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Sagot :
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]
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]
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