## Y y 1 j y c i f

---- = - -\--. Solving for y' gives y = -— , and El^j =

7. (a) Ryx = {x/y)°-] = (y/x)1-" (b) a,, = 1/(1 - a)

9. F(K, N) = y: (K~a + y2N~a)~v/ct, so F'K/F'N = K~a~]/y2N~a-], implying that aKN = 1/(1 + a).

11. z = - In[aK-e + (1 -a)L~c]/q "0/0" as q 0. By l'Hopital's rule, aA"~g In A!" + (1 — a)L~° InL a^-c + (1 - fl)I-e

16.5

1. /(rjc, ry) = (rx)4 + (rx)2(o02 = rV + r2x2r2r = t\x4 + x2y2) = t4f(x, >•), so / is homogeneous of degree 4.

3. /(rx, ty) = (tx)(ty)2 + (tx)3 = r3(x>-2 + x3) = t3f(x, >•), so / is homogeneous of degree 3. [16.19]: x/,'(x, >•) + yf2{x. y) = x(y2 + 3x2) + ylxy = 3x3 + 3xy2 = 3(x3 + xy2) = 3f(x,y). ' [16.20]: It is easy to see that f[{x,y) = y2 + 3x2 and fi(x,y) = 2xy are homogeneous of degree 2. [16.21]: f(x,y) = x3+xy2 = x3[l + (y/x)2] = y3[(x/y)3+x/y]. [16.22]: x2f{\ +2xyf{',+y2f£ = x2(6x)+2xy(2y) + yz(2x) = 6*3+4xy2+2x>-2 = 3 - 2/(x, y).

5. [16.18] requires that t3x3 + t2xy = rA'(x3 + xy) for all t > 0. In particular, for x = y = 1, we must have t3 +t2 = 2tk. For t = 2, we get 12 = 2 • 2k, or 2k = 6. For t = 4, we get 80 = 2 - 4*, or Ak = 40. But 4* should be the square of 2k. So the two values of k must be different, implying that / is not homogeneous of any degree.

7. From [*] with k = 1. we get //, = (-y/x)/," and = (-*/>•)/,",. But

/12 = /i-so //',/£ - (//;)2 = {-y/x)f{'2{-x~ly)f2 - (/£)2 = 0. "

16.6

1. (a) Homogeneous of degree 1. (b) Not homogeneous, (c) Homogeneous of degree —1/2. (d) Homogeneous of degree 1. (e) Not homogeneous, (f) Homogeneous of degree n.

3. Homogeneous of degree 0.136+ (-0.727) +0.914 + 0.816 = 1.139.

5. vj = u\ - «/(*, + • ■ • + *„), so xri = xtf - ^/(x, +

. - • + x„) = a - a x-J{xi H----+ xn) = a - a = 0. Hence, by Euler's theorem, v is homogeneous of degree 0.

7. Let C and D denote the numerator and the denominator respectively in the expression for ayx that is given in Problem 10 of Sec. 16.4. Then, by Euler's theorem, C = -F[F2(xF{ + yF2) = -F[F{F. Using the facts that xF[\ = -vfj; and vF^ = -xF"} = -xF"-,, it follows that D equals xy \{FtfF[\-2F[hF^ + (F[)2F^} = -F", [(yF\)2+2xyF[F{ + {xF[)2} = -F'^xF'.+yF')2 = -F^F2. Thus <r,v = C/D = (-F[F^F)/(-F'^F2) = F[FyFF'{2.

9. Differentiate f(txu tx„) = g(t)f(x],...,xn) w.r.t. t and put t = 1, as in the proof of Theorem 16.1. This yields xif((tx\,..., txn) = g'(\)f(x] xn). Thus, by Euler's theorem, / must be homogeneous of degree ¿'(1). In fact, g(t) = tk where k = ¿'(1).