I am given the demand function $$D(x)=10x^2 + 50x$$ and a total cost of $$C(x) = x^3 + 10x$$ where $x$ is the number of units demanded. I am asked to maximize the profit so what I did is I used the formula $$P(x) = R(x) - C(x)$$ where $R(x) = x D(x)$. Then I took the derivative of $P(x)$ and equated to 0. But I got a negative value of x more so a value less than 1 since $$P'(x) = 27x^2 + 100x -10$$ Am I right or did I do something wrong in between my process? I don't think it's logical to have a quantity which is negative and less than 1 for this kind of problem. Please help.
2026-03-31 17:44:31.1774979071
Maximizing profit function given cost and demand functions
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You should multiply $D(x)$ by the price of a single unit, to get the total revenue.
So if $Q$ is the price of a single unit (which presumably does not depend on $x$), you get
$$P(x) = QD(x) - C(x)$$
After that you are correct,to maximize this value with respect to $x$ you take the derivative and equate it to $0$ (you should show that this is indeed a maximum, and not a minimum or a stationary point. But it's not difficult to do so)