Who knows how to solve max/min application problems?

Since I don't know in which grade you are, I applying 2 different methods:

R(x) = 600x - 0.5.x²; This is a parabola, open downward (a=-0.5<0), with a maximum to be determined:

 1st method: The maximum is when the derivative y' =0, let's calculate the derivative of R(x) = 600x - 0.5.x² →y' = 600 - (2*0.5)x →y' = 600 - x =0
Then x = 600 rooms (for a revenue of 180,000: just plug x value in the equation)

2nd method: the axis of symmetry is - b/2a = (- 600) /[(2)(-0.5)] 
→ -b/2a = 600 ( so x = 600), to calculate y, plug x value in the equation.
And you will find 180,000, so the maximum is (600,180 000), which is the maximum in terms of rfeturn