Question

Find $x$ for which the total revenue function is maximum where $R=2x^3-63x^2+648x+300$

Answer 1

$R=2x^3-63x^2+648x+300$

$\frac{dR}{dx}=6x^2-126x+648$

$\frac{dR}{dx}=0$

$6x^2-126x+648=0$

$x^2-21x+108=0$

$x=\frac{21\pm \sqrt{144-432}}{2}$

$x=\frac{21\pm 3}{2}$

$=12,9$

$x<9$, $dR/dx>0$, $R\left(x\right)$ is increasing

$9<x<12$, $dR/dx<0$; $R\left(x\right)$ is decreasing

$x>12$, $dR/dx>0$, $R\left(x\right)$ is increasing

$\therefore$ At $x=9$ the total revenue function is maximum.