To determine the maximum profit, we will follow several steps:
1. Define the profit function [tex]\( P(x) \)[/tex]:
The profit function [tex]\( P(x) \)[/tex] is the revenue function [tex]\( R(x) \)[/tex] minus the cost function [tex]\( C(x) \)[/tex].
[tex]\[
P(x) = R(x) - C(x)
\][/tex]
Given:
[tex]\[
R(x) = -13.85x^2 + 1660x
\][/tex]
and
[tex]\[
C(x) = 55400 - 279x
\][/tex]
The profit function becomes:
[tex]\[
P(x) = (-13.85x^2 + 1660x) - (55400 - 279x)
\][/tex]
Simplifying [tex]\( P(x) \)[/tex] by combining like terms:
[tex]\[
P(x) = -13.85x^2 + 1660x + 279x - 55400
\][/tex]
[tex]\[
P(x) = -13.85x^2 + 1939x - 55400
\][/tex]
2. Find the first derivative of the profit function [tex]\( P(x) \)[/tex] to identify critical points:
To find the maximum profit, we need to find the derivative of [tex]\( P(x) \)[/tex] and set it to zero:
[tex]\[
P'(x) = \frac{d}{dx}(-13.85x^2 + 1939x - 55400)
\][/tex]
[tex]\[
P'(x) = -27.7x + 1939
\][/tex]
Setting the derivative equal to zero to find critical points:
[tex]\[
-27.7x + 1939 = 0
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
27.7x = 1939
\][/tex]
[tex]\[
x = \frac{1939}{27.7} \approx 70
\][/tex]
3. Evaluate the profit function [tex]\( P(x) \)[/tex] at the critical point:
Substitute [tex]\( x = 70 \)[/tex] back into the profit function to find the profit:
[tex]\[
P(70) = -13.85(70)^2 + 1939(70) - 55400
\][/tex]
Calculating each term:
[tex]\[
-13.85(70)^2 = -13.85 \times 4900 = -67965
\][/tex]
[tex]\[
1939(70) = 1939 \times 70 = 135730
\][/tex]
[tex]\[
P(70) = -67965 + 135730 - 55400
\][/tex]
[tex]\[
P(70) = 67965 - 55400
\][/tex]
[tex]\[
P(70) = 12465
\][/tex]
Thus, the maximum profit that can be made from selling the ovens is [tex]\(\$12,465\)[/tex].
The correct answer is:
A. [tex]\(\$ 12,465\)[/tex]
