Question

A wire of length $36\text{m}$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces, so that the combined area of square and circle is minimum?

Answer 1

Let piece of wire used to make square be $l\text{m}$.
Then, circle is made out of $(36-l)\text{m}$.
Side of square $=\frac{l}{4}$
$\Rightarrow$ Area of square $=\frac{l^2}{4^2}$
Also, $2\pi r=36-l\Rightarrow r=\frac{36-l}{2\pi }$
$\Rightarrow$ Area of circle $=\pi {\left(\frac{36-l}{2\pi }\right)}^2$

Total area $=\frac{l^2}{16}+\pi {\left(\frac{36-l}{2\pi }\right)}^2=\frac{l^2}{16}+\frac{(36-l{)}^2}{4\pi }$

$\frac{dA}{dl}=\frac{l}{8}+\frac{2(36-l)(-1)}{4\pi }=\frac{l}{8}-\frac{36-l}{2\pi }$
$\frac{d^{2}A}{d{l}^2}=\frac{1}{8}+\frac{1}{2\pi }>0$
For $\frac{dA}{dl}=0\Rightarrow \frac{l}{8}-\frac{36-l}{2\pi }=0\Rightarrow \frac{\pi l-4(36-l)}{8\pi }=0$
$\Rightarrow (\pi +4)l-144=0\Rightarrow l=\frac{144}{\pi +4}$

Hence, the combined area is minimum when length of square is $l=\frac{144}{\pi +4}\text{m}$ and that of circle is $(36-l)=\frac{36\pi }{\pi +4}\text{m}$