Question

A wire of length $50m$ is cut into two pieces. One piece of the wire is bent in the shape of a square and the other in the shape of a circle. What should be the length of each piece so that the combined area of the two is minimum?

Answer 1

Length of wire $=50m$ (Given)
Let length of one piece for shape of square $=xm$
$\therefore$ Length of other piece for shape of circle
$=(50-x)m$
Now perimeter of square $=4a=x$
$\Rightarrow a=\frac{x}{4}$
and circumference of circle $=2\pi r=50-x$
$\Rightarrow r=\frac{50-x}{2\pi }$
Combined Area $=a^2+\pi r^2$
$=\frac{x^2}{16}+\pi \cdot {\left(\frac{50-x}{2\pi }\right)}^2$
$=\frac{x^2}{16}+\pi \cdot \frac{(50-x{)}^2}{4{\pi }^2}$
$A=\frac{x^2}{16}+\frac{(50-x{)}^2}{4\pi }$
Differentiating w.r. to $x$, we get
$\frac{dA}{dx}=\frac{2x}{16}+\frac{2(50-x)(-1)}{4\pi }$
$\frac{dA}{dx}=\frac{x}{8}+\frac{(x-50)}{2\pi }$
$=\frac{\pi x+4x-200}{8\pi }$
$=\frac{x(4+\pi )-200}{8\pi }$
For extremum, $\frac{dA}{dx}=0$
$\therefore x(4+\pi )-200=0$
$x=\frac{200}{4+\pi }$
$\frac{d^{2}A}{d{x}^2}>0$
$A$ is minimum at $x=\frac{200}{4+\pi }$
$\therefore$ Length of square of square wire, $x=\frac{200}{4+\pi }m$
and length of circle wire $=50-x$
$=50-\frac{200}{4+\pi }$
$=\frac{50\pi }{4+\pi }m$