Answer :
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.
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.
