Question

Consider that the equation $p=\dfrac{150}{{{q}^{2}}+2}-4$ represents the demand function for a product where p is the price per unit for q units. Determine the marginal revenue.

Answer 1

Hint:To find the marginal revenue, we have to first find the total revenue using the formula $TR\left( q \right)=pq$ , where p is the price per unit and q is the number of units. Marginal revenue can be found by differentiating total revenue with respect to q.

Complete step-by-step solution:
We are given with the demand function $p=\dfrac{150}{{{q}^{2}}+2}-4$ . Let us find the revenue function. We know that revenue is the number of units sold times the price per unit.
$\Rightarrow TR\left( q \right)=pq$
Where p is the price per unit and q is the number of units. p is also the demand function.
$\begin{align}
  & \Rightarrow TR\left( q \right)=\left( \dfrac{150}{{{q}^{2}}+2}-4 \right)q \\
 & \Rightarrow TR\left( q \right)=\left( \dfrac{150-4{{q}^{2}}-8}{{{q}^{2}}+2} \right)q \\
 & \Rightarrow TR\left( q \right)=\left( \dfrac{150q-4{{q}^{3}}-8q}{{{q}^{2}}+2} \right) \\
\end{align}$
Now, let us find the marginal revenue. We know that marginal revenue is change in total revenue divided by change in quantity.
$\begin{align}
  & \Rightarrow \text{Marginal Revenue}\left( \text{MR} \right)=\dfrac{\text{Change in total revenue}}{\text{Change in quantity}} \\
 & \Rightarrow MR=\dfrac{\Delta TR}{\Delta q} \\
\end{align}$
We have to differentiate $TR\left( q \right)$ with respect to q.
$\Rightarrow MR=\dfrac{d}{dq}\left( \dfrac{150q-4{{q}^{3}}-8q}{{{q}^{2}}+2} \right)$
We have to apply the quotient rule which is given by $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ .
$\Rightarrow MR=\dfrac{\left( {{q}^{2}}+2 \right)\dfrac{d}{dq}\left( 150q-4{{q}^{3}}-8q \right)-\left( 150q-4{{q}^{3}}-8q \right)\dfrac{d}{dq}\left( {{q}^{2}}+2 \right)}{{{\left( {{q}^{2}}+2 \right)}^{2}}}$
We know that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ . Therefore, the above equation can be written as
$\begin{align}
  & \Rightarrow MR=\dfrac{\left( {{q}^{2}}+2 \right)\left( 150-12{{q}^{2}}-8 \right)-\left( 150q-4{{q}^{3}}-8q \right)2q}{{{\left( {{q}^{2}}+2 \right)}^{2}}} \\
 & \Rightarrow MR=\dfrac{\left( {{q}^{2}}+2 \right)\left( 142-12{{q}^{2}} \right)-\left( 150q-4{{q}^{3}}-8q \right)2q}{{{\left( {{q}^{2}}+2 \right)}^{2}}} \\
\end{align}$
Let us apply the distributive property.

$\Rightarrow MR=\dfrac{142{{q}^{2}}-12{{q}^{4}}+284-24{{q}^{2}}-300{{q}^{2}}+8{{q}^{4}}+16{{q}^{2}}}{{{\left( {{q}^{2}}+2 \right)}^{2}}}$
We have to add the like terms.
$\Rightarrow MR=\dfrac{-4{{q}^{4}}-166{{q}^{2}}+284}{{{\left( {{q}^{2}}+2 \right)}^{2}}}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ . Therefore, the denominator of the above equation can be written as
$\Rightarrow MR=\dfrac{-4{{q}^{4}}-166{{q}^{2}}+284}{{{q}^{4}}+4{{q}^{2}}+4}$
Let us divide $-4{{q}^{4}}-166{{q}^{2}}+284$ by ${{q}^{4}}+4{{q}^{2}}+4$ .
\[\begin{align}
  & {{q}^{4}}+4{{q}^{2}}+4\overset{-4}{\overline{\left){-4{{q}^{4}}-166{{q}^{2}}+284}\right.}} \\
 & \text{ }\underline{\begin{align}
  & -4{{q}^{4}}-16{{q}^{2}}-16 \\
 & \begin{matrix}
   \left( + \right) & \left( + \right) & \left( + \right) \\
\end{matrix} \\
\end{align}} \\
 & \text{ }-150{{q}^{2}}+300 \\
\end{align}\]
We can write the result in the form $\text{Quotient}+\dfrac{\text{Remainder}}{\text{Divisor}}$ .
$\Rightarrow MR=-4+\dfrac{-150{{q}^{2}}+300}{{{q}^{4}}+4{{q}^{2}}+4}$
Hence, the marginal revenue is $-4+\dfrac{-150{{q}^{2}}+300}{{{q}^{4}}+4{{q}^{2}}+4}$ .

Note: Students must be thorough with the formulas of total revenue and marginal revenue. They have a chance of making a mistake by writing the formula for marginal revenue as the change in quantity divided by total revenue.