Maximum volume of cylinder in sphere

Referring to the figure above, the volume of the cylinder inscribed by in a sphere of radius $r$ is given by:

$\begin{aligned} V & = \pi x^2 \times 2 y & \small \color{#3D99F6} \text{Since }x = r\cos \theta,\ y = r\sin \theta \\ & = 2\pi r^3 \sin \theta \cos^2 \theta \\ & = 2\pi r^3 \sin \theta (1-\sin^2 \theta) \\ & = 2\pi r^3 (\sin \theta -\sin^3 \theta) \end{aligned}$

Since $V$ is maximum when $\dfrac {dV}{d\theta} = 0$ , we have:

$\begin{aligned} \frac {dV}{d\theta} & = 2\pi r^3 (\cos \theta - 3\sin^2 \theta \cos \theta) & \small \color{#3D99F6} \text{Equating }\dfrac {dV}{d\theta} = 0 \\ \cos \theta & = 3\sin^2 \theta \cos \theta & \small \color{#3D99F6} \text{Since} \cos \theta = 0 \text{ when }V \text{ is minimum.} \\ \implies \sin \theta & = \frac 1{\sqrt 3} \end{aligned}$

Since $\dfrac {d^2V}{dx^2} < 0$ , when $\sin \theta = \dfrac 1{\sqrt 3}$ , maximum $V$ is given by:

$\begin{aligned} V_{\text{max}} & = 2\pi r^3 \sin \theta \cos^2 \theta & \small \color{#3D99F6} \text{when }\sin \theta = \frac 1{\sqrt 3} \implies \cos \theta = \sqrt{\frac 23} \\ & = \frac {4\pi r^3}{3\sqrt 3} & \small \color{#3D99F6} \text{Putting }r=5 \\ & \approx \boxed{302.3} \end{aligned}$

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Image credit: Google

This is more of a calculus of variations problem than a simple geometry problem.

First we have to relate the radius of the circle, 5 centimetres, the radius of the cylinder, and the height of the cylinder. This can easily be done by Pythagoras’s Theorem.

25= $r^2$ + $\frac{h^2}{4}$

Where h is the height of the cylinder and r is the radius of the cylinder.

Rearranging this equation gives us

$h=√100-4r^2$

Using this value of h in the equation function of the volume of the sphere,

$V=πr^2√100-4r^2$

Now all that is left to do is finding the extreme value of this function. There can be two possible solutions to the maximum volume: a negative volume, the extraneous solution to the quadratic equation of the radius of the cylinder, and the positive real world volume.

By finding the derivative of the volume function, and at the maximum point the slope is zero, we can find the volume.

$\frac{d}{dr}$ ( $πr^2√100-4r^2$ ) = $\frac{12πr^3+200πr}{√4r^2+100}$

Simplifying this derivative when equating this to zero, you simply get $12πr^3+200πr=0$

Simplifying this further leaves us with a quadratic equation: $-2r^2+r+50=0$

The solutions to this equations are r is 4.1 or -4.1.

Therefore the volume, when the positive and real world value of r, 4.1 is plugged into the volume function, the volume is found to be to the nearest tenth $302.300$