June is selling signed NHL jerseys. Let P represent the profit and let s represent the selling price. The profit can be modelled by the equation: 𝑃

P= −3(𝑠 S− 20)^2 + 588.

 

a) What is the selling price that will maximize profits?

The selling price that will maximize profits is $20 (h-value of equation) since that is when the maximum profit of $588 (k-value of equation) is reached. (20, 588)

 

b)What is the selling price when profits equal zero (also known as the break-even point).

We have to set P=0

P = −3(𝑠 S− 20)^2 + 588

0 = −3(𝑠 S− 20)^2 + 588

-588/-3 = [−3( S− 20)^2]/-3

      196 =       ( S− 20)^2

      196 = S – 20

      196 + 20 = S 

 

 

 

S = 14 + 20

S = 24

(24, 0)  

 

Therefore profits will equal zero when the selling price is $24 and $6.