Aschool typically sells 500 yearbooks each year for $50 each. the economics class does a project and discovers that they can sell 125 more yearbooks for every $5 decrease in price. the revenue for the school sales is r(x)=(500+125x)(50-5x)
to maximize profit, what price should the school charge for the yearbooks?
what is the possible maximum revenue?
if the school attains the maximum revenue, how many yearbooks will they sell?

Simplify first the equation,
                       R(x) = 25000 - 2500x + 6250x - 625x²
Then, we differentiate the equation and equate to zero. 
                      dR(x) = -2500 + 6250 - 1250x = 0
The value of x from the equation is 3. 
(1) The price is equal to 50 - 5(3) = $35.
(2) The possible maximum revenue:
                     R(x) = 2500 - 2500(3) + 6250(3) - 625(3²) = $8125
(3) number of yearbooks sold:
                           500 + 125(3) = 875 yearbooks

1. 0.3 •4 is the same as 4 groups of 0.32. 7•0.8 is the same as 7 groups of 0.83. it increases by 0.5 every time, so 0.5x4 it increases by 0.9 every time, so 0.9x5. 0.9•12=10.86. 0.47•5=2.35i can finish the next ones in a separate post