Question

The demand function is $x=\frac{24-2p}{3}$ where $x$ is the number of units demanded and $p$ is the price per unit. Find:
$\left(i\right)$ The revenue function $R$ in terms of $p$.
$\left(ii\right)$ The price and the number of units demanded for which the revenue is maximum.

Answer 1

Given, $x=\frac{24-2P}{3}$
$P=12-\frac{3}{2}x$
$\left(i\right)$ Revenue function $R\left(P\right)=Px=P\left(\frac{24-2P}{3}\right)$
$=8P-\frac{2}{3}{P}^2$
Revenue function $R\left(x\right)=Px=(12-\frac{3}{2}x)x$
$12x-\frac{3}{2}{x}^2$
$\left(ii\right)$ For maximum revenue, $\frac{dR\left(P\right)}{dP}=8-\frac{4}{3}P=0\Rightarrow P=6$
and $\frac{d^{2}R\left(P\right)}{d{P}^2}=-\frac{4}{3}(<0)$
For maximum revenue, $\frac{dR\left(x\right)}{dx}=12-3x=0$ $\Rightarrow$ $x=4$
and $\frac{d^{2}R\left(x\right)}{d{x}^2}=-3(<0)$
$\therefore$ Price is $6$ and no. of units is $4$ for which the revenue is maximum.