The sum of two non-zero numbers is 8, the minimum value of the sum of the reciprocals is
Question:

The sum of two non-zero numbers is 8, the minimum value of the sum of the reciprocals is

(a) $\frac{1}{4}$

(b) $\frac{1}{2}$

(C) $\frac{1}{8}$

(d) none of these

Solution:

$(b) \frac{1}{2}$

Let the two non-zero numbers be $\mathrm{x}$ and $\mathrm{y}$. Then,

$x+y=8$

$\Rightarrow y=8-x$            …….(1)

Now,

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

$\Rightarrow f(x)=\frac{1}{x}+\frac{1}{8-x}$    [From eq. (1)]

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

For a local minima or a local maxima, we must have

$f^{\prime}(x)=0$

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

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

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

$\Rightarrow 16 x-64=0$

$\Rightarrow x=4$

$f^{\prime \prime}(x)=\frac{2}{x^{3}}-\frac{2}{(8-x)^{3}}$

$\Rightarrow f^{\prime \prime}(4)=\frac{2}{4^{3}}-\frac{2}{(8-4)^{3}}$

$\Rightarrow f^{\prime \prime}(4)=\frac{2}{64}-\frac{2}{64}=0$

$\therefore$ Minimum value $=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$