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 the price and the number of units demanded for which the revenue is maximum.

Price = ₹6, Units = 4

The correct answer is Option (1) → Price = ₹6, Units = 4

The demand function is $x =\frac{24-2p}{3}$

Revenue function $R = px = p ×\frac{24-2p}{3}$

$⇒ R = 8p - \frac{2}{3} p^2$.

To find value of p for which R is maximum, we should find value(s) of p where $\frac{dR}{dp}= 0$ and $\frac{d^2R}{dp^2}<0$

Now $\frac{dR}{dp}=8.1-\frac{2}{3}.2p=8-\frac{4}{3}p$ and $\frac{d^2R}{dp^2}=0-\frac{4}{3}.1=-\frac{4}{3}$.

$\frac{dR}{dp}=0⇒8-\frac{4}{3}p=0⇒p=6$

When $p=6,\frac{d^2R}{dp^2}=-\frac{4}{3}<0$.

∴ The revenue is maximum when $p = 6$ i.e. the price per unit is ₹6.

The number of units demanded is given by $x =\frac{24-2×6}{3}= 4$.

Hence, the revenue is maximum when 4 units are demanded and the price per unit is ₹6.