6 15 points A simplified Cobb-Douglas model for a local economical output P that depends on only two variables r, y is P(x, y) = 18:2/3 yl/3 But there is a constraint on the amount of labor (represented by a) and the amount of capital invested (represented by y). There is no infinite labor available, and no infinite capital to invest either. Assume that the constraint can be modeled as 2 + y = 6. Find the e, y that can maximize the economical output P. This is only a 2-variable optimization problem, same as in previous problem, you can eliminate one variable and reduce it to a single variable optimization problem, then rely on calculus-1 to solve it. But it is no easier than applying the more powerful Lagrange multiplier technique. Using Lagrange multiplier, the solution steps include: step 1: Find the Lagrangian L(x, y, A) = f(x, y) - Ag(x, y). Here f(x, y) = P(x, y) is obvious, and the g(a, y) for the constraint g(x, y) = 0 can beg(x, y) = 2x + y -6. Comment: You can also use g(x, y) = 6 - (2x + y), but it only changes the sign of the final Lagrange multiplier you'll get. It is equivalent to using L(x, y, A) = f(x, y) + Ag(a, y) directly. For grading consistency, use the formulas outside this comment. ) step 2: Find the critical point, which is a solution (i, y, )) for the vector equation VL(x, y, A) = 0. (Or equivalently, Vf(r, y) = >Vg(x, y), plus the constraint) Essentially, you solve L, (r, y, )) = 0, Ly (x, y, A) = 0, Lx(x, y, X) = 0. Lx(x, y, A) = 0 leads to 12x y = 21 , where a = type your answer. type your answer.. Ly(x, y, A) = 0 leads to 6x6yd = > , where c = type your answer. d = type your answer. Lx(x, y, A) = 0 leads to ra + sy = t, where r' = type your answer. type your answer.. type your answer. (input a most simplified quotient if the answer is not integer, or an integer) If you can find thee, y that can solve these equations, you do not need to find > explicitly (unless you want this extra step). For some problems, you would need to find A before you can get the critical point(s), not so here. Eliminate ) and simplify, (that is, you come up with two equations for the two unknowns z and y), you can solve for the critical point (c, y), where = = type your answer. y = type your answer. step 3: Plug the solution , y from step 2 into f(x, y) to get its extreme value. The extreme value of P(x, y) must be type your answer.. step 4: Figure out if the extreme value is a maximum or minimum (or maybe neither if the critical point is a saddle point). Here we cannot use the second order derivative test since it is a constrained optimization problem. But this step is quite obvious for this problem that has some real economical meaning: Clearly : 2 0, y 2 0. So the closed-set that encloses the allowed region of (a, y) turns out to be the line segment of 2:x + y = 6 in the first quadrant. At its boundary (r = 0, y = 6) and (x = 3, y = 0), it is always P(x, y) = type your answer. , which is smaller than the P(a, y) at the critical point, therefore the extreme value must be a maximum, which means we get the optimal output allowed by the budget constraint