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Prove the square is the rectangle with maximal area inscribed in a given circle

Given a circle, prove that the square is the rectangle of maximal area that can be inscribed in the circle.


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Proof. Let x and y denote the lengths of the sides of the inscribed rectangle, and r denote the radius of the circle. Then x^2 + y^2 = 4r^2 implies y = \sqrt{4r^2 - x^2}. Therefore,

    \[ \text{Area} = xy = x \sqrt{4x^2- r^2} \quad \implies \quad f'(x) = 2 \left( \frac{2r^2 - x^2}{\sqrt{4r^2 - x^2}}. \]

We then have f'(x) = 0 when x = \sqrt{2} r and

    \begin{align*}  f'(x) &< 0 & \text{when} && x &< \sqrt{2}r \\  f'(x) &> 0 & \text{when} && x &> \sqrt{2}r. \end{align*}

Hence, f(x) has a maximum at x = \sqrt{2}r. By our equation for y we then have

    \[ y = \sqrt{4r^2 - x^2} = \sqrt{4r^2 - 2r^2} = \sqrt{2r^2} = \sqrt{2} r. \]

Thus, x = y, so the rectangle is a square. \qquad \blacksquare

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