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