Problem Set 3


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1 Problem Set 3 Due Tuesdy, October 18, in clss 1. Lecture Notes Exercise 208: Find R b log(t)dt,where0 <<bre rel numbers. Solution: This cn be solved by integrtion by prts. Let F (t) = t, F 0 (t) =f(t) =1,G(t) =log(t),g 0 (t) =g(t) = 1 t ;then log(t)dt = f(t)g(t)dt =[F (b)g(b) F ()G()] = [b log(b) log() 1 dt = (b log(b) log ) (b ) = b(log(b) 1) (log() 1) F (t)g(t)dt 2. (Sundrm 4.4, pge 110) Find nd clssify ll criticl points (locl mximum, locl minimum, neither) of the following function: f(x, y) = e 2x (x + y 2 +2y). For locl optim tht you find figure out whether they re lso globl optim. ½ e Solution: Criticl points solve 2x (2x +2y 2 +4y +1) = 0 e 2x (2y +2) = 0.From the second eqution, y = 1 nd then from the first eqution µ x = e 0 There is only one criticl point ( 1 2, 1). We hve H( 1 2, 1) =, 0 2e which is positive definite, so it is locl minimum. Moreover, it is globl minimum too, since f(x, y) is bounded from below (why?). 3. (Simon nd Blume 18.7, pge 423): Find the mx nd min of f(x, y, z) = yz + xz subject to y 2 + z 2 =1nd xz =3 Solution: The Lgrngin is L(x, y, z, λ 1,λ 2 )=yz + xz λ 1 (y 2 + z 2 1) λ 2 (xz 3). The first order conditions give us: z λ 2 z = 0 (1) z 2λ 1 y = 0 (2) y + x 2λ 1 z λ 2 x = 0 (3) y 2 + z 2 1 = 0 (4) xz 3 = 0 (5) where the lst two simply repet the constrints. 1
2 (1) tell us tht either z =0or λ 2 =1. But we know from (5) tht z cnnot be zero, so λ 2 =1. Now,plugthisinto(3)ndwefind tht (2) nd (3) tell us tht z 2λ 1 y = 0 y 2λ 1 z = 0 There re two sets of solutions to this pir of equtions  one in which λ 1 = 1 2 nd y = z, ndoneinwhichλ 1 = 1 2 nd y = z. We now plug this into our constrints ((4) nd (5)) to obtin our four criticl points: (3 2, 2 2, 2 2 ); ( 3 2, 2 2, 2 2 ); (3 2, 2 2, 2 2 ); ( 3 2, 2 2, 2 2 ). The first two points re mximizers (yielding vlue of for the objective function), while the second two re minimizers (yielding ). 4. (Simon nd Blume 18.11, pge 434): Mximize f(x, y) =2y 2 x, subject to x 2 + y 2 1,x 0,y 0. Solution: The Lgrngin is L(x, y, λ 1,λ 2,λ 3 )=2y 2 x λ 1 (x 2 + y 2 1)+λ 2 x+λ 3 y. Note tht the terms in the Lgrngin ssocited with the lst two constrints enter positively; those constrints cn be rewritten s x 0 nd y 0. Wehvethefirst order conditions: x = 1 2λ 1 x + λ 2 =0 y = 4y 2λ 1 y + λ 3 =0 λ 1 (x 2 + y 2 1) = 0,λ 2 x =0,λ 3 y =0 λ 1 0,λ 2 0,λ 3 0 x 2 + y 2 1 x 0,y 0 The third row contins the complementryslckness conditions, while the lst two rows give the first order conditions with respect to the Lgrnge multipliers, nd simply restte the intitil constrints. We will begin by showing tht the firstconstrintmustbind. Notetht Weierstrß pplies here; the constrint set is simply the positive qudrnt of the unit circle, which is compct set. Now, suppose tht the first constrint does not bind, so tht x 2 + y 2 < 1. Then λ 1 =0to stisfy the cs condition. This implies, from the first FOC, tht λ 2 =1,whichinturn implies (gin from the cs conditions) tht x =0. But if this is the cse, the problem reduces to tht of mximizing 2y 2, subject to the constrints tht y must be nonnegtive nd tht y 2 < 1 (still ssuming tht the first constrint does not bind). But this clerly hs no mximum, s the function is incresing in y, nd n incresing function hs no mximum onnopenset. Sinceweknowwehvemximum,itcnnotbethe 2
3 cse tht the first constrint is slck when it is chieved. Therefore, it mustbethecsethtx 2 + y 2 =1. But now we re merely looking for the point on positive qudrnt of the circle which mximizes function which is incresing in y nd decresing in x. It is cler tht this occurs t the point (0, 1). 5. Let F (x, y) =2x 2 +2y 2 +8nd G(x, y) =x 2 +2y 2 6x 7. Note: this problem will be much esier nd less tedious if you stop now nd think bout wht these functions look like. () Stte the implicit function theorem. Find ll points on the curve G(x, y) =0round which either y is not expressible s function of x or x is not expressible s function of y. Compute y 0 (x) long the curve when x =2. Solution: This curve is the intersection of the xy plne with the prboloid described by G, which is n ellipse. To see this, note tht we cn write G(x, y) =(x 3) 2 +2y 2 16 = 0, whichclerly defines n ellipse with center (3, 0), nd mjor xis on the xxis, with length 8. According to the implicit function theorem, the curve does not define y(x) where G y =4y =0, or t y =0. Similrly, x(y) is not defined where G x =2x 6=0,orx =3. Notice tht these points re the top, bottom, nd sides of our ellipse, where the curve goes just verticl nd horizontl. The implicit function theorem lso tells us tht, so long s y(x) is defined, y 0 (x) = y x i (x 1,..., G x x (x n )= i 1,...,x n,y ). G Here, we hve y (x 1,...,x n,y ) y 0 (x) = 2x 6 4y = 1 2y when x =2. Plugging x =2into G(x, y) =0, q 15 we find tht y = ± 2, so y0 (x) =± (b) Find ll unconstrined optim of F nd G on R 2.IstheWeierstrß theorem pplicble? Weierstrß does not pply, becuse R 2 is unbounded. It is cler tht neither of these functions hve mxim. Nevertheless, we cn find minim. Observe first tht G(x, y) cn be split into f(x) = x 2 6x nd g(y) =2y 2 7. Both of these functions describe convex prbole (prbols? prboleese? whtever), so it should be cler tht the function will hve no globl mx, tht it will hve globl min, nd tht G s whole describes prboloid in R 3. Tking first order conditions, we find tht 2x 6 = 0 (6) 4y = 0 (7) 3
4 so the only criticl point is (3, 0). Since both second order conditions re positive, we gin see tht the function is convex, nd tht this is therefore globl minimum. Exctly the sme method cn be pplied to F, which yields (0, 0) s the sole criticl point nd globl minimum. (c) Mximize nd minimize d(x, y) = p x 2 + y 2 subject to G(x, y) 0. Does Weierstrß pply? Solution: Weierstrß does pply, becuse the ellipse defined by the constrint is compct set nd the distnce function is continuous. The simplest wy to solve this problem is to tret it first s n equlity constrined problem, then s n inequlity constrined problem. The first is mtter of finding which points on the ellipse described by G =0,ndthesecondoffinding which points on or inside the ellipse, re closest to nd furthest from the origin. The ellipse described by G(x, y) = 0 clerly comes closest to the origin t the point ( 1, 0) ndisfurthestfromitt(7, 0). But we cn lso get the sme nswer using the stndrd Lgrngin method, noting tht the objective function is d(x, y) =. p x 2 + y 2 (the formul for distnce from the origin). L(x, y, λ) = p x 2 + y 2 λ(x 2 +2y 2 6x 7) (8) x = x(x 2 + y 2 ) 1 2 λ(2x 6) = 0 (9) y = y(x 2 + y 2 ) 1 2 4λy =0 (10) λ = x2 +2y 2 6x 7=0 (11) We note tht y =0stsfies the second FOC. Plugging this into the constrint, we find tht two criticl points re ( 1, 0) nd (7, 0). Techniclly, we ve treted x s function of y, nd should therefore be concerned bout the points where this is not defined (i.e., t x =3). This would give us two more criticl points: (3, 2 2) nd (3, 2 2). If we hve drwn the ellipse tht G(x, y) represents we know tht we need not check these points, however  nd indeed, we find tht F ( 1, 0) = 1 (12) F (7, 0) = 49 (13) F (3, 2 2) = F (3, 2 2) = 17 (14) So tht ( 1, 0) is our minimizer nd (7, 0) is our mximizer for the equlity constrined problem. Now we hve to check ll points in the ellipse, in ddition to its boundry. When the constrint doesn t bind, we hve the dditionl criticl point (0, 0)  the origin itself. Thus, the mximum occurs t (7, 0), nd the minimum t (0, 0). 4
5 (d) Mximize nd minimize F (x, y) subject to G(x, y) =0.IstheWeierstrß theorem pplicble? Solution: Note tht F (x, y) =2[d(x, y)] 2 +8, monotonic trnsformtion of d(x, y). Thus, the extreml points must be the sme. Weierstrß is once gin pplicble, nd the mx nd min re t the points we found in prt (c) for the equlity constrined problem. (e) Mximize nd minimize F (x, y) subject to G(x, y) 0. IstheWeierstrß theorem pplicble? Weierstrß is pplicble, nd the mx nd min re the sme points we found in prt (c) for the inequlity constrined problem. (f) Mximize nd minimize F (x, y) subject to G(x, y) 0. IstheWeierstrß theorem pplicble? Weierstrß is no longer pplicble, since the set outside the ellipse is not bounded. There exists no mximum (no point furthest from the origin), but the minimum occurs t ( 1, 0), thepointwhichwe erlier found ws the point on the ellipse closest to the origin. 6. (Simon nd Blume 20.1, pge 493): Which of the following functions re homogeneous? Wht re the degrees of homogeneity of the homogeneous ones? () 3x 5 y +2x 2 y 4 3x 3 y 3 Solution: The function is homogenous of degree six: f(tx, ty) = 3t 6 x 5 y +2t 6 x 2 y 4 3t 6 x 3 y 3 = t 6 f(x, y). (b) x 1/2 y 1/2 +3xy 1 +7 Solution: The function is homogenous of degree zero. (c) x 3/4 y 1/4 +6x Solution: The function is homogenous of degree one. (d) (x2 y 2 ) (x 2 +y 2 ) +3 Solution: The function is homogenous of degree zero. 7. (Simon nd Blume 20.6, pge 493): Prove tht if f nd g re functions on R n tht re homegeneous of different degrees, then f + g is not homogeneous. Solution: Let f be homogenous of degree k; thenf (tx) =t k f(x); similrly, if g is homogenous of degree l 6= k, theng (tx) =t l g(x). Then (f + g)(tx) =f(tx)+g(tx) =t k f(x)+t l g(x), ndsof + g is not homogenous. 8. Mny utility functions we work with exhibit diminishing mrginl returns (i.e., they re concve in ech of their rguments, 2 u < 0). Is this n x 2 i ordinl property? Why or why not? 5
6 Solution: No; it is not invrint under monotonic trnsformtion. For exmple, the function log(x) is concve in x, but the monotonic trnsformtion x 2 =[exp{log(x)}] 2 is convex. 9. Show tht the following functions re homothetic: () e x2y e xy2 Solution: e x2y e xy2 =exp{x 2 y + xy 2 } which is monotonic trnsformtion of the homogeneous (of degree 3) function x 2 y + xy 2. (b) 2logx +3logy Solution: 2logx +3logy =log(x 2 y 3 ) which is monotonic trnsformtion of the homogeneous (of degree 5) function x 2 y 3. (c) x 3 y 6 +15x 2 y 4 +75xy Solution: x 3 y 6 +15x 2 y 4 +75xy = (xy 2 +5) 3,whichis monotonic trnsformtion of xy 2 +5, which is monotonic trnsformtion of the homogenous (of degree 3) functionxy 2. 6
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