Respuesta :
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
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
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, [tex]120+2x[/tex]
Then the number of pairs of shoes sold becomes, [tex]200-2x[/tex]
The total revenue generated is the product of price of shoes and number of pairs of shoes sold, so
[tex]\text{Revenue}=f(x)=(120+2x)\cdot(200-2x)[/tex]
[tex]=120\cdot \:200+120\left(-2x\right)+2x\cdot \:200+2x\left(-2x\right)[/tex]
[tex]=120\cdot \:200-120\cdot \:2x+2\cdot \:200x-2\cdot \:2xx[/tex]
[tex]=-4x^2+160x+24000[/tex]
This is a quadratic function. And the quadratic function is maximized at,
[tex]x=-\dfrac{b}{2a}=-\dfrac{160}{2\times (-4)}=20[/tex]
So the price of each pairs of shoes for maximum revenue is
[tex]=120+2x=120+2(20)=\$160[/tex]
And maximum revenue will be
[tex]=-4(20)^2+160(20)+24000=\$25600[/tex]
