1. Suppose the production function is given by Q(L,K) = L
4
3
K
4
1
. Assuming capital is fixed,
find APL and MPL
.
2. Consider the following short run production function:
2 3 Q 6L 0.4L
a) Find the value of L that maximizes output
b) Find the value of L that maximizes marginal product
c) Find the value of L that maximizes average product
1. Assuming capital is fixed, we can find the average product of labor (APL) and marginal product of labor (MPL) using the given production function Q(L, K) = L^4 K^1.
APL:
APL = Q / L
MPL:
MPL = ∂Q/∂L
To find APL, divide the quantity produced (Q) by the amount of labor (L):
APL = Q(L, K) / L
APL = L^4 K / L
APL = L^3 K
To find MPL, differentiate the production function with respect to L:
MPL = ∂Q/∂L = 4L^3 K
2. The given short-run production function is Q(L) = 6L^2 - 0.4L^3.
a) To find the value of L that maximizes output, we need to find the maximum of the function Q(L).
To do this, take the derivative of Q(L) with respect to L and set it equal to zero, then solve for L:
dQ/dL = 12L - 1.2L^2 = 0
Solving this equation:
12L - 1.2L^2 = 0
L(12 - 1.2L) = 0
L = 0 or 12 - 1.2L = 0
If L = 0, then Q = 0. However, for positive output, we need to consider the second equation:
12 - 1.2L = 0
1.2L = 12
L = 10
Therefore, the value of L that maximizes output is L = 10.
b) To find the value of L that maximizes marginal product, we need to find the maximum of the derivative of Q(L) with respect to L.
To do this, take the second derivative of Q(L) and set it equal to zero, then solve for L:
d^2Q/dL^2 = 12 - 2.4L = 0
Solving this equation:
12 - 2.4L = 0
2.4L = 12
L = 5
Therefore, the value of L that maximizes marginal product is L = 5.
c) To find the value of L that maximizes average product, we need to find the maximum of the function Q(L)/L.
To do this, divide Q(L) by L and find the maximum of the resulting function:
Q(L)/L = (6L^2 - 0.4L^3) / L
= 6L - 0.4L^2
To find the maximum of this function, take the derivative with respect to L and set it equal to zero:
d(Q(L)/L)/dL = 6 - 0.8L = 0
Solving this equation:
6 - 0.8L = 0
0.8L = 6
L = 7.5
Therefore, the value of L that maximizes average product is L = 7.5.