Question

The demand for a certain product is represented by the equation $p = 500 + 25x - \dfrac{{{x^2}}}{3}$ in rupees where $x$ is the number of units and $p$ is the price 3 per unit. Find

1. Marginal revenue function.
2. The marginal revenue when 10 units are sold.

Answer 1

Hint: First, we will find the total revenue $R = 3p \cdot x$ for units and then differentiate it to find the marginal revenue function. Then we will substitute the value of $x$ in the obtained function to find the marginal revenue when 10 units are sold.

Complete step by step solution:
Given the demand function for a certain product is $p = 500 + 25x - \dfrac{{{x^2}}}{3}$, where $p$ is the price 3 per unit.

Assume $R$ be the total revenue for $x$ units, then

$$R = 3p \cdot x \\\ = 3\left( {500x + 25x - \dfrac{{{x^2}}}{3}} \right) \cdot x \\\ = 1500x + 75{x^2} - {x^3} \\\$$

We will find the marginal revenue ${\text{MR}}$, which is the derivative of the total revenue for $x$ units.

$${\text{MR}} = \dfrac{{dR}}{{dx}} \\\ = 1500 + 2 \cdot 75x - 3{x^2} \\\ = 1500 + 150x - 3{x^2} \\\$$

Replacing 10 for $x$ in the above equation to find the marginal revenue when 10 units are sold, we get

$${\left( {{\text{MR}}} \right)_{x = 10}} = 1500 + 150\left( {10} \right) - 3{\left( {10} \right)^2} \\\ = 1500 + 1500 - 300 \\\ = 2700 \\\$$

Therefore, the marginal revenue is $1500 + 150x - 3{x^2}$ and when 10 units are sold is 2700.

Note: We have used that the derivative of the revenue function $R\left( x \right)$ is called marginal revenue with notation $R'\left( x \right) = \dfrac{{dR}}{{dx}}$. In this question, we have also interpreted when the production increases from 10 to 11 units, the revenue increases by 30,000.