Let a rectangle $$ABCD$$ of sides 2 and 4 be inscribed in another rectangle $$PQRS$$ such that the vertices of the rectangle $$ABCD$$ lie on the sides of the rectangle $$PQRS$$. Let $$a$$ and $$b$$ be the sides of the rectangle $$PQRS$$ when its area is maximum. Then $$(a + b)^2$$ is equal to :

Let inner rectangle sides be $$4$$ and $$2$$, inclined at an angle $$\theta$$ to the outer rectangle sides.

Outer dimensions: $$a = 4\cos\theta + 2\sin\theta$$ and $$b = 4\sin\theta + 2\cos\theta$$.

The outer area is maximized when it forms a symmetric configuration around the inner block ($$\theta = 45^\circ$$).

Substituting $$\theta = 45^\circ$$: $$a = b = \frac{4}{\sqrt{2}} + \frac{2}{\sqrt{2}} = 3\sqrt{2}$$.

 $$(a + b)^2 = (3\sqrt{2} + 3\sqrt{2})^2 = (6\sqrt{2})^2 = 72$$.