Jee mains 2016 Mathematics 51

Let us split the wire according to lengths $4x, 2-4x$ for the square and circle respectively. The circle will have a radius: $2\pi r = 2-4x \Rightarrow r = \frac{2-4x}{2\pi}.$ The combined areas can be written as the following function:

$A(x) = x^2 + \pi(\frac{2-4x}{2\pi})^{2}$

such that $A'(x) = 0 \Rightarrow 2x - \frac{2(2-4x)(-4)}{4\pi} = 0 \Rightarrow x = \frac{2}{\pi+4}$ and $A''(\frac{2}{\pi+4}) = 2 + \frac{8}{\pi} > 0$ is a minimum. Substituting this value for $x$ in for the radius $r$ produces:

$2\pi r = 2 - 4x = 2 - 4(\frac{2}{\pi+4}) \Rightarrow r = \frac{1}{\pi+4} = \frac{x}{2} \Rightarrow \boxed{x = 2r}.$