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Sagot :
To solve this problem step-by-step, we need to determine the profit function, find the number of leashes that maximize the profit, and then find the maximum profit.
### 1. Finding the Profit Function
The profit function [tex]\( P(x) \)[/tex] is defined as the revenue minus the cost.
The revenue is given by the price-demand function [tex]\( p(x) \times x \)[/tex], where [tex]\( p(x) \)[/tex] is the price at quantity [tex]\( x \)[/tex] and [tex]\( x \)[/tex] is the number of units sold.
Given:
- Cost function: [tex]\( C(x) = 5x + 9 \)[/tex]
- Price-demand function: [tex]\( p(x) = 77 - 2x \)[/tex]
The revenue [tex]\( R(x) \)[/tex] is:
[tex]\[ R(x) = p(x) \times x = (77 - 2x) \times x = 77x - 2x^2 \][/tex]
The profit function [tex]\( P(x) \)[/tex] is:
[tex]\[ P(x) = R(x) - C(x) = (77x - 2x^2) - (5x + 9) \][/tex]
Simplify the profit function:
[tex]\[ P(x) = 77x - 2x^2 - 5x - 9 \][/tex]
[tex]\[ P(x) = -2x^2 + 72x - 9 \][/tex]
So, the profit function is:
[tex]\[ P(x) = -2x^2 + 72x - 9 \][/tex]
### 2. Finding the Number of Leashes to Maximize Profit
To find the number of leashes that need to be sold to maximize profit, we need to find the critical points of the profit function by taking its derivative and setting it to zero.
The derivative of the profit function [tex]\( P(x) \)[/tex] is:
[tex]\[ P'(x) = \frac{d}{dx} (-2x^2 + 72x - 9) \][/tex]
[tex]\[ P'(x) = -4x + 72 \][/tex]
Set the derivative equal to zero to find the critical points:
[tex]\[ -4x + 72 = 0 \][/tex]
[tex]\[ -4x = -72 \][/tex]
[tex]\[ x = 18 \][/tex]
So, 18 leashes need to be sold to maximize profit.
### 3. Finding the Maximum Profit
To find the maximum profit, we substitute [tex]\( x = 18 \)[/tex] back into the profit function [tex]\( P(x) \)[/tex].
[tex]\[ P(18) = -2(18)^2 + 72(18) - 9 \][/tex]
[tex]\[ P(18) = -2(324) + 1296 - 9 \][/tex]
[tex]\[ P(18) = -648 + 1296 - 9 \][/tex]
[tex]\[ P(18) = 639 \][/tex]
Thus, the maximum profit is $639.
### Summary
- The profit function is:
[tex]\[ P(x) = -2x^2 + 72x - 9 \][/tex]
- The number of leashes which need to be sold to maximize profit is:
[tex]\[ \boxed{18} \][/tex]
- The maximum profit is:
[tex]\[ \boxed{639} \][/tex]
### 1. Finding the Profit Function
The profit function [tex]\( P(x) \)[/tex] is defined as the revenue minus the cost.
The revenue is given by the price-demand function [tex]\( p(x) \times x \)[/tex], where [tex]\( p(x) \)[/tex] is the price at quantity [tex]\( x \)[/tex] and [tex]\( x \)[/tex] is the number of units sold.
Given:
- Cost function: [tex]\( C(x) = 5x + 9 \)[/tex]
- Price-demand function: [tex]\( p(x) = 77 - 2x \)[/tex]
The revenue [tex]\( R(x) \)[/tex] is:
[tex]\[ R(x) = p(x) \times x = (77 - 2x) \times x = 77x - 2x^2 \][/tex]
The profit function [tex]\( P(x) \)[/tex] is:
[tex]\[ P(x) = R(x) - C(x) = (77x - 2x^2) - (5x + 9) \][/tex]
Simplify the profit function:
[tex]\[ P(x) = 77x - 2x^2 - 5x - 9 \][/tex]
[tex]\[ P(x) = -2x^2 + 72x - 9 \][/tex]
So, the profit function is:
[tex]\[ P(x) = -2x^2 + 72x - 9 \][/tex]
### 2. Finding the Number of Leashes to Maximize Profit
To find the number of leashes that need to be sold to maximize profit, we need to find the critical points of the profit function by taking its derivative and setting it to zero.
The derivative of the profit function [tex]\( P(x) \)[/tex] is:
[tex]\[ P'(x) = \frac{d}{dx} (-2x^2 + 72x - 9) \][/tex]
[tex]\[ P'(x) = -4x + 72 \][/tex]
Set the derivative equal to zero to find the critical points:
[tex]\[ -4x + 72 = 0 \][/tex]
[tex]\[ -4x = -72 \][/tex]
[tex]\[ x = 18 \][/tex]
So, 18 leashes need to be sold to maximize profit.
### 3. Finding the Maximum Profit
To find the maximum profit, we substitute [tex]\( x = 18 \)[/tex] back into the profit function [tex]\( P(x) \)[/tex].
[tex]\[ P(18) = -2(18)^2 + 72(18) - 9 \][/tex]
[tex]\[ P(18) = -2(324) + 1296 - 9 \][/tex]
[tex]\[ P(18) = -648 + 1296 - 9 \][/tex]
[tex]\[ P(18) = 639 \][/tex]
Thus, the maximum profit is $639.
### Summary
- The profit function is:
[tex]\[ P(x) = -2x^2 + 72x - 9 \][/tex]
- The number of leashes which need to be sold to maximize profit is:
[tex]\[ \boxed{18} \][/tex]
- The maximum profit is:
[tex]\[ \boxed{639} \][/tex]
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