1. Suppose the production function is given by Q(L,K) = L K . Assuming capital is fixed,
find APL and MPL
.
2. Consider the following short run production function:
Q 6L
2
0.4L
3
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
3. Given a short run cost function as T
2 60 100
3
C
1
Q
3
Q
2
Q
, find the minimum value
of AVC and MC.
1. Assuming capital is fixed, MPL (Marginal Product of Labor) is the derivative of the production function with respect to labor (L). So, taking the derivative of Q(L, K) with respect to L, we get:
MPL = dQ/dL = K
APL (Average Product of Labor) is the total product (Q) divided by the quantity of labor (L). So, APL can be calculated as:
APL = Q/L = LK/L = K
2. a) To find the value of L that maximizes output, we need to find the maximum point of the production function. Taking the derivative of Q with respect to L:
dQ/dL = 12L - 1.2L^2
Setting this equal to zero and solving for L:
12L - 1.2L^2 = 0
L(12 - 1.2L) = 0
L = 0 or L = 10
Since we are looking for a positive value for L, the value that maximizes output is L = 10.
b) To find the value of L that maximizes marginal product, we need to find the value of L at which the derivative of MPL with respect to L is equal to zero. Taking the derivative of MPL (K) with respect to L, we get:
dMPL/dL = dK/dL = 0
Since capital is fixed, the derivative of K with respect to L is zero regardless of the value of L. Therefore, there is no specific value of L that maximizes marginal product.
c) To find the value of L that maximizes average product, we need to find the maximum point of the APL function. APL is given by:
APL = Q/L = (6L^2 - 0.4L^3)/L = 6L - 0.4L^2
Taking the derivative of APL with respect to L:
dAPL/dL = 6 - 0.8L
Setting this equal to zero and solving for L:
6 - 0.8L = 0
0.8L = 6
L = 6/0.8
L = 7.5
Therefore, the value of L that maximizes average product is L = 7.5.
3. To find the minimum value of AVC (Average Variable Cost), we need to find the value of Q (quantity of output) that minimizes AVC. AVC is given by:
AVC = TFC/Q + TVC/Q = (60/Q^2) + 100 - Q + Q/Q^2
Taking the derivative of AVC with respect to Q:
dAVC/dQ = (-120/Q^3) + 1
Setting this equal to zero and solving for Q:
(-120/Q^3) + 1 = 0
120/Q^3 = 1
Q^3 = 120
Q = ∛(120)
Q = 4.932
Therefore, the minimum value of AVC occurs at Q = 4.932.
To find the minimum value of MC (Marginal Cost), we need to find the value of Q that minimizes MC. MC is given by:
MC = dC/dQ = (d/dQ)(T^2/Q^3 - Q^2 + Q)
Taking the derivative of MC with respect to Q:
dMC/dQ = (2T/Q^4) - 2Q + 1
Setting this equal to zero and solving for Q:
(2T/Q^4) - 2Q + 1 = 0
Since T is not given in the equation, we cannot solve for Q and determine the minimum value of MC without additional information.