Question

A wire of length $l$ is cut into two pars. One part is bent into a circle of radius $r$ and the other into a square of side $x$. The sum of the area of circle and square is least, if

• A. $x=r$
• B. $x=\frac{r}{2}$
• C. $x=2r$
• D. $x=\pi r$

Answer 1

The correct option is C $x=2r$

Length of wire $=l$

Let the wire be cut in $d-k$ & $k$ parts.

as per Ques:

$k=4x$ & $l-k=2\pi r$

$\frac{k}{4}=x$ & $r=\frac{l-k}{2\pi }$

Area of square $=x^2={\left(\frac{k}{4}\right)}^2=\frac{k^2}{16}$

Area of circle $=\pi r^2=\pi {\left(\frac{l-k}{2n}\right)}^2$

$=\frac{l^2+k^2-2lk}{4\pi }$

Total area $\Rightarrow \frac{k^2}{16}+\frac{l^2+k^2-2lk}{4\pi }$

$f^{\prime }\left(k\right)=\frac{2k}{16}+\frac{2k-2l}{4\pi }$

$f^{\prime }\left(k\right)=0\Rightarrow \frac{2k}{16}+\frac{2(k-l)}{4\pi }$

$\frac{2k}{16}=\frac{2}{4\pi }(l-k)$

$k\left(\pi \right)=4(l-k)$

$k=\left(\frac{4l}{4+\pi }\right)$

$x=\left(\frac{4l}{4+\pi }\right)\frac{1}{4}=\left(\frac{l}{4+\pi }\right)$

$r=\frac{l-(4l/4+\pi )}{2n}=\frac{l(4+\pi )-4l}{(4+n)2n}$

$r=\frac{l}{(4+\pi )2}$

$\therefore x=2r$