Question

Find the least value of f(x) = \[ax + \frac{b}{x}\], where a > 0, b > 0 and x > 0 .

Answer 1

\[\text { We have }, \]

\[f\left( x \right) = ax + \frac{b}{x}\]

\[ \Rightarrow f'\left( x \right) = a - \frac{b}{x^2}\]

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

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

\[ \Rightarrow a - \frac{b}{x^2} = 0\]

\[ \Rightarrow x^2 = \frac{b}{a}\]

\[ \Rightarrow x = \sqrt{\frac{b}{a}}, - \sqrt{\frac{b}{a}}\]

\[\text { But, }x > 0 \]

\[ \Rightarrow x = \sqrt{\frac{b}{a}}\]

\[\text { Now }, \]

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

\[\text { At }x = \sqrt{\frac{b}{a}} \]

\[f''\left( \sqrt{\frac{b}{a}} \right) = \frac{2b}{\left( \sqrt{\frac{b}{a}} \right)^3} = \frac{2 a^\frac{3}{2}}{b^\frac{1}{2}} > 0 .....................\left[ \because a > 0 \text{ and }b > 0 \right]\]

\[\text { So }, x = \sqrt{\frac{b}{a}} \text { is a point of local minimum }. \]

\[\text { Hence, the least value is }\]

\[f\left( \sqrt{\frac{b}{a}} \right) = a\sqrt{\frac{b}{a}} + \frac{b}{\sqrt{\frac{b}{a}}} = \sqrt{ab} + \sqrt{ab} = 2\sqrt{ab}\]