Determine the number of cars sold in one month to maximize profit

The dealership buys cars for $15000$. When the dealer sells each car for $25000$, she sells $24$ cars per month. For each reduction of $600$ in the selling price, the dealer sells $2$ more cars per month. Determine the number of cars sold in one month to maximize profit.

I know the cost function should be $c(x) = 15000x$ , but what is the revenue function?

The answer is $29$. Thanks for any help.

• Hint: What is the sale price for each car? If $x$ cars are sold, how much does the dealer make? – Sean Roberson Oct 13 '16 at 21:27
• I have come up with something like (24+2x)(25000-600x) for the revenue function, then I do the Profit = revenue - cost, take the derivative to find the critical point, but my finally answer seems bit off. – user97919 Oct 13 '16 at 21:33

2 Answers

The revenue per car (RPC) is

$$\mbox{RPC} (x) = 25000 - 300 (x-24) = 32200 - 300 x$$

and, thus, the profit is

$$\mbox{Profit} (x) = (\mbox{RPC} (x) - 15000) x = 17200 x - 300 x^2$$

Differentiate and find where the derivative vanishes. If the maximiser is not an integer, round up and down. Choose the maximum of the two. That is where profit is maximized.

The number of cars sold is $C=24+2*\frac{25000-p}{600}$. The revenue is $p*C$, where $p$ is the price and $C$ is the number of cars.