If the product of two positive numbers is 9, find the numbers so that the sum of their squares is minimum.

Let numbers be x and $\displaystyle \frac{9}{\mathrm{x}} \therefore$ required sum $\displaystyle =\mathrm{x}^{2}+\frac{81}{\mathrm{x}^{2}}=\mathrm{f}(\mathrm{x})$ say $\displaystyle \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-\frac{162}{\mathrm{x}^{3}}$ $\displaystyle \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow \mathrm{x}=3$