How do you find the quantity that maximizes revenue?

Question

The price, p, and the quantity, x, sold of a certain product obey the demand equation x=-10p+320. What quantity maximizes revenue?

The answer should be 160, but when I tried working on it, I got 16.

Answer 1

We are looking for maximum $px$, so you substitute $x=−10p+320$ into $px$ to obtain a quadratic equation:

Revenue = $p(-10p+320)$

$=(-10)(p^2-32p)$

$=(-10)(p^2-32p+256-256)$

$=(-10)((p-16)^2-256))$

$=(10)(256-(p-16)^2)$.

For revenue to be maximal, $(256-(p-16)^2)$ has to be maximal, so $-(256-(p-16)^2)=(p-16)^2-256$ has to be minimal. This is achieved if $(p-16)^2=0$ (the square of a real number can't be negative) or $p-16=0$, so $p=16$. So, $x=-10(16)+320=160$.