Answer
$x=\frac{15}{2}=7.5$; charge per night $125$ dollars.
Work Step by Step
Step 1. Assuming there are $x$ number of 10-dollar increases, the charge per night is then $50+10x$ and the number of rooms filled will be $800-40x$ (this gives $0\leq x\leq 20$)
Step 2. The revenue can be expressed as $R(x)=(50+10x)(800-40x)$. To find the maximum revenue, let its derivative be zero. We get $R'=10(800-40x)-40(50+10x)=0$ or $6000-800x=0$, which gives $x=\frac{15}{2}=7.5$ and the charge per night will be $50+75=125$ dollars.
Step 3. Check $R''=-800\lt0$ and the region is concave down with a maximum.
