the problem.
2) The manufacturer of a CD player has found that the revenue $R$ (in dollars) is $R(p)=-5 p^{2}+1830 p$, when the unit price is $p$ dollars. If the manufacturer sets the price $p$ to maximize revenue, what is the maximum revenue to the nearest whole dollar?
Answer
The maximum revenue to the nearest whole dollar is \(\boxed{167445}\).
Step 1 ：The problem is asking for the maximum revenue, which can be found by finding the derivative of the revenue function and setting it equal to zero. This will give us the price at which the revenue is maximized.
Step 2 ：Let's denote the price as \(p\) and the revenue as \(R\). The revenue function is given by \(R(p)=-5 p^{2}+1830 p\).
Step 3 ：To find the maximum revenue, we first need to find the derivative of the revenue function. The derivative of \(R(p)\) is \(R'(p) = 1830 - 10p\).
Step 4 ：We then set the derivative equal to zero and solve for \(p\). This gives us \(p_{max} = 183\).
Step 5 ：Substituting \(p_{max}\) back into the revenue function gives us the maximum revenue, \(R_{max} = 167445\).
Step 6 ：The maximum revenue to the nearest whole dollar is \(\boxed{167445}\).
