Question

What are the cost-minimizing values of L and K for the following functions? Note that w=r=2 and the production target is 50

(a) (5) q=f(L,K)=2LK
(b) (5) q=f(L,K)=2L+K
(c) (5) q=f(L,K)=min{2L,K}
(d) (5) q=f(L,K)=2L+√K

Answer 1

To find the cost-minimizing values of L and K, we need to take the derivative of the production function with respect to each variable. In function (a), both L and K are 0, in function (b) there are no cost-minimizing values for L and K, in function (c) there are also no cost-minimizing values for L and K, and in function (d), L is 0 and K is 0.

Step-by-step explanation:

To find the cost-minimizing values of L and K, we need to take the derivative of the production function with respect to each variable. In both functions (a) and (b) the derivative of the production function with respect to L is 2K, whereas in function (c) it is 2, and in function (d) it is 2. Similarly, the derivative with respect to K is 2L in function (a), 1 in function (b), and 0 in function (c) and (d). We can set these derivatives equal to 0 to find the critical points:

1. For function (a): Setting the derivative with respect to L equal to 0 gives 2K = 0, which implies K = 0. Setting the derivative with respect to K equal to 0 gives 2L = 0, which implies L = 0. Therefore, the cost-minimizing values of L and K are both 0.
2. For function (b): Setting the derivative with respect to L equal to 0 gives 2 = 0. This has no solution, so there are no cost-minimizing values for L and K.
3. For function (c): Setting the derivative with respect to L equal to 0 gives 2K = 0, which implies K = 0. Setting the derivative with respect to K equal to 0 gives 2 = 0. This also has no solution, so there are no cost-minimizing values for L and K.
4. For function (d): Setting the derivative with respect to L equal to 0 gives 2 = 0, which has no solution. Setting the derivative with respect to K equal to 0 gives
   `2L + 0.5√K = 0`. This equation simplifies to
   `2L = -0.5√K.`We can square both sides to get
   `(2L)^2 = (-0.5√K)^2`. This yields 4L^2 = 0.25K. Since K cannot be negative, the cost-minimizing value of L is 0 and the cost-minimizing value of K is 0.