Question

A wire of length $l$ is cut into two parts. One part is bent into a circle and other into a square. Show that the sum of areas of the circle and square is the least, if the radius of circle is half the side of the square

Answer 1

Given, length of wire is $l$.
Let length of one part be $x$, then other part be $l-x$.
Let $x$ part be bent into a circle and $l-x$ part be in a square.
$\therefore 2\pi r=x$ and $4s=1-x\Rightarrow s=\frac{l-x}{4}(\because s=$ side $)$
According to question, we have
$r=\frac{s}{2}=\frac{l-x}{8}$
Also, $r=\frac{x}{2\pi }$
$\Rightarrow \frac{l-x}{8}=\frac{x}{2\pi }$
$\Rightarrow l=\frac{8x}{2\pi }+x=\frac{4x}{\pi }+x$
Sum of areas of circle and square,
$f\left(x\right)=\pi {\left(\frac{l-x}{8}\right)}^2+{\left(\frac{l-x}{4}\right)}^2$
$f\left(x\right)=\frac{\pi }{64}\left[\right(l-x{)}^2+4(l-x{)}^2]$
$f\left(x\right)=\frac{\pi }{64}5(l-x{)}^2$
$=\frac{5\pi }{64}{(\frac{4x}{\pi }+x-x)}^2$
$f\left(x\right)=\frac{5\pi }{64}\frac{16x^2}{{\pi }^2}$
$f\left(x\right)=\frac{5x^2}{4\pi }$
$\therefore f^{\prime }\left(x\right)=\frac{5}{4\pi }2x=\frac{5x}{2\pi }$
For minimum value, $f^{\prime }\left(x\right)=0$
$\Rightarrow \frac{5}{2\pi }x=0$
$\Rightarrow x=0$
If $x=o{f}^{\prime }"\left(0\right)=\frac{5}{2\pi }>0$
$\therefore f\left(x\right)$ attains minimum at $x=0$.