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(C) Let the point $P$ on the ellipse be $(5 \cos 	heta, 4 \sin 	heta)$ in the first quadrant.Since the rectangle is symmetric about both axes,its ve

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$4 \sqrt{2}, 5 \sqrt{2}$

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(C) Let the point $P$ on the ellipse be $(5 \cos 	heta, 4 \sin 	heta)$ in the first quadrant.Since the rectangle is symmetric about both axes,its vertices are $(\pm 5 \cos 	heta, \pm 4 \sin 	heta)$.The length of the rectangle is $L = 2(5 \cos 	heta) = 10 \cos 	heta$.The breadth of the rectangle is $B = 2(4 \sin 	heta) = 8 \sin 	heta$.The area $A$ of the rectangle is $A = L 	imes B = (10 \cos 	heta)(8 \sin 	heta) = 80 \sin 	heta \cos 	heta = 40 \sin(2 	heta)$.For the area to be maximum,$\sin(2 	heta)$ must be maximum,i.e.,$\sin(2 	heta) = 1$.This implies $2 	heta = \frac{\pi}{2}$,so $	heta = \frac{\pi}{4}$.Substituting $	heta = \frac{\pi}{4}$ into the expressions for length and breadth:$L = 10 \cos(\frac{\pi}{4}) = 10 	imes \frac{1}{\sqrt{2}} = 5 \sqrt{2}$.$B = 8 \sin(\frac{\pi}{4}) = 8 	imes \frac{1}{\sqrt{2}} = 4 \sqrt{2}$.Thus,the dimensions of the rectangle are $5 \sqrt{2}$ and $4 \sqrt{2}$.

$A$ rectangle of maximum area is inscribed in an ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$. Then its dimensions are:

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