Applications to Business and Economics

The Average Cost Function à c(x) = C(x)/x
The Average Cost Function represents the cost of producing x units of goods or services.
If there is a absolute minimum we can find it by differentiating c(x) = C(x)/x à dc/dx = [x*C’(x)- C(x)]/x^2
i. So When x*C’(x) = C(x) à then dc/dx = 0
ii. The we have that C’(x) = C(x)/x = c(x)
1. Therefore: If the average cost is a minimum, then marginal cost = average cost

Example:
A company estimates that the cost (in dollars) of production x items is
C(x) = 2600+2x+0.001x^2
The company wants to find the cost, average cost, and marginal cost of producing 1000 items, 2000 items, and 3000 items?
They also want to know what production level will the average cost be lowest, and what is the minimum average cost?

Solution:

So we have that the cost function is c(x) = C(x)/x
We were also given that the company found the cost to be
C(x) = 2600+2x+0.001x^2
Therefore we have --> c(x) = 2600+2x+0.001x^2/x
If we use implicit differentiation we get
i. c’(x) = x(2+0.002x)-1(2600+2x +0.001x^2)/x^2
ii. --> c’(x) = 2x+0.002x^2 -2600-2x-0.001x^2/x^2
iii. --> c’(x) = 0.001x^2-2600/x^2
1. c’(x) = 0 when 0.001x^2 = 2600
2. --> x = squrt(2600/0.001) = 1612.45 x gives the number of items that the company should produce in order to minimize cost
3. --> if x = 1612.45 then, we want c(x) to equal C(x) so we get -->
iv. c(1612.45) = [2600+2(1612.45)+0.001(1612.45)^2]/(1612.45) = $5.22 per item
So we have that if they produce about 1612 items of goods and services then the cost per item will be about $5.22 per item.

However, the company wants to find the cost, average cost, and marginal cost of producing 1000 items, 2000 items, and 3000 items?
Because we already found the function c’(x) = 0.001X^2-2600/x^2 is the marginal cost function, and C(x) the average cost function is c(x) = C(x)/x
so we have the following:

Equations used in this solution:
C(x) = 2600+2x+0.001x^2
C’(x) = 2+0.002x
c(x) = C(x)/x = 2600+2x+0.001x^2/x
c’(x) = x*C’(x) – d/dx(x)*C(x)/x^2 = x*C’(x)-C(x)/x^2 = 0.001x^2-2600/x^2
--> our minimum cost is given when c’(x) = 0 in other words when x*C’(x)-C(x) = 0

See the following chart, notice that when C’(x) = C(x) then that is the minimum point.

Considering Marketing, Pricing, and Revenue Associated with Cost
-->The Revenue function is R(x) = x*P(x)
-->The Price function is P(x) = R(x) – C(x)
-->The marginal Revenue is dR/dx
-->The marginal Profit is dP/dx

In order to maximize profit we look for the critical numbers of P
We want to find the critical numbers of P such that P’(x) = R’(x) – C’(x) = 0 thus, R’(x) = C’(x)
So if the profit is a maximum, then marginal revenue = marginal cost

Example:
Determine the production level that will maximize the profit for a company with the given cost and demand functions.
--> total cost for x units is C(x) = 84+1.26x-0.01x^2+0.00007x^3
--> and p(x) = 3.5-0.01x
We want to find the critical numbers of P such that P’(x) = R’(x) – C’(x) = 0

Solution:
-->The Revenue function is given as R(x) = x*p(x) we are given p(x) = 3.5-0.01x
So R(x) = x*(3.5-0.01x) = 3.5x-0.01x^2
So R’(x) = 3.5-.02x
-->We are given C(x) = 84+1.26x-0.01x^2+0.00007x^3
So C’(x) = 1.26-.02x+.00021x^2
Thus marginal revenue R’(x) is equal to marginal cost C’(x) when
3.5- 0.02x = 1.26 -0.02x+.00021x^2
0 = 1.26-0.02+00021x^2+0.02x-3.5
0 = -2.24+00021x^2
Squrt 2.24/00021 = x
x = 103.27 or about 103
Therefore, a production level of 103 units will maximize the profit.