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Prove that for all rectangles of given area, the square admits the smallest circumscribed circle

by RoRi

Consider all rectangles of a fixed area. Prove that the square admits the smallest circumscribed circle.

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Proof. Let x and y denote the lengths of the sides of the rectangle, r the radius of the circumscribed circle, and A = xy the area of the square. (This is similar to the previous exercise, except this time the area of the square is fixed instead of the circle being fixed.) Then,

    \[ A = xy \quad \implies \quad y = \frac{A}{x}. \]

Furthermore, since we have a right triangle whose hypotenuse is the diameter of the circle (therefore is 2r) and whose legs are the sides of the square we have,

    \begin{align*} (2r)^2 = x^2 + y^2 && \implies && r &= \frac{1}{2} \sqrt{x^2+y^2}\\ && \implies && r &= \frac{1}{2} \sqrt{x^2 + \left( \frac{A}{x} \right)^2} \\ && \implies && r &= \frac{\sqrt{x^4 + A^2}}{2x}. \end{align*}

We then want to find the minimum value of this function (since this function gives us the radius of the circle). Calling the function f(x) and taking its derivative we have,

    \[ f'(x) = \frac{(2x)\left( \frac{1}{2} \right)(4x^3)(x^4 + A^2)^{-\frac{1}{2}} - (\sqrt{x^4 + A^2})(2)}{4x^2} = \frac{x^2}{\sqrt{x^4 + A^2}}\, - \, \frac{\sqrt{x^4 + A^2}}{2x^2}. \]

Then, setting this equal to zero,

    \begin{align*} f'(x) = 0 && \implies && \frac{x^2}{\sqrt{x^4 + A^2}} \, - \, \frac{\sqrt{x^4 + A^2}}{2x^2} &= 0 \\ && \implies && \frac{2x^4}{\sqrt{x^4 + A^2}} &= \sqrt{x^4 + A^2} \\ && \implies && 2x^4 &= x^4 + A^2 \\ && \implies && x^4 &= A^2 \\ && \implies && x = \sqrt{A}, \ y &= \sqrt{A}. \end{align*}

This critical point is a minimum since

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

Thus, the radius of the circle is minimized when x = y = \sqrt{A}, so the rectangle is a square. \qquad \blacksquare

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