Answer :
Let's go through each part of the problem step-by-step.
### Part a) Find the demand function [tex]\( p(x) \)[/tex]
The demand function can be considered as a linear relation between the price [tex]\( p \)[/tex] and the quantity sold [tex]\( x \)[/tex].
We have the following information:
- Initially, when [tex]\( x = 1150 \)[/tex], the price [tex]\( p = 360 \)[/tex].
- For every \[tex]$23 rebate, the number of sets sold increases by 230. The general form of the linear demand function is \( p(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. From the information provided: - When the sales increase by 230 units, the price decreases by \$[/tex]23.
We can use these points to find the slope [tex]\( m \)[/tex]:
[tex]\[ m = \frac{-23}{230} = -0.1 \][/tex]
Now, we use the initial condition [tex]\( (1150, 360) \)[/tex] to find the y-intercept [tex]\( b \)[/tex]:
[tex]\[ 360 = -0.1 \times 1150 + b \][/tex]
[tex]\[ 360 = -115 + b \][/tex]
[tex]\[ b = 475 \][/tex]
Thus, the demand function is:
[tex]\[ p(x) = -0.1x + 475 \][/tex]
### Part b) Calculate the rebate to maximize revenue
Revenue [tex]\( R(x) \)[/tex] is given by:
[tex]\[ R(x) = p(x) \cdot x \][/tex]
[tex]\[ R(x) = (-0.1x + 475) \cdot x \][/tex]
[tex]\[ R(x) = -0.1x^2 + 475x \][/tex]
To maximize revenue, we find the critical points by differentiating [tex]\( R(x) \)[/tex] and setting the derivative equal to zero:
[tex]\[ R'(x) = \frac{d}{dx} (-0.1x^2 + 475x) \][/tex]
[tex]\[ R'(x) = -0.2x + 475 \][/tex]
Set the derivative equal to zero to find the critical point:
[tex]\[ -0.2x + 475 = 0 \][/tex]
[tex]\[ 0.2x = 475 \][/tex]
[tex]\[ x = 2375 \][/tex]
The corresponding price [tex]\( p(2375) \)[/tex] is:
[tex]\[ p(2375) = -0.1 \times 2375 + 475 = 237.5 \][/tex]
The rebate required to bring the price down from \[tex]$360 to \$[/tex]237.5 is:
[tex]\[ \text{Rebate} = 360 - 237.5 = \$122.5 \][/tex]
So, the company should offer a rebate of
[tex]\[ \$7.5 \][/tex]
### Part c) Find the rebate to maximize profit
The profit [tex]\( P(x) \)[/tex] is given by the revenue minus the cost:
[tex]\[ P(x) = R(x) - C(x) \][/tex]
[tex]\[ C(x) = 69000 + 120x \][/tex]
[tex]\[ P(x) = (-0.1x^2 + 475x) - (69000 + 120x) \][/tex]
[tex]\[ P(x) = -0.1x^2 + 475x - 69000 - 120x \][/tex]
[tex]\[ P(x) = -0.1x^2 + 355x - 69000 \][/tex]
To maximize profit, we find the critical points by differentiating [tex]\( P(x) \)[/tex] and setting the derivative equal to zero:
[tex]\[ P'(x) = \frac{d}{dx} (-0.1x^2 + 355x - 69000) \][/tex]
[tex]\[ P'(x) = -0.2x + 355 \][/tex]
Set the derivative equal to zero to find the critical point:
[tex]\[ -0.2x + 355 = 0 \][/tex]
[tex]\[ 0.2x = 355 \][/tex]
[tex]\[ x = 1775 \][/tex]
The corresponding price [tex]\( p(1775) \)[/tex] is:
[tex]\[ p(1775) = -0.1 \times 1775 + 475 = 297.5 \][/tex]
The rebate required to achieve this price is:
[tex]\[ \text{Rebate} = 360 - 297.5 = \$62.5 \][/tex]
So, the company should offer a rebate of
[tex]\[ \$-52.5 \][/tex]
### Part a) Find the demand function [tex]\( p(x) \)[/tex]
The demand function can be considered as a linear relation between the price [tex]\( p \)[/tex] and the quantity sold [tex]\( x \)[/tex].
We have the following information:
- Initially, when [tex]\( x = 1150 \)[/tex], the price [tex]\( p = 360 \)[/tex].
- For every \[tex]$23 rebate, the number of sets sold increases by 230. The general form of the linear demand function is \( p(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. From the information provided: - When the sales increase by 230 units, the price decreases by \$[/tex]23.
We can use these points to find the slope [tex]\( m \)[/tex]:
[tex]\[ m = \frac{-23}{230} = -0.1 \][/tex]
Now, we use the initial condition [tex]\( (1150, 360) \)[/tex] to find the y-intercept [tex]\( b \)[/tex]:
[tex]\[ 360 = -0.1 \times 1150 + b \][/tex]
[tex]\[ 360 = -115 + b \][/tex]
[tex]\[ b = 475 \][/tex]
Thus, the demand function is:
[tex]\[ p(x) = -0.1x + 475 \][/tex]
### Part b) Calculate the rebate to maximize revenue
Revenue [tex]\( R(x) \)[/tex] is given by:
[tex]\[ R(x) = p(x) \cdot x \][/tex]
[tex]\[ R(x) = (-0.1x + 475) \cdot x \][/tex]
[tex]\[ R(x) = -0.1x^2 + 475x \][/tex]
To maximize revenue, we find the critical points by differentiating [tex]\( R(x) \)[/tex] and setting the derivative equal to zero:
[tex]\[ R'(x) = \frac{d}{dx} (-0.1x^2 + 475x) \][/tex]
[tex]\[ R'(x) = -0.2x + 475 \][/tex]
Set the derivative equal to zero to find the critical point:
[tex]\[ -0.2x + 475 = 0 \][/tex]
[tex]\[ 0.2x = 475 \][/tex]
[tex]\[ x = 2375 \][/tex]
The corresponding price [tex]\( p(2375) \)[/tex] is:
[tex]\[ p(2375) = -0.1 \times 2375 + 475 = 237.5 \][/tex]
The rebate required to bring the price down from \[tex]$360 to \$[/tex]237.5 is:
[tex]\[ \text{Rebate} = 360 - 237.5 = \$122.5 \][/tex]
So, the company should offer a rebate of
[tex]\[ \$7.5 \][/tex]
### Part c) Find the rebate to maximize profit
The profit [tex]\( P(x) \)[/tex] is given by the revenue minus the cost:
[tex]\[ P(x) = R(x) - C(x) \][/tex]
[tex]\[ C(x) = 69000 + 120x \][/tex]
[tex]\[ P(x) = (-0.1x^2 + 475x) - (69000 + 120x) \][/tex]
[tex]\[ P(x) = -0.1x^2 + 475x - 69000 - 120x \][/tex]
[tex]\[ P(x) = -0.1x^2 + 355x - 69000 \][/tex]
To maximize profit, we find the critical points by differentiating [tex]\( P(x) \)[/tex] and setting the derivative equal to zero:
[tex]\[ P'(x) = \frac{d}{dx} (-0.1x^2 + 355x - 69000) \][/tex]
[tex]\[ P'(x) = -0.2x + 355 \][/tex]
Set the derivative equal to zero to find the critical point:
[tex]\[ -0.2x + 355 = 0 \][/tex]
[tex]\[ 0.2x = 355 \][/tex]
[tex]\[ x = 1775 \][/tex]
The corresponding price [tex]\( p(1775) \)[/tex] is:
[tex]\[ p(1775) = -0.1 \times 1775 + 475 = 297.5 \][/tex]
The rebate required to achieve this price is:
[tex]\[ \text{Rebate} = 360 - 297.5 = \$62.5 \][/tex]
So, the company should offer a rebate of
[tex]\[ \$-52.5 \][/tex]
