Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ f\left(x\right)=ax+\frac{b}{x} \\ \Rightarrow f\text{'}\left(x\right)=a-\frac{b}{x^2} \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow a-\frac{b}{x^2}=0 \\ \Rightarrow x^2=\frac{b}{a} \\ \Rightarrow x=\sqrt{\frac{b}{a}},-\sqrt{\frac{b}{a}} \\ \mathrm{But},x>0 \\ \Rightarrow x=\sqrt{\frac{b}{a}} \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=\frac{2b}{x^3} \\ \mathrm{At}x=\sqrt{\frac{b}{a}} \\ f\text{'}\text{'}\left(\sqrt{\frac{b}{a}}\right)=\frac{2b}{{\left(\sqrt{{\displaystyle \frac{b}{a}}}\right)}^3}=\frac{2a^{{\displaystyle \frac{3}{2}}}}{b^{{\displaystyle \frac{1}{2}}}}>0\left[\because a>0\mathrm{and}b>0\right] \\ \mathrm{So},x=\sqrt{\frac{b}{a}}\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{of}\mathrm{local}\mathrm{minimum}. \\ \mathrm{Hence},\mathrm{the}\mathrm{least}\mathrm{value}\mathrm{is} \\ f\left(\sqrt{\frac{b}{a}}\right)=a\sqrt{\frac{b}{a}}+\frac{b}{\sqrt{{\displaystyle \frac{b}{a}}}}=\sqrt{ab}+\sqrt{ab}=2\sqrt{ab} \end{gathered}$$