Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
To solve the given problem, let's break it down into a series of steps:
1. Find the Profit Function [tex]\( P(x) \)[/tex]:
We know that the profit function [tex]\( P(x) \)[/tex] is given by the difference between the revenue function [tex]\( R(x) \)[/tex] and the cost function [tex]\( C(x) \)[/tex].
- The cost function is: [tex]\( C(x) = 5x + 7 \)[/tex]
The revenue function [tex]\( R(x) \)[/tex] is the product of the price-demand function and the number of vats sold.
- The price-demand function is: [tex]\( p(x) = 71 - 3x \)[/tex]
Therefore, the revenue function is: [tex]\( R(x) = p(x) \cdot x \)[/tex]
[tex]\( R(x) = (71 - 3x) \cdot x \)[/tex]
[tex]\( R(x) = 71x - 3x^2 \)[/tex]
Now, the profit function is:
[tex]\( P(x) = R(x) - C(x) \)[/tex]
[tex]\( P(x) = (71x - 3x^2) - (5x + 7) \)[/tex]
[tex]\( P(x) = 71x - 3x^2 - 5x - 7 \)[/tex]
[tex]\( P(x) = -3x^2 + 66x - 7 \)[/tex]
So, the profit function [tex]\( P(x) = -3x^2 + 66x - 7 \)[/tex].
2. Find the Number of Vats to Maximize Profit:
To maximize the profit, we need to find the critical points by taking the derivative of the profit function and setting it to zero.
The derivative of [tex]\( P(x) \)[/tex] is:
[tex]\( P'(x) = d(-3x^2 + 66x - 7) / dx \)[/tex]
[tex]\( P'(x) = -6x + 66 \)[/tex]
Set the derivative to zero to find the critical points:
[tex]\[ -6x + 66 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -6x = -66 \][/tex]
[tex]\[ x = 11 \][/tex]
Therefore, 11 vats of ice cream need to be sold to maximize the profit.
3. Find the Maximum Profit:
Substitute [tex]\( x = 11 \)[/tex] back into the profit function [tex]\( P(x) \)[/tex] to find the maximum profit:
[tex]\( P(11) = -3(11)^2 + 66(11) - 7 \)[/tex]
[tex]\( P(11) = -3(121) + 726 - 7 \)[/tex]
[tex]\( P(11) = -363 + 726 - 7 \)[/tex]
[tex]\( P(11) = 356 \)[/tex]
The maximum profit is [tex]$356. 4. Find the Price to Charge Per Vat to Maximize Profit: Substitute \( x = 11 \) back into the price-demand function \( p(x) \): \[ p(11) = 71 - 3(11) \] \[ p(11) = 71 - 33 \] \[ p(11) = 38 \] The price to charge per vat to maximize profit is $[/tex]38.
To summarize:
1. The profit function is: [tex]\( P(x) = -3x^2 + 66x - 7 \)[/tex].
2. The number of vats to be sold to maximize the profit is 11.
3. The maximum profit is [tex]$356. 4. The price to charge per vat to maximize profit is $[/tex]38.
1. Find the Profit Function [tex]\( P(x) \)[/tex]:
We know that the profit function [tex]\( P(x) \)[/tex] is given by the difference between the revenue function [tex]\( R(x) \)[/tex] and the cost function [tex]\( C(x) \)[/tex].
- The cost function is: [tex]\( C(x) = 5x + 7 \)[/tex]
The revenue function [tex]\( R(x) \)[/tex] is the product of the price-demand function and the number of vats sold.
- The price-demand function is: [tex]\( p(x) = 71 - 3x \)[/tex]
Therefore, the revenue function is: [tex]\( R(x) = p(x) \cdot x \)[/tex]
[tex]\( R(x) = (71 - 3x) \cdot x \)[/tex]
[tex]\( R(x) = 71x - 3x^2 \)[/tex]
Now, the profit function is:
[tex]\( P(x) = R(x) - C(x) \)[/tex]
[tex]\( P(x) = (71x - 3x^2) - (5x + 7) \)[/tex]
[tex]\( P(x) = 71x - 3x^2 - 5x - 7 \)[/tex]
[tex]\( P(x) = -3x^2 + 66x - 7 \)[/tex]
So, the profit function [tex]\( P(x) = -3x^2 + 66x - 7 \)[/tex].
2. Find the Number of Vats to Maximize Profit:
To maximize the profit, we need to find the critical points by taking the derivative of the profit function and setting it to zero.
The derivative of [tex]\( P(x) \)[/tex] is:
[tex]\( P'(x) = d(-3x^2 + 66x - 7) / dx \)[/tex]
[tex]\( P'(x) = -6x + 66 \)[/tex]
Set the derivative to zero to find the critical points:
[tex]\[ -6x + 66 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -6x = -66 \][/tex]
[tex]\[ x = 11 \][/tex]
Therefore, 11 vats of ice cream need to be sold to maximize the profit.
3. Find the Maximum Profit:
Substitute [tex]\( x = 11 \)[/tex] back into the profit function [tex]\( P(x) \)[/tex] to find the maximum profit:
[tex]\( P(11) = -3(11)^2 + 66(11) - 7 \)[/tex]
[tex]\( P(11) = -3(121) + 726 - 7 \)[/tex]
[tex]\( P(11) = -363 + 726 - 7 \)[/tex]
[tex]\( P(11) = 356 \)[/tex]
The maximum profit is [tex]$356. 4. Find the Price to Charge Per Vat to Maximize Profit: Substitute \( x = 11 \) back into the price-demand function \( p(x) \): \[ p(11) = 71 - 3(11) \] \[ p(11) = 71 - 33 \] \[ p(11) = 38 \] The price to charge per vat to maximize profit is $[/tex]38.
To summarize:
1. The profit function is: [tex]\( P(x) = -3x^2 + 66x - 7 \)[/tex].
2. The number of vats to be sold to maximize the profit is 11.
3. The maximum profit is [tex]$356. 4. The price to charge per vat to maximize profit is $[/tex]38.
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
