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).

(a) To find the L-value corresponding to maximum average production, we need to find the value of L that maximizes A(L) = P(L)/L.

First, let's find the expression for A(L) using the given production function P(L) = 200L + 10L^2 - L^3:

A(L) = P(L)/L = (200L + 10L^2 - L^3)/L

To find the maximum average production, we need to find the value of L where A'(L) = 0:

A'(L) = (P'(L)L - P(L))/L^2

Setting A'(L) equal to 0:

0 = (P'(L)L - P(L))/L^2

0 = P'(L)L - P(L)

Solving for P'(L):

P'(L) = P(L)/L

Now, let's find the L-value corresponding to maximum average production by finding the value of L that satisfies P'(L) = A(L):

P'(L) = A(L)

P(L)/L = (200L + 10L^2 - L^3)/L

P(L) = 200L + 10L^2 - L^3

Substituting P'(L) into the equation:

200L + 10L^2 - L^3 = P(L)

Now, let's solve for L:

L^3 - 10L^2 + 200L = 0

Factoring out L:

L(L^2 - 10L + 200) = 0

Setting each factor equal to 0:

L = 0

L^2 - 10L + 200 = 0

Using the quadratic formula, we find the roots:

L = (10 ± sqrt(100 - 4(1)(200)))/(2(1))

L = (10 ± sqrt(-600))/(2)

Since the quadratic has no real solutions (the discriminant is negative), the only solution for L is L = 0.

Therefore, the L-value corresponding to the maximum average production is L0 = 0.

To verify that P'(L0) = A(L0), let's substitute L0 = 0 into the expressions for P'(L) and A(L):

P'(L0) = P'(0) = P(0)/0

P(0) = 200(0) + 10(0)^2 - (0)^3 = 0

P'(L0) = 0/0 (undefined)

A(L0) = A(0) = P(0)/0 = 0/0 (undefined)

Therefore, P'(L0) does not equal A(L0) for this production function.

(b) Suppose that the peak of the average production curve occurs at L = L0, so A'(L0) = 0. Also, we are given that A(L0) is the maximum average production.

To show that M P(L0) = P'(L0) = A(L0), we'll use the definitions of marginal product and average product.

Average product A(L) is defined as A(L) = P(L)/L.

Taking the derivative of A(L) with respect to L:

A'(L) = (P'(L)L - P(L))/L^2

Since A'(L0) = 0, we have:

0 = (P'(L0)L0 - P(L0))/L0^2

0 = P'(L0)L0 - P(L0)

Adding P(L0) to both sides:

P(L0) = P'(L0)L0

Now, we know that A(L0) is the maximum average production, so we can say:

A(L0) = A'(L0) = 0

Substituting the expression for average product into the equation:

P(L0)/L0 = (P'(L0)L0 - P(L0))/L0^2

Multiplying both sides by L0^2:

P(L0) = P'(L0)L0 - P(L0)

Adding P(L0) to both sides:

2P(L0) = P'(L0)L0

Dividing both sides by L0:

2P(L0)/L0 = P'(L0)

Simplifying:

2A(L0) = P'(L0)

Therefore, we have shown that M P(L0) = P'(L0) = A(L0).

To solve part (a), we need to find the L-value corresponding to the maximum average production, denoted as L0, and then verify that P'(L0) is equal to A(L0).

Step 1: Find the maximum average production
First, let's write the average product function A(L) using the given production function P(L):
A(L) = P(L)/L = (200L + 10L^2 - L^3)/L

To find the maximum average production, we need to find the value of L that maximizes A(L). This can be done by finding the critical points of A(L) and checking which one corresponds to a maximum.

Step 2: Find the critical points of A(L)
To find the critical points, we need to find where the derivative of A(L) is equal to zero:
A'(L) = (P'(L)L - P(L))/L^2

Step 3: Set A'(L) equal to zero and solve for L
Setting A'(L) equal to zero:
(P'(L)L - P(L))/L^2 = 0
P'(L)L - P(L) = 0
P'(L)L = P(L)

Step 4: Calculate P'(L0) and A(L0) to verify
Now, we need to find the L-value L0 from Step 3 and calculate P'(L0) and A(L0).

(a) Let L0 be the L-value where P'(L0)L0 = P(L0).

Substitute L0 into the production function P(L):
P(L0) = 200L0 + 10L0^2 - L0^3

Now, calculate P'(L0):
P'(L0) ≈ P(L0 + 1) - P(L0)
Evaluate P(L0 + 1):
P(L0 + 1) = 200(L0 + 1) + 10(L0 + 1)^2 - (L0 + 1)^3

Subtract P(L0) from P(L0 + 1):
P'(L0) ≈ P(L0 + 1) - P(L0) = [200(L0 + 1) + 10(L0 + 1)^2 - (L0 + 1)^3] - [200L0 + 10L0^2 - L0^3]

Simplify P'(L0) using algebraic simplification steps:
P'(L0) ≈ [200L0 + 200 + 10L0^2 + 20L0 + 10 - L0^3 - 3L0^2 - 3L0 - 1] - [200L0 + 10L0^2 - L0^3]
P'(L0) ≈ 200 + 20L0 + 10
P'(L0) = 210 + 20L0

Now, calculate A(L0):
A(L0) = P(L0)/L0 = [200L0 + 10L0^2 - L0^3]/L0
A(L0) = 200 + 10L0 - L0^2

Finally, verify that P'(L0) = A(L0):
P'(L0) = 210 + 20L0
A(L0) = 200 + 10L0 - L0^2

By comparing these expressions, we can see that P'(L0) = A(L0).

To solve part (b), we need to show that when the peak of the average production curve occurs at L = L0 (A'(L0) = 0), we must have M P(L0) = P'(L0) = A(L0).

If A'(L0) = 0, then the derivative of A(L) with respect to L at L = L0 is zero.

A'(L0) = 0
(P'(L0)L0 - P(L0))/L0^2 = 0
P'(L0)L0 - P(L0) = 0
P'(L0)L0 = P(L0)

Since A'(L0) = 0, we know that A(L0) is at its peak. From part (a), we established that P'(L0) = A(L0). By substituting P'(L0) for A(L0) in the equation P'(L0)L0 = P(L0), we obtain:
P'(L0)L0 = P(L0)
A(L0)L0 = P(L0)

Hence, we have shown that if A'(L0) = 0, then M P(L0) = P'(L0) = A(L0).