#### 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

\[(b) \frac{1}{2}\]

\[\text { Let the two non - zero numbers be x and y . Then,} \]

\[x + y = 8\]

\[ \Rightarrow y = 8 - x . . . \left( 1 \right)\]

\[\text { Now,} \]

\[f\left( x \right) = \frac{1}{x} + \frac{1}{y}\]

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

\[ \Rightarrow f'\left( x \right) = \frac{- 1}{x^2} + \frac{1}{\left( 8 - x \right)^2}\]

\[\text { For a local minima or a local maxima, we must have } \]

\[f'\left( x \right) = 0\]

\[ \Rightarrow \frac{- 1}{x^2} + \frac{1}{\left( 8 - x \right)^2} = 0\]

\[ \Rightarrow \frac{- \left( 8 - x \right)^2 + x^2}{\left( x \right)^2 \left( 8 - x \right)^2} = 0\]

\[ \Rightarrow - 64 - x^2 + 16x + x^2 = 0\]

\[ \Rightarrow 16x - 64 = 0\]

\[ \Rightarrow x = 4\]

\[f''\left( x \right) = \frac{2}{x^3} - \frac{2}{\left( 8 - x \right)^3}\]

\[ \Rightarrow f''\left( 4 \right) = \frac{2}{4^3} - \frac{2}{\left( 8 - 4 \right)^3}\]

\[ \Rightarrow f''\left( 4 \right) = \frac{2}{64} - \frac{2}{64} = 0\]

\[ \therefore \text { Minimum value }= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\]