To find the maximum profit and the optimal selling price for the roll of chicken wire, follow these steps:
1. Define the revenue and cost functions:
[tex]\[ R(x) = -8.25x^2 + 560x \][/tex]
[tex]\[ C(x) = -17.5x + 8250 \][/tex]
2. Express the profit function [tex]\( P(x) \)[/tex] as the difference between revenue and cost:
[tex]\[ P(x) = R(x) - C(x) \][/tex]
[tex]\[ P(x) = (-8.25x^2 + 560x) - (-17.5x + 8250) \][/tex]
[tex]\[ P(x) = -8.25x^2 + 560x + 17.5x - 8250 \][/tex]
[tex]\[ P(x) = -8.25x^2 + 577.5x - 8250 \][/tex]
3. To maximize the profit, take the derivative of the profit function [tex]\( P(x) \)[/tex] and set it to zero to find the critical points:
[tex]\[ P'(x) = \frac{d}{dx} (-8.25x^2 + 577.5x - 8250) \][/tex]
[tex]\[ P'(x) = -16.5x + 577.5 \][/tex]
[tex]\[ 0 = -16.5x + 577.5 \][/tex]
[tex]\[ 16.5x = 577.5 \][/tex]
[tex]\[ x = \frac{577.5}{16.5} \][/tex]
[tex]\[ x = 35 \][/tex]
4. Substitute [tex]\( x = 35 \)[/tex] back into the profit function to find the maximum profit:
[tex]\[ P(35) = -8.25(35)^2 + 577.5(35) - 8250 \][/tex]
[tex]\[ P(35) = -8.25 \cdot 1225 + 577.5 \cdot 35 - 8250 \][/tex]
[tex]\[ P(35) = -10106.25 + 20212.5 - 8250 \][/tex]
[tex]\[ P(35) = 1856.25 \][/tex]
5. To find the optimal selling price, which is the price per unit that maximizes profit, substitute [tex]\( x = 35 \)[/tex] into the revenue function and divide by [tex]\( x \)[/tex]:
[tex]\[ R(35) = -8.25 \cdot 35^2 + 560 \cdot 35 \][/tex]
[tex]\[ R(35) = -8.25 \cdot 1225 + 560 \cdot 35 \][/tex]
[tex]\[ R(35) = -10106.25 + 19600 \][/tex]
[tex]\[ R(35) = 9493.75 \][/tex]
Divide this by [tex]\( x \)[/tex] (35) to get the selling price per unit:
[tex]\[ \text{Optimal Selling Price} = \frac{R(35)}{35} = \frac{9493.75}{35} = 271.25 \][/tex]
Thus, the maximum profit of \[tex]$1856.25 can be made when the selling price of the chicken wire is set to \$[/tex]271.25.
So the correct answers are:
- Maximum profit: [tex]\(\$1856\)[/tex]
- Optimal selling price: [tex]\(\$271\)[/tex]
