Question

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

Answer 1

Here we have to find the revenue function in terms of p. As x is the number of units demanded and p is the price per unit given in the question. Revenue function will be equal to the product of the number of units demanded and price per unit. From there, we will get the revenue function in terms of p. Then we will find the value of p and x at which the revenue is maximum, we will differentiate the revenue function and equate it with zero to get the value of p and x. Also we will find the second order derivative of revenue function to be negative.

Complete step-by-step answer:
Given:
$x=\dfrac{24-2p}{3}$, here x is the number of units demanded and p is the price per unit.
We know the revenue function R is the product of the number of units demanded and price per unit.
Therefore,
$R=p.x$
Now, we will put the value of x here.
$R\left( p \right)=p.\left( \dfrac{24-2p}{3} \right)$
Multiplying the terms, we get
$\Rightarrow$ $R\left( p \right)=\dfrac{24p-2{{p}^{2}}}{3}$
We can also write the revenue function as;
$\Rightarrow$ $R\left( p \right)=8p-\dfrac{2}{3}{{p}^{2}}$
This is the required revenue function in terms of p.
Now, we have to find the value of price per unit i.e. p and number of units demanded i.e. x for which the revenue is maximum.
We will first differentiate the revenue function with respect to p.
$\Rightarrow$ $\dfrac{dR\left( p \right)}{dp}=\dfrac{d\left( 8p-\dfrac{2}{3}{{p}^{2}} \right)}{dp}$
$\Rightarrow$ $\dfrac{dR\left( p \right)}{dp}=8-\dfrac{4}{3}p$
Again differentiating with respect to p, we get
$\Rightarrow$ $\dfrac{d}{dp}\left( \dfrac{dR\left( p \right)}{dp} \right)=\dfrac{d}{dp}\left( 8-\dfrac{4}{3}p \right)$
$\Rightarrow$ $\dfrac{{{d}^{2}}R\left( p \right)}{d{{p}^{2}}}=-\dfrac{4}{3}<0$
Thus, revenue is maximum as the second order derivative of revenue is negative.
We will find the value of p now by equating the first order derivative of revenue with zero.
$\Rightarrow$ $\dfrac{dR\left( p \right)}{dp}=0$
$\Rightarrow$ $8-\dfrac{4}{3}p=0$
On further simplification, we get
$\Rightarrow$ $p=6$
We will put the value of p in the demand function.
$\Rightarrow$ $x=\dfrac{24-2\times 6}{3}$
Further simplifying the fraction, we get
$\Rightarrow$ $x=4$
Therefore, the price and the number of units demanded are 6 and 4 respectively for which the revenue is maximum.

Note: We have found the revenue function here. Revenue is defined as the total amount earned by the business entities by selling their products. Revenue is also defined as the sales by the business.
Revenue is equal to the product of the number of units sold and the price per unit.