 1/2, we havex1= x2= y/3. The cost function is thenc(w1, w2, y) = minw1, w2, (w1+ w2)/3y.5.5 The input requirement set is not convex.Sincey= maxx1, x2,the rm will use whichever factor is cheaper; hence the cost function isc(w1, w2, y) = minw1, w2y. The factor demand function for factor 1 hasthe form(yif w1 w25.6 We havea= 1/2 and c=1/2 by homogeneity, andb= 3 sincex1/w2= x2/w1.5.7 Set up the minimization problemmin x1+ x2x1x2= y.Substitute to get the unconstrained minimization problemminx1+ y/x1.The rst-order condition is1 y/x21,which impliesx1=y.By symmetry,x2=y.We are given that2y= 4, so y= 2, from which it follows that y = 4.Ch. 5 COST FUNCTION115.8 Ifp= 2, the rm will produce 1 unit of output. Ifp= 1, the rst-order condition suggestsy= 1/2, but this yields negative prots. Therm can get zero prots by choosingy = 0. The prot function is (p) =maxp2/4 1, 0.5.9.a d/d = py 0.5.9.b dy/d = p/c00(y) 0.5.9.c p0() = ny + p/c00/D0(p) n/c00 0.5.10 Let y(p, w) be the supply function. Totally dierentiating, we havedy =ny(p, w)wii=1dwi= nxi(p, w)pi=1dwi= nxi(w, y) y(p, w)ypi=1dwi.The rst equality is a denition; the second uses the symmetry of thesubstitution matrix; the third uses the chain rule and the fact that theunconditional factor demand, xi(p, w), and the conditional factor demand,xi(w, y), satisfy the identity xi(w, y(p, w) = xi(p, w). The last expressionon the right shows that if there are no inferior factors then the output ofthe rm must increase.5.11.a x = (1, 1, 0, 0).5.11.b minw1+ w2, w3+ w4y.5.11.c Constant returns to scale.5.11.d x = (1, 0, 1, 0).5.11.e c(w, y) = minw1, w2 + minw3, w4y.5.11.f Constant.5.12.a The diagram is the same as the diagram for an inferior good inconsumer theory.5.12.b If the technology is CRS, then conditional factor demands take theformxi(w, 1)y. Hence the de