(a)OR(b)

The median of an equilateral triangle is increasing at the rate of $\displaystyle 2 \sqrt{3} \mathrm{~cm} / \mathrm{s}$. Find the rate at which its side is increasing.

In an equilateral triangle, median is same as altitude. Let 'h' denote the length of the median (or altitude) and ' $\displaystyle \mathrm{x}$ ' be the side of $\displaystyle \triangle \mathrm{ABC}$. Then, $\displaystyle \mathrm{h}=\frac{\sqrt{3}}{2} \mathrm{x}$ or $\displaystyle \mathrm{x}=\frac{2 \mathrm{~h}}{\sqrt{3}}$ $\displaystyle \_\_\_\_$ (i) Let the two numbers be x and y . Then, $\displaystyle \mathrm{x}+\mathrm{y}=5$ or $\displaystyle \mathrm{y}=5-\mathrm{x}$ Let S denote the sum of the cubes of these numbers. ThenS=\mathrm{x}^{3}+\mathrm{y}^{3}=\mathrm{x}^{3}+(5-\mathrm{x})^{3} \frac{d S}{d \mathrm{x}}=3 \mathrm{x}^{2}-3(5-\mathrm{x})^{2}=15(2 \mathrm{x}-5)Now $\displaystyle \frac{\mathrm{dS}}{\mathrm{dx}}=0$, gives $\displaystyle \mathrm{x}=\frac{5}{2}$ Showing S is minimum at $\displaystyle \mathrm{x}=\frac{5}{2}$ So, the two numbers are $\displaystyle \frac{5}{2}$ and\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=\frac{25}{4}+\frac{25}{4}=\frac{25}{2} $$