Problem
Express the revenue function R in terms of x: The CarryluAl mini-van, a popular vehicle among soccer morns, obeys the demand equation $p=-\frac{1}{40} x+14,000$. The cost of producing $x$ vans is given by the function $C(x)=10100 x+10,000$. a) Express the revenue $R$ as a function of $x$. b) Express the profit $P$ as a function of $x$ c) Find the value of $x$ that maximizes profit. What is the maximum profit? d) What price should be charged in order to maximize profit? a) $R(x)=-0.025 x^{2}+14000 x$ (Simplify your answer. Do not factor.) b) $P(x)=\square$ (Simplify your answer, Do not factoc)
Solution
Step 1 ：Given the demand equation for the CarryluAl mini-van is \(p=-\frac{1}{40} x+14,000\), and the cost of producing \(x\) vans is given by the function \(C(x)=10100 x+10,000\).
Step 2 ：We are asked to express the revenue \(R\) as a function of \(x\). The revenue function \(R\) is the product of the price \(p\) and the quantity \(x\).
Step 3 ：Substitute the given price \(p\) into the equation for \(R\), we get \(R(x) = x*(14000 - 0.025*x)\).
Step 4 ：Simplify the equation to get the final revenue function: \(R(x) = \boxed{14000x - 0.025x^2}\).
