To determine the maximum profit and the corresponding selling price for the dog food, let's break down the problem into a series of steps.
1. Understand the Functions:
- The revenue function, [tex]\( R(x) \)[/tex], represents how much money is made from sales:
[tex]\[
R(x) = -31.72x^2 + 2030x
\][/tex]
- The cost function, [tex]\( C(x) \)[/tex], represents the total cost of producing and selling the x units:
[tex]\[
C(x) = -126.96x + 26391
\][/tex]
2. Calculate the Profit Function:
- The profit function, [tex]\( P(x) \)[/tex], is the revenue function minus the cost function:
[tex]\[
P(x) = R(x) - C(x)
\][/tex]
- Substituting the given functions:
[tex]\[
P(x) = (-31.72x^2 + 2030x) - (-126.96x + 26391)
\][/tex]
Simplify to get:
[tex]\[
P(x) = -31.72x^2 + 2030x + 126.96x - 26391
\][/tex]
[tex]\[
P(x) = -31.72x^2 + 2156.96x - 26391
\][/tex]
3. Find the Critical Points:
- To find the maximum profit, we need to find the critical points by taking the derivative of [tex]\( P(x) \)[/tex] and setting it to zero:
[tex]\[
P'(x) = \frac{d}{dx}(-31.72x^2 + 2156.96x - 26391)
\][/tex]
[tex]\[
P'(x) = -63.44x + 2156.96
\][/tex]
Set the derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
-63.44x + 2156.96 = 0
\][/tex]
[tex]\[
x = \frac{2156.96}{63.44}
\][/tex]
[tex]\[
x \approx 34
\][/tex]
4. Calculate the Maximum Profit:
- Substitute [tex]\( x = 34 \)[/tex] back into the profit function to find the maximum profit:
[tex]\[
P(34) = -31.72(34)^2 + 2156.96(34) - 26391
\][/tex]
Calculate each term:
[tex]\[
-31.72 \times 1156 + 2156.96 \times 34 - 26391
\][/tex]
Simplify the terms:
[tex]\[
-36704.32 + 73337 - 26391
\][/tex]
[tex]\[
P(34) \approx 10277
\][/tex]
Therefore, the maximum profit of [tex]$10,277 can be made when the selling price of the dog food is set to $[/tex]34 per bag.
The correct answers to fill in the blanks are:
- The maximum profit of \[tex]$10,277 can be made when the selling price of the dog food is set to \$[/tex]34 per bag.
