#### Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise Multiple choice question, question 9.

Answer: option (b) $\frac{1}{2}$

Hint: For local maxima or minima, we must have f'(x) =0.

Given: $f(x)=\frac{1}{x}+\frac{1}{y}$

Solution:

Let the two non-zero numbers be x and y.

x + y =8

y = 8 -x                                                                                                         ...(i)

We have,

$f(x)=\frac{1}{x}+\frac{1}{y}$

$f(x)=\frac{1}{x}+\frac{1}{(8-x)}$                                                  …(ii)                   From Equation (i)

On differentiating w.r.t x

$f'(x)=\frac{-1}{x^2}+\frac{1}{(8-x)^2}$
for maxima and minima$f'(x)=0$

$\Rightarrow \frac{-1}{x^2}+\frac{1}{(8-x)^2}=0$

$\Rightarrow \frac{-(8-x)^2+x^2}{x^2(8-x)^2}=0$

$\Rightarrow -64-x^2+16x+x^2=0$

$\Rightarrow 16x=64$

$\Rightarrow x=4$

Now, again differentiating (ii) w.r.t  x

\begin{aligned} &f^{\prime \prime}(x)=\frac{2}{x^{3}}-\frac{2}{(8-x)^{3}} \\ &f^{\prime \prime}(4)=\frac{2}{4^{3}}-\frac{2}{(8-4)^{3}} \\ &f^{\prime \prime}(4)=\frac{2}{64}-\frac{2}{64}=0 \end{aligned}

Minimum Value =$\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$