**Hint:** Let $n$ be the number of one-dollar increases in the room rate. Then the room rate is $$80+n$$ and the number of rooms rented is $$100-n.$$
You should be able to find the value of $n$ that maximizes the revenue function $$R(n)=(80+n)(100-n)=8000+20n-n^2.$$ Once the optimal $n$ is determined, the corresponding price is $80+n$.
**Hint 2:** Complete the square to rewrite the revenue function as $$R(n) = 8100-(n-10)^2. $$
From this, you can see that the revenue is maximized when $n-10$ is zero, because if it is anything other than zero you will be subtracting it from $8100$ to get the revenue, so the revenue will be strictly less than $8100$ if $n-10$ is not zero.