To determine the maximum profit and the optimal selling price per bag, we need to follow these steps:
1. Define the Revenue and Cost Functions:
- The revenue function [tex]\( R(x) \)[/tex] is given by [tex]\( R(x) = -31.72x^2 + 2030x \)[/tex], where [tex]\( x \)[/tex] represents the number of bags sold.
- The cost function [tex]\( C(x) \)[/tex] is given by [tex]\( C(x) = -126.96x + 26391 \)[/tex].
2. Define the Profit Function:
- The profit function [tex]\( P(x) \)[/tex] is the difference between the revenue and cost functions:
[tex]\[
P(x) = R(x) - C(x) = (-31.72x^2 + 2030x) - (-126.96x + 26391)
\][/tex]
Simplifying, we get:
[tex]\[
P(x) = -31.72x^2 + 2156.96x - 26391
\][/tex]
3. Finding the Maximum Profit:
- To find the maximum profit, we need to determine the critical points by setting the derivative of [tex]\( P(x) \)[/tex] with respect to [tex]\( x \)[/tex] to zero and solving for [tex]\( x \)[/tex].
- The derivative of [tex]\( P(x) \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[
P'(x) = \frac{d}{dx}(-31.72x^2 + 2156.96x - 26391)
\][/tex]
Simplifying, we get:
[tex]\[
P'(x) = -63.44x + 2156.96
\][/tex]
- Setting this equal to zero and solving for [tex]\( x \)[/tex]:
[tex]\[
-63.44x + 2156.96 = 0
\][/tex]
Solving for [tex]\( x \)[/tex], we get:
[tex]\[
x = \frac{2156.96}{63.44} = 34
\][/tex]
4. Evaluating the Profit Function at the Critical Points:
- We evaluate [tex]\( P(x) \)[/tex] at [tex]\( x = 34 \)[/tex] to find the maximum profit:
[tex]\[
P(34) = -31.72(34)^2 + 2156.96(34) - 26391
\][/tex]
Simplifying this, we get:
[tex]\[
P(34) = 10277
\][/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.
Hence, the completed statement is:
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.
