Q.Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\frac{4r}{3}$.

Let the cone have altitude $h$, base radius $R$, inscribed in a sphere of radius $r$.

From the sphere's geometry: $R^2+(h-r)^2=r^2 \implies R^2=2hr-h^2$.

Volume: $V(h)=\dfrac13\pi R^2h=\dfrac{\pi}{3}(2rh^2-h^3)$.

$\frac{dV}{dh}=\frac{\pi}{3}h(4r-3h)=0 \implies h=\frac{4r}{3}\ \text{(rejecting }h=0\text{)}$ …

Setting up the geometry

Let the sphere have fixed radius $r$, and let the inscribed cone have altitude $h$ and base radius $R$. Place the sphere's centre at $O$ and the cone's axis along a diameter, with the apex of the cone on the sphere.

If $C$ is the centre of the cone's circular base, then $OC=h-r$ (the base is a distance $h-r$ from the sphere's centre, measured along the axis — signed so this works whether $h$ is less than or greater than $r$). Since the rim of the base lies on the sphere, the right triangle with legs $R$ and $|h-r|$ and hypotenuse $r$ gives:

Expanding:

Writing volume as a function of $h$ alone

Differentiating

Set $\dfrac{dV}{dh}=0$: …

Method: Optimizing a Solid Inscribed in a Sphere

This method finds the maximum (or minimum) volume of a solid — a cone here — whose vertex and base circle both touch a sphere of fixed radius, by turning two unknowns into one using the sphere's own geometry.

Steps

Step 1: Draw the axial cross-section and write the sphere constraint

Slice the sphere and the inscribed cone through the common axis. If the cone has base radius $R$ and height $h$, and the sphere has radius $r$, placing the sphere's centre at the origin gives the rim of the cone's base lying on the sphere:

This single equation is what lets you eliminate one variable — always look for the right-triangle or circle relation the picture gives you before touching calculus.

Step 2: Write the volume in one variable

Substitute the constraint into the volume formula $V = \frac13\pi R^2 h$ so that $V$ depends only on $h$:

Step 3: Differentiate and solve for critical points

Setting this to zero gives $h=0$ (a degenerate cone with zero volume — always discard it) or $h = \frac{4r}{3}$.

Step 4: Confirm it is a maximum …

Common Mistakes

Mistake 1: Writing the sphere constraint with the wrong sign or missing the shift

A student often writes $R^2 + h^2 = r^2$, forgetting that the base circle's plane is not through the sphere's centre — it is offset by $(h-r)$ from the centre. Why it's wrong: this drops the geometry that actually makes the cone's apex and base both touch the sphere. Correct approach: draw the axial cross-section explicitly and use $R^2 + (h-r)^2 = r^2$.

Mistake 2: Reporting $h=0$ as a valid critical point

Factoring $\frac{dV}{dh} = \frac13\pi h(4r-3h) = 0$ gives two roots, $h=0$ and $h=\frac{4r}{3}$. Why it's wrong: $h=0$ makes the cone degenerate (zero volume), so it can never be the maximum — reporting both roots without discarding this one is an incomplete answer. Correct approach: always interpret each critical point physically before keeping it. …