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A company has established that the revenue function in dollars is $R(x) = 2x^3 +40x^2 +8x$ and the cost function in dollars is $C(x) = 3x^3 + 19x^2 + 80x − 800$. Find the price per unit to maximize the profit.

Here's my problem.

I'm unsure of where to start, I want hints only, so I can solve it myself.

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    $\begingroup$ I am amazed by how nonsensical are these functions. Why bother doing anything when $C(0) = -800$? $\endgroup$ – svavil Oct 12 '16 at 10:59
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Note that the profit is the revenue minus the cost. In mathematical notation, $P(x) = R(x) - C(x)$. Then, you can find the maximum of that profit function by performing the derivative tests.

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  • $\begingroup$ Thats what I though too, but it asks for the price per unit to maximise the profit $\endgroup$ – SomebodyNeedingHelp Oct 9 '16 at 5:09
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    $\begingroup$ @SomebodyNeedingHelp In this case, revenue is the total price of all the products sold. In other words, revenue equals price per unit times the quantity of products sold. $\endgroup$ – W. Zhu Oct 9 '16 at 5:32

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