If \(l\) and \(b\) are respectively the length and breadth of the rectangle of greatest area that can be…
If \(l\) and \(b\) are respectively the length and breadth of the rectangle of greatest area that can be inscribed in the ellipse \(x^2+4 y^2=64\), then \((l, b)=\)
\((16 \sqrt{2}, 4 \sqrt{2})\)
\((8 \sqrt{2}, 6 \sqrt{2})\)
\((8 \sqrt{2}, 4 \sqrt{2})\)
\((6 \sqrt{2}, 4 \sqrt{2})\)
Solution
Let \(P Q R S\) be a rectangle inscribed in the ellipse \(\frac{x^2}{64}+\frac{y^2}{16}=1\). Let the coordinate of \(P\) be \((8 \cos \theta, 4 \sin \theta)\). Then, the coordinates of \(Q, R\) and \(S\) are \((-8 \cos \theta, 4 \sin \theta),(-8 \cos \theta, 4 \sin \theta)\) and \((8 \cos \theta,-4 \sin \theta)\) respectively.
Let \(A\) be the area of rectangle \(P Q R S\). Then
\(\begin{gathered}
A=P Q \times R S \Rightarrow A=16 \cos \theta \times 8 \sin \theta \\
=64 \sin 20 \\
\Rightarrow \frac{d A}{d \theta}=128 \cos 2 \theta \text { and } \frac{d^2 A}{d \theta^2}=-256 \sin 2 \theta
\end{gathered}\)
The critical numbers of \(A\) are given by \(\frac{d A}{d \theta}=0\)
\(\therefore \quad \frac{d A}{d \theta}=0 \Rightarrow 4 a b \cos 2 \theta=0\)
\(\Rightarrow \quad \cos 2 \theta=0 \Rightarrow 2 \theta=\frac{\pi}{2}\) or \(\frac{3 \pi}{2}\)
\(\Rightarrow \quad \theta=\frac{\pi}{4}\) or \(\frac{3 \pi}{4}\)
Clearly, \(\left(\frac{d^2 A}{d \theta^2}\right)_{\theta=\frac{\pi}{4}}=-256 \sin \frac{\pi}{2}=-256 < 0\)
So, \(A\) is maximum when \(\theta=\frac{\pi}{4}\)
\(\therefore \quad P Q=l=16 \cos \theta=16 \cos \frac{\pi}{4}=16 \times \frac{1}{\sqrt{2}}=8 \sqrt{2}\)
and \(P S=b=8 \sin \theta=8 \sin \frac{\pi}{4}=8 \times \frac{1}{\sqrt{2}}=4 \sqrt{2}\)
Hence, \((l, b)=(8 \sqrt{2}, 4 \sqrt{2})\)