Answer

Verified

348k+ views

**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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Change the following sentences into negative and interrogative class 10 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE