Suppose, the sides of the rectangle are $x$ and $y$ and radius of circle = $R$ unit then

$$\max 2(x+y)$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; subject \;\;to \;\; x^2 + y^2 = (2R)^2$$

By Lagrangian Multiplier,

Lagrangian function $\mathcal{L} = 2(x+y) – \lambda(x^2 + y^2 – (2R)^2)$

Given $R=1$

$\mathcal{L} = 2(x+y) – \lambda (x^2 + y^2 – 2^2)$

By first order optimality condition,

$\frac{\partial \mathcal{L}}{\partial x} = 0 \Rightarrow 2 - \lambda \times 2x = 0 \Rightarrow \lambda = \frac{1}{x}$ and

$\frac{\partial \mathcal{L}}{\partial y} = 0 \Rightarrow 2 - \lambda \times 2y = 0 \Rightarrow \lambda = \frac{1}{y}$

So, $x=y \rightarrow (1)$

$\frac{\partial \mathcal{L}}{\partial \lambda} = 0 \Rightarrow x^2 + y^2 – 2^2 = 0 \Rightarrow x^2 + y^2 = 2^2 \rightarrow (2)$

From $(1)$ and $(2)$

$2x^2 = 2^2 \Rightarrow x = \pm \sqrt{2}$

So, critical points are : $(\sqrt{2},\sqrt{2}), (-\sqrt{2},-\sqrt{2})$

For $(\sqrt{2},\sqrt{2}),$ $2(x+y)$ will give maximum value.

Hence, Area of rectangle = $\sqrt{2} \times \sqrt{2} = 2$

$\textbf{Answer: (B)}$