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I have meet following question: Find the maximum average revenue if the demand equation is $P=500 + 10x - x^2$

I know that average revenue is equal total revenue/number of item, so I have divided it by $x$ and got $P=\frac{500}{x}+10-x,$ then I took the derivative of this new given function and got $\frac{\textrm{d}P}{\textrm{d}x}=-\frac{500}{x^2}-1.$ But when I set it to $0$ I get $x^2=-500$ which only has a complex solution. I was thinking that I dont need to divide it by $x$, just take simple derivative of given function, so that $\frac{\textrm{d}P}{\textrm{d}x}=10-2x$ set it zero and get $x=5$, then put into original equation and finally I have got $500+50-25=525$, but I am not sure that it is right because they asking me average revenue. Please help.

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    Now that we know what the answer is, someone should post it as an answer, just to tidy things up.2011-08-09

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As $P$ is the average revenue, we can set the derivative $10-2x$ to zero and find $x=5$. (posted as Gerry Myerson asked)