Chiang/Wainwright: Fundamental Methods of Mathematical Economics

Transcription

1 CHAPTER 12 Eercise (a) Z = y + (2 2y). The necessary condition is: Z =2 2y =0 Z = y =0 Z y = 2 =0 Thus 2, =1, y 2 yielding z 2. (b) Z = y +4 + (8 y). The necessary condition is: Z =8 y =0 Z = y +4 =0 Z y = =0 Thus =6, =6, y =2 yielding z =36. (c) Z = 3y y + (6 y). The necessary condition is: Z =6 y =0 Z =1 y =0 Z y = 3 =0 Thus = 4, =1, y =5 yielding z 9. (d) Z =7 y ( y). The necessary condition is: Z = y =0 Z =2 =0 Z y =0 Thus, = 1 2, y 2 yielding z = (a) Increase; at the rate dz dc = 2. (b) Increase; dz dc =6. (c) Decrease; dz dc (d) Decrease; dz dc = 4 3. (a) Z = +2y +3w + y yw + (10 y 2w). Hence: Z =10 y 2w =0 Z =1+y =0 Z y =2+ w =0 Z w =3 y 2 =0 76

2 (b) Z = 2 +2y + yw y w 2 + (8 w). Thus Z =24 2 y w 2 =0 Z =8 w =0 Z =2 +2y 2 =0 Z y =2 + w 2 =0 Z w =2yw 2w =0 4. Z = f(, y)+ [0 G(, y)] = f(, y) G(, y). The rst-order condition becomes: Z = G(, y) =0 Z = f G =0 Z y = f y G y =0 5. Since the constraint g = c is to prevail at all times in this constrained optimization problem, the equation takes on the sense of an identity, and it follows that dg must be zero. Then it follows that d 2 g must be zero, too. In contrast, the equation dz =0is in the nature of a rst-order condition dz is not identically zero, but is being set equal to zero to locate the critical values of the choice variables. Thus d 2 z does not have to be zero as a matter of course. 6. No, the sign of will be changed. The new is the negative of the old. Eercise (a) Since H = (b) Since H = (c) Since H = (d) Since H = =4,z 2 is a maimum =2, z =36is a maimum = 2, z 9 is a minimum = 2, z =6 3 4 is a minimum

3 2. H 1 = 0 g 1 g 1 Z 11 = g2 1 < 0 3. The zero can be made the last (instead of the rst) element in the principal diagonal, with g 1, g 2 and g 3 (in that order appearing in the last column and in the last row. 0 0 g 1 1 g2 1 g3 1 g g 1 2 g2 2 g3 2 g g H = 1 1 g 2 1 Z 11 Z 12 Z 13 Z 14 g 2 1 g2 2 Z 21 Z 22 Z 23 Z 24 g 3 1 g3 2 Z 31 Z 32 Z 33 Z 34 g4 1 g4 2 Z 41 Z 42 Z 43 Z 44 A sucient condition for maimum z is H3 < 0 and H4 = H > 0. A sucient condition for minimum z is H3 > 0 and H > 0. Eercise Eamples of acceptable curves are: 78

4 2. (a) Quasiconcave, but not strictly so. This is because f(v) =f(u) =a,andthusf [u +(1)v] = a, which is equal to (not greater than) f(u). (b) Quasiconcave, and strictly so. In the present case, f (v) f(u) means that a+bv a+bu, or v u. Moreover, to have u and v distinct, we must actually have v>u. Since f [u +(1)v] = a + b [u +(1)v] = a + b [u +(1)v]+(bu bu) = a + bu + b (1 )(v u) = f (u) b (1 )(v u) =f (u)+some positive term it follows that f [u +(1 ) v] >f(u). Hence f () =a + b, (b >0), strictlyquasiconcave. (c) Quasiconcave, and strictly so. Here, f(v) f (u) means a + cv 2 a + cu 2,orv 2 u 2 (since c<0). For nonnegative distinct values of u and v, this in turn means v<u. Now we have f [u +(1)v] = a + c [u +(1) v] 2 + cu 2 cu 2 = a + cu 2 + c n[u +(1) v] 2 u 2o Using the identity y 2 2 (y + )(y ), we can rewrite the above epression as a + cu 2 + c [u +(1) v + u][u +(1) v u] = f(u)+c [(1 + ) u +(1) v][(1 )(v u)] 79

5 = f (u)+ some positive term >f(u) Hence f () =a + c 2, (c <0), is strictly quasiconcave. 3. Both f () and g () are monotonic, and thus quasiconcave. However, f ()+g() displays both a hill and a valley. If we pick k =5 1 2, for instance, neither S nor S will be a conve set. Therefore f () +g () is not quasiconcave. (a) This cubic function has a graph similar to Fig. 2.8c, with a hill in the second quadrant and valley in the fourth. If we pick k =0, neither S nor S is a conve set. The function is neither quasiconcave nor quasiconve. (b) This function is linear, and hence both quasiconcave and quasiconve. (c) Setting 2 ln 1 = k, and solving for 2, we get the isovalue equation 2 =ln 1 + k. In the 1 2 plane, this plots for each value of k as a log curve shifted upward vertically by the amount of k. The set S = {( 1, 2 ) f ( 1, 2 ) k} the set of points on or below the isovalue curve is a conve set. Thus the function is quasiconve. (but not quasiconcave). 4. (a) A cubic curve contains two bends, and would thus violate both parts of (12.21). (b) From the discussion of the cubic total-cost function in Sec. 9.4, we know that if a, c, d>0, b<0, andb 2 < 3ac, then the cubic function will be upward-sloping for nonnegative. Then, by (12.21), it is both quasiconcave and quasiconve. 5. et u and v be two values of, andletf (v) =v 2 f (u) =u 2, which implies v u. Since f 0 () =2, we nd that 80

6 f 0 (u)(v u) =2u (v u) 0 f 0 (v)(v u) =2v (v u) 0 Thus, by (12.22), the function is both quasiconcave and quasiconve., conrming the conclusion in Eample The set S, involving the inequality y k, consists of the points lying on or below a rectangular hyperbola not a conve set. Hence the function is quasiconve by (12.21). Alternatively, since f = y, f y =, f =0, f y =1,andf yy =0,wehave B 1 = y 2 0 and B 2 =2y 0, which violates the necessary condition (12.25 ) for quasiconveity. 7. (a) Since f = 2, f y = 2y, f = 2, f y =0, f yy = 2, wehave B 1 = 4 2 < 0 B 2 =8 2 + y 2 > 0 By (12.26), the function is quasiconcave. (b) Since f = 2( +1), f y = 2(y +2), f = 2, f y =0, f yy = 2, wehave B 1 = 4( +1) 2 < 0 B 2 =8( +1) 2 +8(y +2) 2 > 0 By (12.26), the function is quasiconcave. Eercise (a) Z =( +2)(y +1)+ ( y) (b) The rst-order condition requires that Z y =0, Z = y +14 =0, Z y = +26 =0 Thus we have =3, =16,andy =11. (c) H = =48> 0. Hence utility is maimized (d) No. 81

7 2. (a) Z =( +2)(y +1)+ (B P yp y ) (b) As the necessary condition for etremum, we have Z = B P yp y =0 or P P y y = B Z = y +1 P =0 Z y = +2 P y =0 P + y P y + = 2 By Cramer s Rule, we can nd that = B+2P+Py 2P P y = B2P+Py 2P y = B+2PPy 2P y (c) 0 P P y H = P 0 1 =2P P y > 0. Utility is maimized. P y 1 0 (d) When P =4,P y =6,andB 30, weget =3, =16and y =11. These check with the preceding problem. 3. Yes. B 2P > 0, P = B+Py 2P < 0, 2 P y 2P > 0, y B 2P y > 0, y P P y > 0, y P y = B+2P 2P < 0. y 2 An increase in income B raises the level of optimal purchases of and y both; an increase in the price of one commodity reduces the optimal purchase of that commodity itself, but raises the optimal purchase of the other commodity. 4. We have U = U yy =0,U y = U y =1, J = H =2P P y. = (B2P +P y ) 2P,and = (B+2P +P y ) 2P P y. Thus: (a) (b) B P 2P,and = (B+P y) 2P,and 2 y B 2P y. y P P y. These answers check with the preceding problem. 5. A negative sign for that derivative can mean either that the income eect (T 1 ) and the substitution eect (T 2 ) in (12.33 ) are both negative (normal good), or that the income eect is positive (inferior good) but is overshadowed by the negative substitution eect. The statement is not valid. 82

8 6. The optimal utility level can be epressed as U = U (,y ). Thus du = U d + U y dy, where U and U y are evaluated at the optimum. When U is constant, we have du =0,or U d + U y dy =0. From (12.42 ), we have U U y = P P y at the optimum. Thus we can also epress du =0by P d + P y dy =0,or-P d P y dy =0. 7. (a) No; diminishing marginal utility means only that U and U yy are negative, but says nothing about U y. Therefore we cannot be sure that H > 0 in (12.32) and d 2 y d 2 > 0 in (12.33 ). (b) No; if dy2 d > 0, and hence H > 0, nothing denite be said about the sign of 2 U and U yy, because U y also appears in H. Eercise (a) p (j)(jy)=j = y; homogeneous of degree one. h (b) (j) 2 (jy) 2i 1 2 = j 2 y ; homogeneous of degree one. (c) Not homogeneous. (d) 2j + jy +3 p (j)(jy) =j 2 + y +3 y ; homogeneous of degree one. (e) (j)(jy)2 jw +2(j)(jw) =j 2 y 2 w +2w ; homogeneous of degree two. (f) (j) 4 5(jy)(jw) 3 = j 4 4 5yw 3 ; homogeneous of degree four. 2. et j k,then Q K = f K K, K = f l, K = K. Thus Q = K K. (a) When MPP K =0,wehave Q Q = Q, or = Q,orMPP = AP P. (b) When MPP =0,wehaveK Q Q K = Q, or K = Q K,orMPP K = AP P K. 3. Yes, they are true: 4. (a) AP P = (k); hence AP P indeed can be plotted against k. (b) MPP K = 0 (k) =slope of AP P. (c) AP P K = (k) k AP P curve = AP P k = ordinate of a point on the AP P curve abscissa of that point =slope of radius vector to the 83

9 (d) MPP = (k) k 0 (k) =AP P k MPP K 5. b. AP P = Ak,thustheslopeofAP P = Ak 1 = MPP K. c. Slope of a radius vector = Ak k = Ak 1 = AP P K. d. AP P k MPP K = Ak kak 1 = Ak Ak = A (1 ) k = MPP. 6. (a) Since the function is homogeneous of degree ( + ), if+ >1, the value of the function will increase more than j-fold when K and are increased j-fold, implying increasing returns to scale. (b) If + <1, the value of the function will increase less than j-fold when K and are increased j-fold, implying decreasing returns to scale. (c) Taking the natural log of both sides of the function, we have ln Q =lna + ln K + ln Thus ² QK = (ln Q) (ln K) = and ² (ln Q) Q = (ln ) = 7. (a) A (jk) (j) b (jn) c = j a+b+c AK b N c = j a+b+c Q; homogeneous of degree a + b + c. (b) a + b + c =1implies constant returns to scale; a + b + c>1 implies increasing returns to scale. (c) Share for N = N ( Q N ) Q = NAKa b cn c1 = c AK a b N c 8. (a) j 2 Q = g (jk,j) (b) et j. Q = g K 2, 1 = K Then the equation in (a) yields: (c) MPP K = Q k = 2 0 (k) k K well as k. = (k). This implies that Q = 2 (k). = 2 0 (k) 1 = 0 (k). Now MPP K depends on as (d) If K and are both increased j-fold in the MPP K epression in (c), we get (j) 0 jk j = j 0 K = j 0 (k) =j MPP K Thus MPP K is homogeneous of degree one in K and. 84

10 Eercise (a) inear homogeneity implies that the output levels of the isoquants are in the ratio of 1:2:3(from southwest to northeast). (b) With second-degree homogeneity, the output levels are in the ratio of 1: Since 1:4:9. b a = :3 2,or P a P b, it will plot as a straight line passing through the origin, with a (positive) slope equal to. This result does not depend on the assumption + =1. The elasticity of substitution is merely the elasticity of this line, which can be read (by the method of Fig. 8.2) to be unity at all points. 3. Yes, because Q and Q KK have both been found to be negative. 4. On the basis of (12.66), we have K 1+ i 5. d 2 K d 2 = d d h 1 (1 + ) K (1 + ) K d K d 1 2 dk d K > 0 because dk d < 0 (a) abor share Capital share = f Kf k the labor share. K. A larger implies a larger capital share in relation to (b) No; no. 6. = (1) = constant. The isoquants would be downward- (a) If, (12.66) yields dk d sloping straight lines. (b) By (12.68), is not dened for. But as 1,. (c) inear isoquants and innite elasticity of substitution both imply that the two inputs are perfect substitutes. 85

11 7. If both K and are changed j-fold, output will change from Q to: h A (jk) +(1)(j) i r = A {j [K +(1) ]} r =(j ) r Q = j r Q Hence r denotes the degree of homogeneity. With r > 1(r < 1), wehaveincreasing(decreas- ing) returns to scale. 8. By Hôpital s Rule, we have: 9. (a) lim =lim =7 (b) lim 0 e 1 = lim 0 e 1 =1 5 (c) lim e 0 ln (d) lim = lim 0 5 ln 5e 1 =ln5 1 = lim 1 1 =0 (a) By successive applications of the rule, we nd that lim n e = lim n n1 e (b) By taking m () =ln, andn (),wehave lim 0 + ln 1 = lim 0 + = lim n(n1) n2 e =... = lim n! e = = lim 0 + =0 (c) Since = ep (ln )=ep(ln ), and since, from (b) above, the epression ln tends to zero as 0 +, must tend to e 0 =1as