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.