Q.Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is $\frac{2R}{\sqrt{3}}$. Also find the maximum volume.

Idea: Use the sphere to write the cylinder's volume in one variable (its height), then maximise with a single derivative.

Let the inscribed cylinder have radius $r$ and height $h$. Its rim lies on the sphere, so from the cross-section

Volume:

Differentiate and set to zero:

Since $V''(h)=\pi\left(-\tfrac{3h}{2}\right)<0$, this is a maximum. …

The geometry

A right circular cylinder is inscribed in a sphere of radius $R$, with its axis through the centre. Let its radius be $r$ and height $h$. Take the plane cross-section through the axis: the sphere becomes a circle of radius $R$, and the cylinder becomes a rectangle of width $2r$ and height $h$ inscribed in it. From the centre to a top corner, the Pythagorean theorem gives

This is the single constraint linking $r$ and $h$.

Reduce to one variable

The quantity to maximise is

From $(1)$, $r^2=R^2-\dfrac{h^2}{4}$. Substitute:

Maximise

Differentiate with respect to $h$:

Set $V'(h)=0$:

(We take the positive root.) The second derivative

confirms a maximum. (It also makes sense: $V\to0$ as $h\to0$ or $h\to2R$, so the single interior critical point is the peak.)

Radius and maximum volume …

Method: Optimizing a Cylinder Inscribed in a Sphere

This method finds the maximum volume of a cylinder whose top and bottom circular rims both touch a sphere of fixed radius, using the same "constraint from the cross-section, reduce to one variable" pattern as any solid-in-a-sphere optimisation.

Steps

Step 1: Draw the axial cross-section and find the constraint

Slicing through the axis turns the sphere into a circle of radius $R$ and the cylinder into an inscribed rectangle of width $2r$ and height $h$. The diagonal from the centre to a top corner has length $R$, so by the Pythagorean theorem:

Note the $h/2$, not $h$ — it is half the height that forms the right triangle with $r$ and $R$, because the cylinder is centred on the sphere's centre.

Step 2: Write the volume in one variable

From the constraint, $r^2 = R^2 - \frac{h^2}{4}$. Substitute into $V=\pi r^2 h$:

Step 3: Differentiate and solve $V'(h)=0$

$V'(h) = \pi\left(R^2 - \frac{3h^2}{4}\right) = 0 \quad\Longrightarrow\quad h = \frac{2R}{\sqrt3}\ \text{(taking the positive root)}.$ …

Common Mistakes

Mistake 1: Writing the constraint as $r^2+h^2=R^2$, dropping the factor of $\tfrac12$

Why it's wrong: the right triangle in the cross-section is formed by $r$, half the height $h/2$, and the sphere's radius $R$ — using the full height overstates the constraint and produces a wrong critical value. Correct approach: always redraw the cross-section and identify exactly which segment forms the triangle before writing the equation.

Mistake 2: Stopping at $h=\frac{2R}{\sqrt3}$ when the question also asks for the maximum volume …