Question

The price p (in dollars) and the quantity x sold of a certain product obey the demand equation p = -8x + 600. What quantity x maximizes revenue (R= xp)? What is the maximum revenue? What price should the company charge to maximize revenue?

Answer 1

Given:

Demand equation is

`\text{ p}=-8x+600.`

where p is the price(in dollars) and x is the quantity of the product.

Required:

We have to find the quantity x that is, the value of x for which revenue is maximized, the maximum revenue, and the price should the company charge to maximize the revenue.

Step-by-step explanation:

The formula for the revenue function is

`\begin{gathered} R(x)=x*\text{ p} \\ \Rightarrow R(x)=x(-8x+600) \end{gathered}`
`\Rightarrow R(x)=-8x^2+600x.`

Comparing this equation with

`y=ax^2+bx+c`

we get

`a=-8\text{ and }b=600.`

The value of x for maximum revenue is

`x=(-b)/(2a)=(-600)/(2(-8))=(-600)/(-16)=37.5`

So we take the value of x approximately 38 because it cannot be decimal or fraction.

Therefore, the maximum revenue is

`\begin{gathered} -8*(38)+600*(38) \\ =-304+22800 \end{gathered}`
`=22496`

To maximize the revenue company should charge

`\begin{gathered} \text{ p}=-8*38+600 \\ \Rightarrow\text{ p}=-304+600 \end{gathered}`
`\Rightarrow p=296`

Final answer:

Hence the final answer is:

The quantity x maximizes revenue is

`38.`

The maximum revenue is

`22496.`

The price should the company charge to maximize revenue is

`296.`