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Find the cylinder of largest lateral surface area that can be inscribed in a given sphere

by RoRi

Consider a sphere with given radius R. Find the values for the radius and altitude (r and h, respectively) of the right circular cylinder of maximum lateral surface area (given by 2 \pi r h) that can be inscribed in the sphere.

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First, we want to write h as a function of r. From the diagram, we see that we have a right triangle with hypotenuse of length R and legs of lengths r and \frac{h}{2} (since the second leg only goes to the center of the sphere, not the full length of the cylinder). Therefore,

    \[ R^2 = r^2 + \frac{h^2}{4} \quad \implies \quad h = 2 \sqrt{R^2 - r^2}. \]

Then, we are given that the lateral surface area, A, is given by the formula

    \[ A = 2 \pi r h = 2 \pi r (2 \sqrt{R^2 -r^2}) = 4 \pi r \sqrt{R^2 - r^2}. \]

Calling this function f(r) and differentiating we find,

    \[ f'(r) = 4 \pi \sqrt{R^2 - r^2} - \frac{4 \pi r^2}{\sqrt{R^2 - r^2}}. \]

Setting this equal to 0 we have,

    \begin{align*} f'(r) = 0 && \implies && \frac{4 \pi r^2}{\sqrt{R^2 - r^2}} &= 4 \pi \sqrt{R^2 - r^2} \\ && \implies && r^2 &= \left( \sqrt{R^2 - r^2} \right)^2 \\ && \implies && 2r^2 &= R^2 \\ && \implies && r &= \frac{R}{\sqrt{2}} \\ && \implies && \frac{h}{2} &= \sqrt{R^2 - \frac{R^2}{2}} = \frac{R}{\sqrt{2}}. \end{align*}

Since

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

this point is a minimum. Hence, the lateral surface area is minimal when

    \[ r = \frac{h}{2} = \frac{R}{\sqrt{2}}. \]

One comment

  1. Artem says:

    We need to find the cylinder with the MAXIMUM lateral surface. Your derivative sign is incorrect for the intervals to the left and to the right of the critical point.

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