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# Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius $r$.

Let us take the circle to be a circle with center (0,0) and radius r .
Let P(x,y) be the vertex of the rectangle that lies on the first quadrant.
Let $\theta$ be the angle made y OP with the x-axis.
Then $x=r \cos \theta, y= r \sin \theta$
The dimensions of the rectangle are $2x=2r \cos \theta,2y =2r \sin \theta$
Area of the rectangle are $A=4r^2 \sin \theta \cos \theta$
$A(\theta)= 2r^2 \sin 2 \theta$
$A'(\theta)= 4r^2 \cos 2 \theta=0$
$A''(\theta)= -8r^2 \sin 2 \theta=0$
For max/ min $A'(\theta)=0 => 4r^2 \cos 2 \theta$
=> $\cos 2 \theta=0 => 2 \theta = \large\frac{\pi}{2}$
=> $\theta =\large\frac{\pi}{4}$
when $\theta =\large\frac{\pi}{4}$
$A''(\theta)= -8r^2 \sin 2 \bigg (\large\frac{\pi}{4} \bigg)$
$\qquad= -8r^2 \sin 2 \bigg (\large\frac{\pi}{2} \bigg)$
$\qquad = -8r^2 < 0$
$\therefore$ A is larger
when $\theta =\large\frac{\pi}{4}$
$2x= 2r \times \large\frac{1}{\sqrt 2}$$=\sqrt 2 r 2y=2r \times \large\frac{1}{\sqrt 2}$$=\sqrt 2 r$
The dimensions of the rectangle are $\sqrt 2r$ and $\sqrt 2r$
The rectangle is square also.