Answer :
Let's go through each part of the problem step-by-step.
### Part (a): Finding the Demand Function [tex]\( p(x) \)[/tex]
Given:
1. Initial price [tex]\( P_0 = \$360 \)[/tex]
2. Initial sales [tex]\( x_0 = 1150 \)[/tex] units
3. Rebate per item [tex]\( r = \$23 \)[/tex]
4. Sales increase per rebate [tex]\( \Delta x = 230 \)[/tex] units
The demand function [tex]\( p(x) \)[/tex] is linear and can be expressed as:
[tex]\[ p(x) = initial\_price - b \cdot (x - initial\_sales) \][/tex]
First, we need to determine the slope [tex]\( b \)[/tex]:
The slope [tex]\( b \)[/tex] is the rate of decrease in price for each additional unit sold, calculated as:
[tex]\[ b = \frac{\text{rebate per item}}{\text{increase in sales per rebate}} \][/tex]
[tex]\[ b = \frac{23}{230} = 0.1 \][/tex]
Thus, the demand function [tex]\( p(x) \)[/tex] is:
[tex]\[ p(x) = -0.1x + (initial\_price + b \cdot initial\_sales) \][/tex]
[tex]\[ p(x) = -0.1x + (360 + 0.1 \cdot 1150) \][/tex]
[tex]\[ p(x) = -0.1x + 475 \][/tex]
So, the demand function is:
[tex]\[ p(x) = -0.1x + 475 \][/tex]
### Part (b): Maximizing Revenue
Revenue [tex]\( R(x) \)[/tex] is given by:
[tex]\[ R(x) = x \cdot p(x) \][/tex]
[tex]\[ R(x) = x \cdot (-0.1x + 475) \][/tex]
To maximize revenue, we take the derivative of [tex]\( R(x) \)[/tex] with respect to [tex]\( x \)[/tex] and set it to zero:
[tex]\[ \frac{dR(x)}{dx} = \frac{d}{dx} \left(x \cdot (-0.1x + 475)\right) \][/tex]
[tex]\[ \frac{dR(x)}{dx} = -0.2x + 475 \][/tex]
Setting the derivative equal to zero to find the critical points:
[tex]\[ -0.2x + 475 = 0 \][/tex]
[tex]\[ 0.2x = 475 \][/tex]
[tex]\[ x = 2375 \][/tex]
To find the rebate required to achieve the maximum revenue:
[tex]\[ \text{Maximum Revenue Price} = p(2375) \][/tex]
[tex]\[ p(2375) = -0.1 \cdot 2375 + 475 \][/tex]
[tex]\[ p(2375) = 237.5 \][/tex]
The rebate needed is:
[tex]\[ \text{Rebate} = Initial Price - Maximum Revenue Price \][/tex]
[tex]\[ \text{Rebate} = 360 - 237.5 \][/tex]
[tex]\[ \text{Rebate} = 122.5 \][/tex]
So, the company should offer a rebate of:
[tex]\[ \$122.5 \][/tex]
### Part (c): Maximizing Profit
Profit [tex]\( \Pi(x) \)[/tex] is given by:
[tex]\[ \Pi(x) = R(x) - C(x) \][/tex]
where [tex]\( C(x) \)[/tex] is the cost function:
[tex]\[ C(x) = 69000 + 120x \][/tex]
Therefore, the profit function is:
[tex]\[ \Pi(x) = x \cdot (-0.1x + 475) - (69000 + 120x) \][/tex]
[tex]\[ \Pi(x) = -0.1x^2 + 475x - 69000 - 120x \][/tex]
[tex]\[ \Pi(x) = -0.1x^2 + 355x - 69000 \][/tex]
To maximize profit, we take the derivative of [tex]\( \Pi(x) \)[/tex] with respect to [tex]\( x \)[/tex] and set it to zero:
[tex]\[ \frac{d\Pi(x)}{dx} = -0.2x + 355 \][/tex]
Setting the derivative equal to zero to find the critical points:
[tex]\[ -0.2x + 355 = 0 \][/tex]
[tex]\[ 0.2x = 355 \][/tex]
[tex]\[ x = 1775 \][/tex]
To find the rebate required to maximize the profit:
[tex]\[ \text{Maximum Profit Price} = p(1775) \][/tex]
[tex]\[ p(1775) = -0.1 \cdot 1775 + 475 \][/tex]
[tex]\[ p(1775) = 297.5 \][/tex]
The rebate needed is:
[tex]\[ \text{Rebate} = Initial Price - Maximum Profit Price \][/tex]
[tex]\[ \text{Rebate} = 360 - 297.5 \][/tex]
[tex]\[ \text{Rebate} = 62.5 \][/tex]
So, the company should offer a rebate of:
[tex]\[ \$62.5 \][/tex]
To summarize:
- The demand function is [tex]\( p(x) = -0.1x + 475 \)[/tex].
- To maximize revenue, a rebate of \[tex]$122.5 should be offered. - To maximize profit, a rebate of \$[/tex]62.5 should be set.
### Part (a): Finding the Demand Function [tex]\( p(x) \)[/tex]
Given:
1. Initial price [tex]\( P_0 = \$360 \)[/tex]
2. Initial sales [tex]\( x_0 = 1150 \)[/tex] units
3. Rebate per item [tex]\( r = \$23 \)[/tex]
4. Sales increase per rebate [tex]\( \Delta x = 230 \)[/tex] units
The demand function [tex]\( p(x) \)[/tex] is linear and can be expressed as:
[tex]\[ p(x) = initial\_price - b \cdot (x - initial\_sales) \][/tex]
First, we need to determine the slope [tex]\( b \)[/tex]:
The slope [tex]\( b \)[/tex] is the rate of decrease in price for each additional unit sold, calculated as:
[tex]\[ b = \frac{\text{rebate per item}}{\text{increase in sales per rebate}} \][/tex]
[tex]\[ b = \frac{23}{230} = 0.1 \][/tex]
Thus, the demand function [tex]\( p(x) \)[/tex] is:
[tex]\[ p(x) = -0.1x + (initial\_price + b \cdot initial\_sales) \][/tex]
[tex]\[ p(x) = -0.1x + (360 + 0.1 \cdot 1150) \][/tex]
[tex]\[ p(x) = -0.1x + 475 \][/tex]
So, the demand function is:
[tex]\[ p(x) = -0.1x + 475 \][/tex]
### Part (b): Maximizing Revenue
Revenue [tex]\( R(x) \)[/tex] is given by:
[tex]\[ R(x) = x \cdot p(x) \][/tex]
[tex]\[ R(x) = x \cdot (-0.1x + 475) \][/tex]
To maximize revenue, we take the derivative of [tex]\( R(x) \)[/tex] with respect to [tex]\( x \)[/tex] and set it to zero:
[tex]\[ \frac{dR(x)}{dx} = \frac{d}{dx} \left(x \cdot (-0.1x + 475)\right) \][/tex]
[tex]\[ \frac{dR(x)}{dx} = -0.2x + 475 \][/tex]
Setting the derivative equal to zero to find the critical points:
[tex]\[ -0.2x + 475 = 0 \][/tex]
[tex]\[ 0.2x = 475 \][/tex]
[tex]\[ x = 2375 \][/tex]
To find the rebate required to achieve the maximum revenue:
[tex]\[ \text{Maximum Revenue Price} = p(2375) \][/tex]
[tex]\[ p(2375) = -0.1 \cdot 2375 + 475 \][/tex]
[tex]\[ p(2375) = 237.5 \][/tex]
The rebate needed is:
[tex]\[ \text{Rebate} = Initial Price - Maximum Revenue Price \][/tex]
[tex]\[ \text{Rebate} = 360 - 237.5 \][/tex]
[tex]\[ \text{Rebate} = 122.5 \][/tex]
So, the company should offer a rebate of:
[tex]\[ \$122.5 \][/tex]
### Part (c): Maximizing Profit
Profit [tex]\( \Pi(x) \)[/tex] is given by:
[tex]\[ \Pi(x) = R(x) - C(x) \][/tex]
where [tex]\( C(x) \)[/tex] is the cost function:
[tex]\[ C(x) = 69000 + 120x \][/tex]
Therefore, the profit function is:
[tex]\[ \Pi(x) = x \cdot (-0.1x + 475) - (69000 + 120x) \][/tex]
[tex]\[ \Pi(x) = -0.1x^2 + 475x - 69000 - 120x \][/tex]
[tex]\[ \Pi(x) = -0.1x^2 + 355x - 69000 \][/tex]
To maximize profit, we take the derivative of [tex]\( \Pi(x) \)[/tex] with respect to [tex]\( x \)[/tex] and set it to zero:
[tex]\[ \frac{d\Pi(x)}{dx} = -0.2x + 355 \][/tex]
Setting the derivative equal to zero to find the critical points:
[tex]\[ -0.2x + 355 = 0 \][/tex]
[tex]\[ 0.2x = 355 \][/tex]
[tex]\[ x = 1775 \][/tex]
To find the rebate required to maximize the profit:
[tex]\[ \text{Maximum Profit Price} = p(1775) \][/tex]
[tex]\[ p(1775) = -0.1 \cdot 1775 + 475 \][/tex]
[tex]\[ p(1775) = 297.5 \][/tex]
The rebate needed is:
[tex]\[ \text{Rebate} = Initial Price - Maximum Profit Price \][/tex]
[tex]\[ \text{Rebate} = 360 - 297.5 \][/tex]
[tex]\[ \text{Rebate} = 62.5 \][/tex]
So, the company should offer a rebate of:
[tex]\[ \$62.5 \][/tex]
To summarize:
- The demand function is [tex]\( p(x) = -0.1x + 475 \)[/tex].
- To maximize revenue, a rebate of \[tex]$122.5 should be offered. - To maximize profit, a rebate of \$[/tex]62.5 should be set.
