The total revenue for Joe's Realty is given as the function R(x)=200x−0.2x2, where x is the number of rooms filled. What number of rooms filled produces the maximum revenue?

Answer 1

R(x) =200x -0.2x^2

take the derivative and set equal to zero

R'(x) = 200 - .4x = 0

.4x = 200

x= 200/.4 = 500 = revenue maximizing number of rooms

R(500) = 200(500) - .2(500)^2 = 100,000-50,000=50,000