Maximizing revenue from a linear demand function

First expand the revenue function: $R(p) = 1200p - 40p^2$. This is a downward-opening parabola in $p$, so it has a single global maximum at its critical point. Differentiate to get $R'(p) = 1200 - 80p$.

Setting $R'(p) = 0$ gives $1200 - 80p = 0$, so $p = 15$. Since $R''(p) = -80 < 0$, this confirms $p = 15$ is a maximum. Substituting back, $R(15) = 15(1200 - 600) = 15(600) = 9000$, so the maximum revenue is $9000 dollars, achieved at a price of$15 per item.