Wire of length 2 units is cut into two parts which are bent respectively to form a square of side $x$ units and a circle of radius $r$ units. If the sum of the areas of the square and the circle so formed is minimum, then

Let $A$be the sum of the areas of the square and $t$ circle.

$\begin{array}{l}A=x^2+\pi r^2,\text{ where }4x+2\pi r=2\\ \Rightarrow A={\left(\frac{1}{2}-\pi r\right)}^2+\pi r^2\\ \Rightarrow A=\frac{1}{4}(1-\pi r{)}^2+\pi r^2\\ \Rightarrow \frac{dA}{dr}=-\frac{\pi }{2}(1-\pi r)+2\pi r\text{ and }\frac{d^{2}A}{d{r}^2}=\frac{{\pi }^2}{2}+2\pi \end{array}$

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

$\frac{dA}{dr}=0\Rightarrow \frac{1}{2}(1-\pi r)=2r\Rightarrow 4r=1-\pi r\Rightarrow r=\frac{1}{\pi +4}$

Clearly, $\frac{d^{2}A}{d{r}^2}>0$ for all $r$. So, $A$ is minimum when $r=\frac{1}{\pi +4}$

Putting $r=\frac{1}{\pi +4}\text{ is }4x+2\pi r=2$, we get

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

Clearly, $x = 2r.$