Question

an athletic store sells about 200 pairs of basketball shoes per month when it charges $120 per pair. for each $2 increase in price, the store sells two fewer pairs of shoes. how much should the store charge to maximize monthly revenue? what is the maximum monthly revenue?

Answer 1

The store should charge $160 to maximize monthly revenue and the maximum monthly revenue is $25,600

Explanation:

Let us assume that, x is the number of $2 increases in price.

So price of the shoes becomes,
`120+2x`

Then the number of pairs of shoes sold becomes,
`200-2x`

The total revenue generated is the product of price of shoes and number of pairs of shoes sold, so

`\text{Revenue}=f(x)=(120+2x)\cdot(200-2x)`

`=120\cdot \:200+120\left(-2x\right)+2x\cdot \:200+2x\left(-2x\right)`

`=120\cdot \:200-120\cdot \:2x+2\cdot \:200x-2\cdot \:2xx`

`=-4x^2+160x+24000`

This is a quadratic function. And the quadratic function is maximized at,

`x=-(b)/(2a)=-(160)/(2* (-4))=20`

So the price of each pairs of shoes for maximum revenue is

`=120+2x=120+2(20)=\$160`

And maximum revenue will be

`=-4(20)^2+160(20)+24000=\$25600`