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Find the smallest circumscribed sphere about cylinders of a fixed lateral surface area

by RoRi

Consider all right circular cylinders of a fixed lateral surface area (the lateral surface area of a right circular cylinder is given by the formula 2 \pi r h where r is the radius of the base and h is the altitude of the cylinder). Prove that the smallest sphere that can be circumscribed about any of them is the sphere with radius R = \sqrt{2} r, where r is the radius of the cylinder.

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Proof. The lateral surface area A = 2 \pi r h is a constant, and from the picture we see

    \[ R^2 = r^2 + \frac{h^2}{4}. \]

So, we have the following,

    \[ A = 2 \pi r h \quad \implies \quad h = \frac{A}{2 \pi r}. \]

Therefore,

    \begin{align*} R^2 &= r^2 + \frac{h^2}{4} = r^2 + \frac{A^2}{16 \pi^2 r^2} \\ \implies \quad R &= \sqrt{r^2 + \frac{A^2}{16 \pi^2 r^2}}. \end{align*}

Calling this function f(r) we take the derivative,

    \[ f'(r) = \frac{1}{2} \left( r^2 + \frac{A^2}{16 \pi^2 r^2} \right)^{-\frac{1}{2}} \left( 2r - \frac{2A^2}{16 \pi^2 r^3} \right). \]

Then, setting f'(r) = 0 we have,

    \begin{align*} &&\frac{1}{2} \left( r^2 + \frac{A^2}{16 \pi^2 r^2} \right)^{-\frac{1}{2}} \left( 2r - \frac{2A^2}{16 \pi^2 r^3} \right) &= 0 \\ \implies && 2r &= \frac{A^2}{8 \pi^2 r^3} \\ \implies && r^4 &= \frac{A^2}{16 \pi^2} \\ \implies && r &= \frac{1}{2} \sqrt{ \frac{A}{\pi}}. \end{align*}

Since

    \begin{align*} f'(r) &< 0 & \text{when} && r &< \frac{1}{2} \sqrt{ \frac{A}{\pi}} \\ f'(r) &> 0 & \text{when} && r &> \frac{1}{2} \sqrt{ \frac{A}{\pi}} \end{align*}

we have that f(r) has a minimum r = \frac{1}{2} \sqrt{\frac{A}{\pi}}.
Therefore, the value of this minimal radius R of the sphere at r = \frac{1}{2} \sqrt{\frac{A}{\pi}} is given by

    \[ R = \sqrt{r^2 + \frac{A^2}{16 \pi^2 r^2}} = \sqrt{ \frac{2A}{2 \pi}} = \sqrt{2} r. \qquad \blacksquare \]

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