Step 1 :The problem is asking for the price that will maximize the revenue. The revenue is the product of the price and the number of cars rented.
Step 2 :We know that the number of cars rented decreases by 2 for every $5 increase in price. We can express the number of cars rented as a function of the price, and then express the revenue as a function of the price.
Step 3 :Let's denote the price as \(p\) and the number of cars as \(c\). We can express \(c\) as a function of \(p\): \(c = 64 - 2\frac{p}{5}\).
Step 4 :Then, the revenue \(r\) can be expressed as a function of \(p\): \(r = p(64 - 2\frac{p}{5})\).
Step 5 :To find the maximum of this function, we can take the derivative of \(r\) with respect to \(p\) and set it equal to zero: \(r' = 64 - 4\frac{p}{5} = 0\).
Step 6 :Solving this equation gives us the price that maximizes the revenue: \(p = 80\).
Step 7 :Substituting \(p = 80\) into the equation for \(c\) gives us the number of cars that will be rented at this price: \(c = 64 - 2\frac{80}{5} = 32\).
Step 8 :Final Answer: The price that should be charged to maximize its revenue is \(\boxed{\$80}\). The number of cars it will rent at this price is \(\boxed{32}\).