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Explain Solution for RD Sharma Maths Class 12 Chapter 17 Maxima and Minima Exercise Fill in the blanks Question 14 maths textbook solution

Answers (1)

Answer:

              1

Hint:

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

Given:

Sum of two non-zero numbers is 4

Solution:

Let us consider

              x+y=4

Or          y=4-x

That is

\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}

Or f\left ( x \right )=\frac{4}{xy}=\frac{4}{x\left ( 4-x \right )}          

Now

\begin{aligned} &f(x)=\frac{4}{\left(4 x-x^{2}\right)} \\ &f^{\prime}(x)=\left(\frac{-4}{\left(4 x-x^{2}\right)^{2}}\right)(4-2 x) \end{aligned}

Substituting {f}'\left ( x \right )  we get 4-2x=0

{f}'\left ( x\therefore x=2,y=2 \right )

Therefore

\begin{aligned} &\min \left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{2}+\frac{1}{2} \\ &\min \left(\frac{1}{x}+\frac{1}{y}\right)=1 \end{aligned}

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