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Find the maximum area of a square circumscribed about a given square

Given a square with edges each having length L, find the area of the largest square that can be circumscribed about the given square.


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Let e be the length of the edge of the circumscribed square. Then e = x + y. Furthermore, if L is the length of the edge of the given square, we have

    \[ x^2 + y^2 = L^2 \quad \implies \quad y = \sqrt{L^2 - x^2}. \]

Then,

    \begin{align*}  \text{Area} = (x+y)^2 &= (x + \sqrt{L^2 - x^2})^2 \\  &= x^2 + 2x \sqrt{L^2 - x^2} + L^2 - x^2 \\  &= 2x \sqrt{L^2 - x^2} + L^2 \end{align*}

Let this function be f(x), and take the derivative,

    \[ f'(x) = 2 \sqrt{L^2 - x^2} - \frac{2x}{\sqrt{L^2 - x^2}} = 2\frac{L^2 - x^2 - x}{\sqrt{L^2-x^2}}. \]

Thus, the derivative is zero when x = \frac{L}{\sqrt{2}}. Furthermore,

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

Therefore, f(x) is increasing when x < \frac{L}{sqrt{2}} and decreasing when x > \frac{L}{sqrt{2}}; hence, f(x) has a maximum at x = \frac{L}{\sqrt{2}}. Solving for y we then have

    \[ y = \sqrt{L^2 - \frac{L^2}{2}} = \sqrt{\frac{L^2}{2}} = \frac{L}{\sqrt{2}}. \]

Finally, since e = x + y we then have the area of the circumscribed square given by

    \[ \text{Area} = e^2 = (x+y)^2 = \left( \frac{2L}{\sqrt{2}} \right)^2 = 2L^2. \]

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