## Info

1. (a) y) = x + y - A(x2 + y - 1). The equations C\ = 1 - 2Ajc = 0, = 1 — A = 0, and x2 + y = 1 have the solution x = 1/2, y = 3/4, and A = 1. (b) The solution is illustrated in Fig. 49. The minimization problem has no solution.

(c) x = 0.5 and y = 0.85. The change in the value function is /*(1.1) -/*(1) = (0.5+ 0.85)-(0.5+ 0.75) = 0.1. Because X = 1 ,X-dc= 1-0.1 = 0.1. So, in this case, [18.8] is satisfied with equality.

3. (a) Solution: x = 50 and y = 50, with X = 250. To see that this solves the problem, let x = 50 + h and y = 50 + k. Inserting these values of x and y into the constraint yields 50 + h + 50 + k = 100, that is, k = —h. Then x2 + 3xy + y2 = (50 + h)2 + 3(50 + h)(50 + fc) + (50 + k)2 = 12,500- h2, using k — —h and simplifying. Now, 12,500—h2 has a maximum for h = 0, that is, for x = 50, and then y = 50. (b) Solution: x = 8/3 and y = 1, with X = 4. From the constraint equation, y = 3 — 3x/4. Because we must have y > 0, so x < 4. Now, if we let h(x) = \2xJZ - 3x/4, then h'(x) = (72 - 21x)/2^/3 - 3x/4, and the sign variation of this derivative reveals that h is maximized at x = 8/3. 5. (a) £(x.y) = 10xI/2y1/3 - X(2x + 4y - m). The two first-order conditions C\ = 5x-^-yxr> - 2A = 0 and C2 = {\0/3)x^~y-2'3 - 4X = 0 imply that (10/3)x1/2y"2/3 = lOx-'^y1/3, so jc = 3y. Substituting this into the constraint 2x + Ay = m gives y = m/10 and so x = 3m/10, with A = 2.5(10/27m)1/6.

(b) f*{m) = 10I/631/2™5/6 and df*(m)/dm = X = 2.5 • 10I/63"1/2tfr1/6.

7. (a) (2,1) (b) With C(x, y)=x + 2y - A[p(*2 + y2) + x2y2 - 4], equating the first-order partials to 0 yields C\ = 1 - 2Xpx - 2Xxy2 = 0 and = 2-2Xpy-2Xx2y = 0. Hence, 2Ax(p+y2) = 1 and 2xy(p+o:2) = 2. EUm-inating A yields the first equality in [**]. The second is just the constraint.

(c) Differentiating [**] w.r.L p, with x and y as functions of p, yields

 2x + 2px' - y - py' + Zx'y2 + *xyy' - 2xx'y - x2y' = 0 and

 x2 + 2pxx'+y2 + 2pyy'+2xx'y2 + 2x2yy' = 0. Letting p = 0, recalling that x(O) = 2 and y(0) = 1, yields the equations -2x'(0)+4/(0) = -3 and FIGURE 49

FIGURE 49

4r/(0) + 8/(0) = -5, with the solution x'(O) = 1/8 and /(0) = -11/16. (d) h(p) = x{p) + 2y(p), so h'(0) = x'(0) + 2/(0) = -5/4.

18.3

1. The problem with systems of three equations and two unknowns is that they are usually inconsistent (have no solutions), not that they are difficult to solve. The equations f'x(x, y) = f'y{x, y) = 0 are not valid at the optimal point.

3. x = -1 and y = 0 solves the problem, with /(—1,0) = 1. (Actually, this problem is quite tricky. The only stationary point of the Lagrangean is (0, 0), with X = -4, and with /(0,0) = 4. The point is that at (-1,0) both gi(—1,0) and g'2(—1,0) are 0, so the Lagrangean is not necessarily stationary at this point. The problem is to minimize the (square of the) distance from (-2,0) to a point on the graph of g(jc. y) = 0. But the graph consists of the single point (—1,0) and a nice curve, as illustrated in Fig. 50.) FIGURE 50

18.4

1. The Lagrangean £(x, y) = 10:c1/2yly3-X(2jc+4y—m) is concave as a sum of two concave functions (see Example 17.18 in Section 17.8), so Theorem 18.2 applies.

18.5

1. (a) C(x, >-, 7) = x2 + y2 + -2 _ + y + z _ jYie only solution of the necessary conditions'is (1/3, 1/3. 1/3). (b) The problem is to find the shortest distance from the origin to a point in the plane x + y + z = 1. The corresponding maximization problem has no solution.

3. (a) jc = ct(wL+m)/p, y = 0(wL+m)/q, and / = (a+B)L-m(l-ct-P)/w. The condition given is equivalent to / > 0. (b) The solution is I = 0, with x = am/(a + 0)p and y = £m/(a +P)q. (In this case, unearned income is so high that it is optimal for the individual not to work at all.)

5. With linear equality constraints, we shall not resist the temptation to eliminate variables. In fact, adding the constraints gives 3x = 6 and so x = 2. Thus y = -(l + r). Substituting for x and y in the objective function reduces it to 2(1 + z)2 + z2 + z = 3z2 + 5z + 2. This quadratic polynomial has a minimum when z = -5/6. Then y = -1/6. The solution is (*,y.z) = (2,-1/6, -5/6).

7. TheLagrangean is £ = x+y-X{x2+2y2+z2n{x+y+z-\), which is stationary when  Cx = 1 - 2Xx - fx = 0;  £'v = 1 - 4Ay - fu. = 0;  C. = -2Xz - ii = 0. From  and , 2X(x - 2y) = 0. If A = 0, then

 and  yield fM = 1 and fx = 0. Therefore x — 2y instead. Substituting this value for x in the constraints gives 6y2 + z2 = 1, 3y + z = 1. Thus, z = 1 - 3y and 1 = 6y2 + (1 - 3y)2 = 15y2 - 6y + 1. Hence y = 0 or y = 2/5, implying that x = 0 or 4/5, and that z = 1 or -1/5. The only two solution candidates, are (*, y, z) = (0, 0, 1) with A = -1/2, fx = 1, and (x, y, z) = (4/5, 2/5, -1/5) with X = 1/2, M = 1/5- Because x + y is 0 at (0,0,1) and 6/5 at (4/5, 2/5, -1/5), these are respectively the minimum and the maximum. (The constraints determine geometrically the curve which is the intersection of an ellipsoid (see Fig. 15.12 in Section 15.2) and a plane. The continuous function x + y does attain a maximum and a minimum over this closed bounded set.)

9. (a) Here £ = (y+z-3)2-A(x2+y+z-2)-Ai(x+y2+2z-2), which is stationary when  C'x = -2Xx - fx = 0;  £'v = 2(y + z - 3) - A - 2fiy = 0;

 £'. = 2(y + z - 3) - A - 2(u. = 0. From (2) and (3), X + 2fxy = X + i(x, so fi(y - 1) = 0. If ^ = 0, then from  and , Ajc = 0 and 2(y + z - 3) = X. Now X = 0 would imply that y + z = 3 and so x2 = — 1 from the first constraint. Thus X ^ 0 and so p. = 0 x =0. Then the two constraints yield y — 2 — z and y2 = 2(1 - z). so that y2 — 2y + 2 = 0, which also has no real root. We conclude that // # 0 and so y = 1. The two constraints now imply that x2 + z = 1 and x + 2z = 1- Hence x2 + — x) = 1, which has roots x = 1 and x — —1/2. The only two solution candidates are (;c,y,z) = (1,1,0) and (x,y,z) = (—1/2, 1,3/4). The corresponding values of (X, are (4/3, -8/3) and (-5/6, -5/6) respectively. Because (y + z - 3)2 is 4 at (1, 1.0) and 25/16 at (-1/2, 1,3/4), the latter is the appropriate solution. (The method used in Problem 4 can be applied to show that this gives the minimum.) (b) The second solution, which is (1,1,0), gives /(1,1,0) = 4. But (-2,-2,0), for example, satisfies both constraints, and gives /(-2, -2,0) = 25.

11. Differentiating the constraint w.r.t. x} yields g', -I-g'^idx^/dx]) = 0, or

(1) dx3/dx, = -g\/g'y Similarly, (2) dx3/dx2 = -g'2/g'y Now, the first-order conditions for the maximization of z = f(x\,x2, x3), where x3 is a function of (xux2), are (3) Bz/dx^ = f[ + /¡(dx3/dx{) = 0 and

 dz/dx2 = f{ + fi(dx3/dx2) = 0. Substitute from (1) and (2) into (3) and (4), letting /. = f^/g'y This yields the equations f[ - Xg\ = 0 and fi ~ Hi = and by the definition of a, /3' - Xg~ = 0. These are the conditions in (18.17) for n = 3.

18.6

l. x = -l^b, y = -4-A z = -\-Jb, X = —3/yfb, rib) = -6Vb, df*/db = -3 ¡Jb = A. 3. (a)  1 - f A = imx2x3x4,  1 - }A = fix 1*3*4;  1 - |A = /xxix2x4;  1 _ = {¿X1X2X3, together with , , and . (b) Note that /z = 0 would give 1 — |A = 1 - |A = 0, which is impossible. Because xi,..., x4 are all nonzero, it follows from  and  that X] = x2, and from  and  that X3 = x4. Then  and  imply that 1 — jA = ¿¿xixf, whereas  and  imply that 1 - |A = ¿¿xfx3, and also xfxj = 144 by . Hence, X1X3 = 12 and also |xi + \x3 = 3, from the first constraint. These last two equations have two solutions, which are (xi,x3) = (3,4) and (3/2, 8). Maximizing xi +x2+x3+x4 = 2(xi +x3) requires choosing the latter. Thus, (xi,x2,x3,x4) = (3/2, 3/2, 8, 8) must solve the problem, with A = 13 and H = -5/144.