# Finding the maximum average revenue

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.

-
You're supposed to divide the total revenue by the number of items, but you divided $P$ by the number of items. $P$ isn't the total revenue, is it? – Gerry Myerson Aug 8 '11 at 12:13
P yes you are right is not total revenue ,so i think that i should us formula R=P*x?(where R is total revenue, and P-demand function,x number of items)? – dato datuashvili Aug 8 '11 at 12:23
@user3196: right. Which means $P$ is the average revenue. – Ross Millikan Aug 8 '11 at 13:40
so as i concluded if i multiply demand by x get total revenue and then divide it by number of items to get avarage revenue it means that demand is the same as avarage revenue yes?so just i should take derivatives set it so zero and put critical point into my demand function?am i right? – dato datuashvili Aug 8 '11 at 13:54
@user 3196: You are right. – André Nicolas Aug 8 '11 at 14:35

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