A wire of length ‘$l$’ is cut in to two parts which are bent respectively in the form of a square and a circle. What are the length of the pieces of the wire so that the sum of the areas is the least ?

Let ‘$x$’ and ‘$l-x$’ bethe parts made as a square and a circle let ‘$y$’ be the side of square and ‘$r$’ be the radius of the circle

$x=4yl-x=2\pi r$

$y=\frac{x}{4}r=\frac{l-x}{2\pi }$

Sum of area $\left(\mathrm{A}\right)=y^2+\pi r^2$

$=\frac{x^2}{16}+\pi {\left(\frac{l-x}{2\pi }\right)}^2=\frac{x^2}{16}+\frac{\pi }{4{\pi }^2}[l-x{]}^2$

Let $f\left(x\right)=\frac{x^2}{16}+\frac{1}{4\pi }(l-x{)}^2$

$f^{\mathrm{\prime }}\left(x\right)=\frac{2x}{16}+\frac{1}{4\pi }2(l-x)(-1)=\frac{x}{8}-\frac{1}{2\pi }(l-x)$

$f^{\prime \prime }\left(x\right)=\frac{1}{8}-\frac{1}{2\pi }(-1)=\frac{1}{8}+\frac{1}{2\pi }>0(min)$

Clearly it has min area

consider $f^{\mathrm{\prime }}\left(x\right)=0\Rightarrow \frac{x}{8}-\frac{(l-x)}{2\pi }=0$

$\frac{x}{8}=\frac{l-x}{2\pi }\Rightarrow \pi x=4l-4x\Rightarrow \pi x+4x=4l$

$x(\pi +4)=4l\Rightarrow x=\frac{4l}{\pi +4}$

Length of pieces are $x,l-x$

$\frac{4l}{\pi +4},l-\frac{4l}{\pi +4}\Rightarrow \frac{4l}{\pi +4},\frac{\pi l+4l-4l}{\pi +4}=\frac{4l}{\pi +4},\frac{\pi l}{\pi +4}$