# A wire of length 28 m is to be cut into two pieces.

Question:

A wire of length 28 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 lengths of the two pieces so that the combined area of the circle and the square is minimum?

Solution:

Suppose the wire, which is to be made into a square and a circle, is cut into two pieces of length $x$ m and $y$ m, respectively. Then,

$x+y=28$       ......(1)

Perimeter of square, 4 (side) $=x$

$\Rightarrow$ Side $=\frac{x}{4}$

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

Circumference of circle, $2 \pi r=y$

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

Area of circle $=\pi r^{2}=\pi\left(\frac{y}{2 \pi}\right)^{2}=\frac{y^{2}}{4 \pi}$

Now,

$z=$ Area of square $+$ Area of circle

$\Rightarrow z=\frac{x^{2}}{16}+\frac{y^{2}}{4 \pi}$

$\Rightarrow z=\frac{x^{2}}{16}+\frac{(28-x)^{2}}{4 \pi}$

$\Rightarrow \frac{d z}{d x}=\frac{2 x}{16}-\frac{2(28-x)}{4 \pi}$

For maximum or minimum values of $z$, we must have

$\frac{d z}{d x}=0$

$\Rightarrow \frac{2 x}{16}-\frac{2(28-x)}{4 \pi}=0$       [From eq. (1)]

$\Rightarrow \frac{x}{4}=\frac{(28-x)}{\pi}$

$\Rightarrow \frac{x \pi}{4}+x=28$

$\Rightarrow x\left(\frac{\pi}{4}+1\right)=28$

$\Rightarrow x=\frac{28}{\left(\frac{\pi}{4}+1\right)}$

$\Rightarrow x=\frac{112}{\pi+4}$

$\Rightarrow y=28-\frac{112}{\pi+4}$              [From eq. (1)]

$\Rightarrow y=\frac{28 \pi}{\pi+4}$

$\frac{d^{2} z}{d x^{2}}=\frac{1}{8}+\frac{1}{2 \pi}>0$

Thus, $z$ is minimum when $x=\frac{112}{\pi+4}$ and $y=\frac{28 \pi}{\pi+4}$.

Hence, the length of the two pieces of wire are $\frac{112}{\pi+4} \mathrm{~m}$ and $\frac{28 \pi}{\pi+4} \mathrm{~m}$ respectively.