Question

A wire of length $25cm$ is to be cut off into two pieces. One piece is to be made onto a circle and other into a square. What should be the lengths of pieces so that combined area of circle and square is minimum?

Answer 1

Let $x$ be the length of the first part.

Length if other part $=25-x$

Given that one part is converted into a circle.

Let $r$ be the radius of circle.

$2\pi r=x$

$\Rightarrow r=\frac{x}{2\pi }$

Also given that another part is converted into a square.

Let '$a$' be the side of circle.

$4a=25-x$

$a=\frac{25-x}{4}$

Total area $\left(A\right)=$ Area of circle $+$ Area of square

$\therefore A=\pi r^2+a^2$

$\Rightarrow A=\pi {\left(\frac{x}{2\pi }\right)}^2+{\left(\frac{25-x}{4}\right)}^2$

$\Rightarrow A=\frac{x^2}{4\pi }+\frac{{(25-x)}^2}{16}.....\left(1\right)$

Differentiating above equation w.r.t. $x$, we get

$\frac{dA}{dx}=\frac{x}{2\pi }+\frac{x-25}{8}.....\left(2\right)$

Putting $\frac{dA}{dx}=0$, we have

$\frac{x}{2\pi }+\frac{x-25}{8}=0$

$\frac{4x+\pi x-25\pi }{8\pi }=0$

$\Rightarrow x=\frac{25\pi }{4+\pi }$

Now, differentiating equation $\left(2\right)$, we get

$\frac{d^{2}A}{d{t}^2}=\frac{1}{2\pi }+\frac{1}{8}>0$

Hence at $x=\frac{25\pi }{4+\pi }$, total area will be minimum.

Therefore,

Length of pieces-

$x=\frac{25\pi }{4+\pi }$

$25-x=25-\frac{25\pi }{4+\pi }=\frac{100-25\pi +25\pi }{4+\pi }=\frac{100}{4+\pi }$

Hence the length of the pieces are $\frac{25\pi }{4+\pi }cm$ and $\frac{100}{4+\pi }cm$ respectively.