main algebraic solution

Mathematics
Tutor: None Selected Time limit: 1 Day

May 30th, 2015

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 = 550 - y = 550 - 2x 
Price charged per shoe = $99 + $x 

Revenue = (200 - 2x)(99 + x) = -2x^2 + 550x - 198x + 19800 
Revenue = -2x^2 + 352x + 19800 

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

x - -352 / (2*-2) = $88 

Price charged per show = $99 + $x = $99 + $20 = $119. 

Maximum revenue = -2x^2 + 352x + 19800 (evaluated at x = $88) 


May 30th, 2015

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