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

Answer:

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

Step-by-step 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,

$x=-\dfrac{b}{2a}=-\dfrac{160}{2\times (-4)}=20$

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

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

And maximum revenue will be

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

Answer 2

 Let x = dollar increase in price 
Let y = fewer number of pairs sold 

Since 2 fewer shoes are sold for each 1 dollar (factor of 2) 

y = 2x 

Revenue = Number of shoes sold * Price charged per shoe 

Number of shoes sold = 200 - y = 200 - 2x 
Price charged per shoe = $60 + $x 

Revenue = (200 - 2x)(60 + x) = -2x^2 + 200x - 120x + 12000 
Revenue = -2x^2 + 80x + 12000 

In a quadratic equation, Revenue is maximized when x = -b / 2a. In this case: 

x - -80 / (2*-2) = $20 

Price charged per show = $60 + $x = $60 + $20 = $80. 

Maximum revenue = -2x^2 + 80x + 12000 (evaluated at x = $20) 

Maximum revenue = -2(20^2) + 80*20 + 12000 = $12800