Let a rectangle $A B C D$ of sides 2 and 4 be inscribed in another rectangle $P Q R S$ such that the vertices of the rectangle $A B C D$ lie on the sides of the rectangle $P Q R S$. Let $a$ and $b$ be the sides of the rectangle $P Q R S$ when its area is maximum. Then $(a+b)^2$ is equal to :

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Area }=(4 \cos 	heta+2 \sin 	heta)(2 \cos 	heta+4 \sin 	heta) \ & =8 \cos ^2 	heta+16 \sin 	heta \cos 	heta+4 \sin 	heta \cos 	heta+8 \sin ^2 	heta \ & =8+20 \sin 	heta \cos 	heta \ & =8+10 \sin 2 	heta \ & 	ext { Max Area }=8+10=18(\sin 2 	heta=1) 	heta=45^{\circ} \ & (a+b)^2=(4 \cos 	heta+2 \sin 	heta+2 \cos 	heta+4 \sin 	heta)^2 \ & =(6 \cos 	heta+6 \sin 	heta)^2 \ & =36(\sin 	heta+\cos 	heta)^2 \ & =36(\sqrt{2})^2\end{aligned}$ $=72$

Let a rectangle $A B C D$ of sides 2 and 4 be inscribed in another rectangle $P Q R S$ such that the…

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