Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the…

Question

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 ...

Accepted Answer

$\begin{gathered} PD=4\cos 	heta \Rightarrow PS=4\cos 	heta +2\sin 	heta \ DS=2\sin 	heta \ AP=4\sin 	heta \ QA=2\cos 	heta \Rightarrow PQ=2\cos 	heta +4\sin 	heta \ \Rightarrow PQRS=4(2\cos 	heta +\sin 	heta )(\cos 	heta +2\sin 	heta ) \ =4[2{\cos }^{2}	heta +2{\sin }^{2}	heta +5\sin 	heta \cos 	heta ] \ =8+10\sin 2	heta \end{gathered}$

Area is maximum when $\sin 2	heta =1\Rightarrow 	heta ={45}^{\circ }$

$\Rightarrow$ Maximum area $=8+10=18$

$	herefore PS=4	imes \frac{1}{\sqrt{2}}+2	imes \frac{1}{\sqrt{2}}=\frac{6}{\sqrt{2}}$

$\begin{gathered} PQ=2	imes \frac{1}{\sqrt{2}}+4	imes \frac{1}{\sqrt{2}}=\frac{6}{\sqrt{2}} \ 	herefore \, (a+b{)}^{2}={(\frac{6}{\sqrt{2}}+\frac{6}{\sqrt{2}})}^{2}=(\frac{12}{\sqrt{2}})=(6\sqrt{2}{)}^{2}=72 \end{gathered}$

Solutions · 1