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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.
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