A cylinder is inscribed in a sphere of radius R — find max volume of cylinder

Question

A cylinder of maximum volume is inscribed in a sphere of radius $R$ . Find the maximum volume of the cylinder.

Solution — Step by Step

Let the cylinder have radius $r$ and height $h$ . The cylinder is inscribed in the sphere of radius $R$ , so all points of the cylinder (including the top and bottom circular edges) lie on the sphere.

The relationship between $r$ , $h$ , and $R$ : if we draw a cross-section through the centre, the diagonal from the centre of the sphere to a top corner of the cylinder is the sphere’s radius $R$ .

Using Pythagoras (with half-height as the vertical component):

r^2 + \left(\frac{h}{2}\right)^2 = R^2

Therefore: $r^2 = R^2 - \dfrac{h^2}{4}$

Volume of cylinder:

V = \pi r^2 h = \pi \left(R^2 - \frac{h^2}{4}\right) h = \pi\left(R^2 h - \frac{h^3}{4}\right)

Now $V$ is a function of $h$ alone.

\frac{dV}{dh} = \pi\left(R^2 - \frac{3h^2}{4}\right)

Set $\dfrac{dV}{dh} = 0$ :

R^2 - \frac{3h^2}{4} = 0

h^2 = \frac{4R^2}{3} \implies h = \frac{2R}{\sqrt{3}}

At $h = \dfrac{2R}{\sqrt{3}}$ :

r^2 = R^2 - \frac{h^2}{4} = R^2 - \frac{4R^2/3}{4} = R^2 - \frac{R^2}{3} = \frac{2R^2}{3}

V_{\max} = \pi r^2 h = \pi \cdot \frac{2R^2}{3} \cdot \frac{2R}{\sqrt{3}} = \frac{4\pi R^3}{3\sqrt{3}} = \frac{4\pi R^3\sqrt{3}}{9}

\boxed{V_{\max} = \frac{4\pi R^3}{3\sqrt{3}} = \frac{4\sqrt{3}\pi R^3}{9}}

Verify it’s a maximum: $\dfrac{d^2V}{dh^2} = -\dfrac{3\pi h}{2} < 0$ ✓ (negative, so maximum)

Why This Works

The cylinder’s volume depends on both $r$ and $h$ , but the sphere constraint ties them together: making the cylinder taller reduces its radius (since it must stay inside the sphere), and vice versa. There’s an optimal trade-off.

By expressing $r^2$ in terms of $h$ using the sphere constraint, we reduced a two-variable optimisation to a single-variable problem. Then standard calculus (first derivative = 0, second derivative negative) finds the maximum.

The optimal height $h = 2R/\sqrt{3}$ is about 1.15R — the cylinder is actually shorter than the sphere’s diameter (which is 2R). This is because making the cylinder taller would require a much narrower radius, reducing volume more than the height gain adds.

Alternative Method — Substitution with Angle

Let $h/2 = R\sin\theta$ where $\theta$ is the angle from the equator. Then $r = R\cos\theta$ .

$V = \pi r^2 h = \pi R^2\cos^2\theta \cdot 2R\sin\theta = 2\pi R^3 \cos^2\theta \sin\theta$

Differentiate with respect to $\theta$ and set to zero: this gives $\tan\theta = 1/\sqrt{2}$ , which yields the same result.

Common Mistake

Setting up the Pythagorean constraint incorrectly. The sphere radius connects the centre to a top rim of the cylinder. The centre of the sphere is at height 0 (centred), so the vertical distance to the top of the cylinder is $h/2$ , not $h$ . Writing $r^2 + h^2 = R^2$ (using full height $h$ instead of $h/2$ ) gives the wrong answer.

Question

Solution — Step by Step

Why This Works

Alternative Method — Substitution with Angle

Common Mistake

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