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(B) The equation of the ellipse is $x^2+4y^2=64$,which can be written as $\frac{x^2}{64} + \frac{y^2}{16} = 1$.This is in the form $\frac{x^2}{a^2} +

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$8\sqrt{2}, 4\sqrt{2}$

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(B) The equation of the ellipse is $x^2+4y^2=64$,which can be written as $\frac{x^2}{64} + \frac{y^2}{16} = 1$.This is in the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a^2=64$ $(a=8)$ and $b^2=16$ $(b=4)$.Let a vertex of the rectangle in the first quadrant be $(x, y) = (a \cos 	heta, b \sin 	heta) = (8 \cos 	heta, 4 \sin 	heta)$.The sides of the rectangle are $2x$ and $2y$,so the area $A = (2x)(2y) = 4xy = 4(8 \cos 	heta)(4 \sin 	heta) = 128 \sin 	heta \cos 	heta = 64 \sin(2	heta)$.The area is maximum when $\sin(2	heta) = 1$,i.e.,$2	heta = 90^\circ$ or $	heta = 45^\circ$.Substituting $	heta = 45^\circ$,we get $x = 8 \cos 45^\circ = 8 	imes \frac{1}{\sqrt{2}} = 4\sqrt{2}$ and $y = 4 \sin 45^\circ = 4 	imes \frac{1}{\sqrt{2}} = 2\sqrt{2}$.The sides of the rectangle are $2x = 8\sqrt{2}$ and $2y = 4\sqrt{2}$.

The sides of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=64$ are:

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