The sum of two nonzero numbers is 8. The minimum value of the sum of their reciprocals is :

Solution :
$$\eqalign{ & {\text{Here, }}a + b = 8. \cr & {\text{Let }}y = \frac{1}{a} + \frac{1}{b} = \frac{1}{a} + \frac{1}{{8 - a}} = \frac{8}{{a\left( {8 - a} \right)}} \cr & \therefore \frac{{dy}}{{da}} = \frac{{ - 8}}{{{a^2}{{\left( {8 - a} \right)}^2}}}\left\{ {8 - 2a} \right\} \cr & \therefore \frac{{dy}}{{da}} = 0{\text{ if }}8 - 2a = 0,\,{\text{i}}{\text{.e}}{\text{., }}a = 4 \cr & {\text{Also}},{\left. {{\text{ }}\frac{{dy}}{{dx}}} \right)_{4 - \in }} = \frac{{ - 8}}{{{{\left( {4 - \in } \right)}^2}.{{\left( {4 + \in } \right)}^2}}}\left( {2 \in } \right) < 0 \cr & {\left. {{\text{ }}\frac{{dy}}{{dx}}} \right)_{4 + \in }} = \frac{{ - 8}}{{{{\left( {4 + \in } \right)}^2}.{{\left( {4 - \in } \right)}^2}}}\left( { - 2 \in } \right) > 0 \cr & \therefore a = 4,\,y{\text{ is minimum}}{\text{.}} \cr & {\text{So min }}y = \frac{1}{4} + \frac{1}{{8 - 4}} = \frac{1}{2} \cr} $$