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?
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.