Z, = d - h(x, xz, . . .. x,,) = 0 Z; = fi - 181 - uh; =0 (i = 1, 2, . .., ") These should normally enable us to solve for all the x, as well as > and u. As before, first two equations of the necessary condition represent essentially a mere restatement the two constraints. ERCISE 12.2 1. Use the Lagrange-multiplier method to find the stationary values of z: (a) z = xy, subject to x + 2y = 2. (b) z= x(y + 4), subject to x + y = 8. (c) z = x - 3y - xy, subject to x + y = 6. (d) z= 7 - y+ x2, subject to x + y =0. 2. In Prob. 1, find whether a slight relaxation of the constraint will increase or decrease the optimal value of z. At what rate? 3. Write the Lagrangian function and the first-order condition for stationary values (with- out solving the equations) for each of the following: ( a) z = x + 2y + 3w+ xy- yw, subject to x + y+ 2w=10. (b) z = x2 + 2xy + yw2, subject to 2x + y + w= 24 and x + w=8. 4. If, instead of g(x, y) = c, the constraint is written in the form of G(x, y) = 0, how should the Lagrangian function and the first-order condition be modified as a consequence? 5. In discussing the total-differential approach, it was pointed out that, given the con- straint g(x, y) = c, we may deduce that dg = 0. By the same token, we can further deduce that d'g = d(dg) = d(0) = 0. Yet, in our earlier discussion of the unconstrained extremum of a function z = f(x, y), we had a situation where dz = 0 is accompanied by either a positive definite or a negative definite d2z, rather than d2z = 0. How would you account for this disparity of treatment in the two cases? 6. If the Lagrangian function is written as Z = f(x, y) + >[g(x, y) - c] rather than as in (12.7), can we still interpret the Lagrange multiplier as in (12.16)? Give the new inter- pretation, if any