Aschool typically sells 500 yearbookseach year for $50 each. the econmics class does a project and discovers that they can sell 125 more yearbooks for every $5 decrease in price. the revenue for yearbook sales is r(x)=500+125x)(50-5x) to maximize profit, what price should the school charge for the yearbooks? a. 35 b. 40 c. 45 d. 50 what is the possible maximum revenue? a. 25,000 b. 30,625 c. 43,750 if the school attains the maximum revenue, how many yearbooks will they sell? a. 500 b. 625 c. 750 d. 875 try to explain a little, i'm confused on how to do this?

Simplify first the equation,                       R(x) = 25000 - 2500x + 6250x - 625x²Then, we differentiate the equation and equate to zero.                       dR(x) = -2500 + 6250 - 1250x = 0The 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
                                                        

                               Another way you can do this...

A is 100 because its the additional copies per each $5 decrease.
B is 5 because it represents the $5 decrease from 50.

If you need more help I will be glad to help!:D
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The decimal expansion would be .27 repeated