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(A) Let the rectangle $ABCD$ be inscribed in a circle of radius $r$ centered at the origin. Let the coordinates of vertex $B$ be $(r \cos 	heta, r \s

Vedclass · Accepted Answer

$2 r^2$ sq. units

Vedclass · Answer

(A) Let the rectangle $ABCD$ be inscribed in a circle of radius $r$ centered at the origin. Let the coordinates of vertex $B$ be $(r \cos 	heta, r \sin 	heta)$.Then,the length $AB = 2r \cos 	heta$ and the width $BC = 2r \sin 	heta$.The area of the rectangle $A(	heta) = AB 	imes BC = (2r \cos 	heta)(2r \sin 	heta) = 2r^2 \sin 2	heta$.To find the maximum area,we differentiate $A(	heta)$ with respect to $	heta$:$A'(	heta) = 4r^2 \cos 2	heta$.Setting $A'(	heta) = 0$,we get $\cos 2	heta = 0$,which implies $2	heta = \frac{\pi}{2}$,so $	heta = \frac{\pi}{4}$.Checking the second derivative: $A''(	heta) = -8r^2 \sin 2	heta$.At $	heta = \frac{\pi}{4}$,$A''(\frac{\pi}{4}) = -8r^2 \sin(\frac{\pi}{2}) = -8r^2 Since the second derivative is negative,the area is maximum at $	heta = \frac{\pi}{4}$.Substituting $	heta = \frac{\pi}{4}$ into the area formula:Maximum Area $= 2r^2 \sin(2 	imes \frac{\pi}{4}) = 2r^2 \sin(\frac{\pi}{2}) = 2r^2(1) = 2r^2$ sq. units.

The maximum area of the rectangle that can be inscribed in a circle of radius $r$ is

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