JEE 2016 Problem

The sum of the perimeters of the two shapes is: $Π=4x+2πr$ . This must be equal to 2 units since it's the lenght of the wire. So: $4x+2πr=2\Rightarrow r=\frac{1-2x}{π}$ . The sum of the areas of the two shapes is: $E=x^{2}+πr^{2}\Rightarrow E(x)=x^{2}+π\frac{(1-2x)^{2}}{π^{2}}\Rightarrow E(x)=\frac{(4+π)x^{2}-4x+1}{π}$ . Now we take the derivative of E(x) in terms of x: $E'(x)=\frac{2(4+π)x-4}{π}$ . Since E(x) is minimum, we set $E'(x)=0\Rightarrow 2(4+π)x=4\Rightarrow x=\frac{2}{4+π}$ . We substitute this value of x in the equation $r=\frac{1-2x}{π}$ to get: $r=\frac{1}{4+π}$ from which we derive that : $\boxed{x=2r}$