A 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 of r units. If the sum of the areas of square and the circle so formed is minimum, then

Perimeter of the square \(=4 x\)
Perimeter of the circle \(=2 \pi \mathrm{r}\)
\(\begin{array}{ll}
\therefore & 4 x+2 \pi r=2 \\
\therefore & 2 x+\pi r=1 \Rightarrow r=\frac{1-2 x}{\pi}
\end{array}...(i)\)
Sum of the areas \((\mathrm{A})=x^2+\pi \mathrm{r}^2\)
\(\therefore \quad \mathrm{A}=x^2+\pi\left(\frac{1-2 x}{\pi}\right)^2\)
...[From (i)
\(\therefore \quad \mathrm{A}=x^2+\frac{1}{\pi}(1-2 x)^2\)
Differentiating A w.r.t. \(x\), we get
\(\begin{aligned}
& \frac{\mathrm{dA}}{\mathrm{~d} x}=2 x+\frac{2}{\pi}(1-2 x)(-2), \frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{~d} x^2}=2+\frac{8}{\pi} \gt 0 \\
& \frac{\mathrm{dA}}{\mathrm{~d} x}=0 \Rightarrow 2 x-\frac{4}{\pi}+\frac{8 x}{\pi}=0 \\
& \Rightarrow(2 \pi+8) x=4 \\
& \Rightarrow(\pi+4) x=2 \\
& \Rightarrow x=\frac{2}{\pi+4}
\end{aligned}\)
\(\therefore \quad\) Area is minimum when \(x=\frac{2}{\pi+4}\)
Substituting \(x=\frac{2}{\pi+4}\) in equation (i), we get
\(\begin{aligned}
& \mathrm{r}=\frac{1}{\pi+4} \\
& \Rightarrow x=2 \mathrm{r}
\end{aligned}\)