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  • Explain Solution for RD Sharma Maths Class 12 Chapter 17 Maxima and Minima Exercise Fill in the blanks Question 18 maths textbook solution

Explain Solution for RD Sharma Maths Class 12 Chapter 17 Maxima and Minima Exercise Fill in the blanks Question 18 maths textbook solution    

Answers (1)

Answer:

              75

Hint:

For maxima or minima we must have {f}'\left ( x \right )=0

Given:

\begin{aligned} &f^{\prime}(x)=\left(x^{2}-\frac{250}{x^{2}}\right) \\ \end{aligned}

Solution:

Let

\begin{aligned} &f^{\prime}(x)=\left(x^{2}-\frac{250}{x^{2}}\right) \\ \end{aligned}

Then

\begin{aligned} &f^{\prime}(x)=\left(2 x-\frac{250}{x^{2}}\right) \\ &f^{\prime \prime}(x)=\left(2+\frac{500}{x^{3}}\right) \\ &f^{\prime}(x)=0 \Rightarrow 2 x^{3}-250=0 \\ &2 x^{3}=250 \\ &x^{3}=125 \\ &x=5 \end{aligned}

\begin{aligned} &{{f}'}'(5)=\left(2+\frac{500}{125}\right)=6> 0 \\ \end{aligned} 

\therefore f\left ( x \right ) is minimum at x=5 and minimum value

=\left ( 25+\frac{250}{x} \right )=75

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