Economists use production functions to describe how output of a system varies with another variable such as labour or capital. For example, the production function P(L) = 200L + 10L^2 - L^3 goves the output of a system as a function of the number of labourers. The average product A(L) is the average output per labourer when L labourers are working: A(L) = P(L)/L. The marginal product M P (L) is the approximate change in output when one additional labourer is added to L labourers; that is, M P(L) = P'(L) ~ P(L + 1) - P(L).
(a) For the production function P(L) = 200L + 10L^2 - L^3, find the L-value corresponding to maximum average production, and call this value L0. Verify that P'(L0) = A(L0).
(b) Now let P(L) be any general production function (not just the one in part (a)), and suppose that the peak of the average production curve occurs at L = L0, so that A'(L0) = 0. Show that we must have M P(L0) = P'(L0) = A(L0).
P(L) = 200L + 10L^2 - L^3
P'(L) = 200 + 20L - 3L^2
A(L) = 200 + 10L - L^2
A' = 10 - 2L
Max avg P is where L=5 That is L0
A(5) = 200+50-25 = 225
P'(5) = 200 + 100 - 75 = 225 = A(5)
A'(L0) = 0 means (P(L)/L)' = 0 at L0
(P'*L - P)/L^2 = 0
P'*L - P = 0
P' = P/L = A
To find the L-value corresponding to the maximum average production (L0) for the production function P(L) = 200L + 10L^2 - L^3, we need to follow these steps:
(a) First, find the expression for the average product A(L) by dividing the total output (P(L)) by the number of laborers (L):
A(L) = P(L) / L
Substitute the given production function into the equation:
A(L) = (200L + 10L^2 - L^3) / L
Simplify the equation:
A(L) = 200 + 10L - L^2
Next, we need to find the derivative of A(L) with respect to L to find the maximum average production point:
A'(L) = 10 - 2L
Set A'(L) = 0 to find the critical point:
10 - 2L = 0
2L = 10
L = 5
So, L0 = 5 represents the L-value corresponding to the maximum average production for this production function.
To verify that P'(L0) = A(L0), we need to find the marginal product at L = L0:
P'(L0) = P(L0 + 1) - P(L0)
Substitute the production function into the equation:
P'(L0) = (200(L0 + 1) + 10(L0 + 1)^2 - (L0 + 1)^3) - (200L0 + 10L0^2 - L0^3)
Simplify the equation:
P'(L0) = 200 + 10(2L0 + 1) - 3(L0^2 + L0 + 1)
P'(L0) = 200 + 20L0 + 10 - 3L0^2 - 3L0 - 3
P'(L0) = 10 + 17L0 - 3L0^2
Now substitute L0 = 5 into the equation:
P'(L0) = 10 + 17(5) - 3(5^2)
P'(L0) = 10 + 85 - 75
P'(L0) = 20
Now, substitute L0 = 5 into the average product equation:
A(L0) = 200 + 10(5) - (5^2)
A(L0) = 200 + 50 - 25
A(L0) = 225
Therefore, we can verify that P'(L0) = A(L0) = 20 = 225.
(b) Now suppose the peak of the average production curve occurs at L = L0, so that A'(L0) = 0.
We can start by finding the expression for A'(L) by taking the derivative of A(L):
A'(L) = d/dL (P(L)/L) = (P'(L)L - P(L))/L^2
Since A'(L0) = 0, we can write:
(P'(L0)L0 - P(L0))/L0^2 = 0
P'(L0)L0 - P(L0) = 0
Now, to show that M P (L0) = P'(L0) = A(L0), we need to prove that the marginal product (M P(L0)) is equal to P'(L0) and A(L0).
First, let's find the expression for M P(L):
M P(L) = P'(L) ~ P(L + 1) - P(L)
Substitute P(L) = P(L + 1) into the equation:
M P(L) = P'(L) ~ P(L) - P(L) = P'(L)
Therefore, we can conclude that M P(L) = P'(L).
Substituting L0 into the equation:
M P(L0) = P'(L0)
Since we showed earlier that A(L0) = P'(L0), it implies that:
M P(L0) = P'(L0) = A(L0)
So, we have successfully shown that when the peak of the average production curve occurs at L0 and A'(L0) = 0, then M P(L0) = P'(L0) = A(L0).