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.

Explanation:

The formula for the revenue function is

$\begin{gathered} R(x)=x\times\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

$x=\frac{-b}{2a}=\frac{-600}{2(-8)}=\frac{-600}{-16}=37.5$

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

Therefore, the maximum revenue is

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

To maximize the revenue company should charge

$\begin{gathered} \text{ p}=-8\times38+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

$296.$