A school typically sells 500 yearbooks each year for 50 dollars each. The economic calls does a project and discovers that they can sell 100 more yearbooks for every $5 decrease in price. The revenue for yearbook sales is equal to the number of yearbooks sold times the price of the yearbook. Let x represent the number of $5 decrease in price. If the expression that represents the revenue is written in the form R(x)=(500+ax)(50-bx). Find the value of a and b

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