Question

A wire $10$ meter long is not into two parts. One part is bent into the shape of circle and the other into the shape of an equilateral triangle. How should the wire be cut so that the combined area of the two figures is as small as possible?

Answer 1

Total length of wire$=10m$

Let 'a' be length of part cut and bent into circle

$\therefore 2\pi r=a$

$r=\frac{a}{2\pi }$

Let $(10-a)$ be length of part bent into equilateral triangle

$\therefore 3S=(10-a)$

$S=\left(\frac{10-a}{3}\right)$

Combined area of two figures$=\pi r^2+\frac{\sqrt{3}}{4}{S}^2$

$A=\pi \left(\frac{a^2}{4{\pi }^2}\right)+\frac{\sqrt{3}}{4}{\left(\frac{10-a}{3}\right)}^2$

$A=\frac{a^2}{4\pi }+\frac{\sqrt{3}}{36}(a^2-20a+100)$

$A=(\frac{1}{4\pi }+\frac{\sqrt{3}}{36})a^2-\frac{5\sqrt{3}}{9}a+\frac{25\sqrt{3}}{9}$

$A^{\prime }=2(\frac{1}{4\pi }+\frac{\sqrt{3}}{36})a-\frac{5\sqrt{3}}{9}=0$

$\Rightarrow a=\frac{\frac{5\sqrt{3}}{9}}{\frac{1}{2\pi }+\frac{\sqrt{3}}{18}}=\frac{5\sqrt{3}}{(\frac{9}{2\pi }+\frac{\sqrt{3}}{2})}$

$\therefore a=\frac{10\pi \sqrt{3}}{9+\pi \sqrt{3}},10-a=\frac{90}{9+\pi \sqrt{3}}$

Wire must be cut in ration, $a:10-a=\pi :3\sqrt{3}$