Question

At what output is total revenue a maximum? $y=48x-2{{x}^{2}}$ , where $y$ is the total revenue and x is the output.
(a) 2
(b) 12
(c) 48
(d) 4

Answer 1

Hint: To find the output at which the total revenue will be maximum, we have to differentiate the given equation for the revenue with respect to x. Then, we have to equate this derivative to 0 and find the value of x. Then, we have to differentiate the given equation twice and substitute the value of x in it and find the corresponding values of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ . We have to look for larger value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ if there are more than one value of x. The corresponding x value will be the required answer. Else, we will look for the greatest value among the first derivative and the second derivative.

Complete step by step solution:
We are given that $y=48x-2{{x}^{2}}$ . We have to find the output at which the total revenue will be maximum. Let us differentiate the given equation for total revenue with respect to x.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( 48x-2{{x}^{2}} \right)$
We know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ and $\dfrac{d}{dx}\left( a{{x}^{n}} \right)=a\times n{{x}^{n-1}}$ . Therefore, we can write the above derivative as
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( 48x-2{{x}^{2}} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=48-2\times 2x \\
 & \Rightarrow \dfrac{dy}{dx}=48-4x...\left( i \right) \\
\end{align}$
Let us equate $\dfrac{dy}{dx}=0$ .
$\Rightarrow 0=48-4x$
We have to take -4x to the LHS.
$\Rightarrow 4x=48$
Let us take the coefficient of x to the RHS.
$\begin{align}
  & \Rightarrow x=\dfrac{48}{4} \\
 & \Rightarrow x=12 \\
\end{align}$
Now, we have to differentiate (i) with respect to x.
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 48-4x \right)$
We know that $\dfrac{d}{dx}\left( \text{constant} \right)=0$ . Therefore, the above derivative becomes
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-4$
We can see that the second derivative is negative. Therefore, maximum revenue occurs at the output of 12.

So, the correct answer is “Option b”.

Note: The given question is an application of derivatives. We can find the revenue at $x=12$ by substituting this value in the given equation for revenue.
$\begin{align}
  & \Rightarrow y=48\times 12-2{{\left( 12 \right)}^{2}} \\
 & \Rightarrow y=576-288 \\
 & \Rightarrow y=288 \\
\end{align}$
Therefore, maximum revenue is 288.