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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.

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  • $\begingroup$ Hint: What is the sale price for each car? If $x$ cars are sold, how much does the dealer make? $\endgroup$ – Sean Roberson Oct 13 '16 at 21:27
  • $\begingroup$ 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. $\endgroup$ – user97919 Oct 13 '16 at 21:33
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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.

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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.

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