Q.Show that the height of the right circular cylinder of maximum volume inscribed in a right circular cone of radius $r$ and height $h$ is one-third of the height of the cone and the maximum volume of the cylinder is $\frac{4}{9}$ of the volume of the cone.

Concept: Lagrange Multipliers (or single-variable optimisation using similar triangles).

Let the cone have radius $r$ and height $h$. Inscribe a cylinder of radius $x$ and height $y$. By similar triangles,

$\frac{x}{r} = \frac{h - y}{h} \quad\Rightarrow\quad x = r\left(1 - \frac{y}{h}\right)$.

Volume of cylinder: $V = \pi x^2 y = \pi r^2 \left(1 - \frac{y}{h}\right)^2 y$.

Maximise $V(y)$ for $0 < y < h$. Differentiate:

$\frac{dV}{dy} = \pi r^2 \left[ \left(1 - \frac{y}{h}\right)^2 - \frac{2y}{h}\left(1 - \frac{y}{h}\right) \right] = 0$.

Factor: $\left(1 - \frac{y}{h}\right)\left(1 - \frac{y}{h} - \frac{2y}{h}\right) = 0$.

The non-zero solution: $1 - \frac{3y}{h} = 0 \quad\Rightarrow\quad y = \frac{h}{3}$.

Then $x = r\left(1 - \frac{1}{3}\right) = \frac{2r}{3}$, so …

We have a right circular cone of base radius $r$ and height $h$. Inside it we inscribe a right circular cylinder of radius $x$ and height $y$, with its axis along the cone’s axis. The cylinder touches the cone’s lateral surface — that is, the top rim of the cylinder lies on the cone’s sloping side. This gives a geometric relation between $x$ and $y$.

The problem is a constrained optimisation: maximise the cylinder’s volume $V = \pi x^2 y$ subject to the condition that the cylinder fits exactly inside the cone. That condition is linear: from similar triangles, the height from the apex down to the top of the cylinder is proportional to the radius at that level.

1. Set up the constraint

Take a vertical cross-section through the axis of the cone. The cone appears as an isosceles triangle of height $h$ and base width $2r$. The cylinder appears as a rectangle of width $2x$ and height $y$, sitting on the base.

From similar triangles: the distance from the apex to the top of the cylinder is $h - y$, and the radius at that height is $x$. The full height $h$ corresponds to the full radius $r$, so

Rearranging:

This is the constraint. It is linear in $x$ and $y$, and it tells us that as $x$ increases, $y$ decreases — a trade-off.

2. Express volume in one variable

Substitute the constraint into $V = \pi x^2 y$:

Now it’s a single-variable optimisation problem on $0 < x < r$ (the cylinder radius must be positive and less than the cone’s radius).

3. Differentiate and find critical point

Differentiate $V(x)$ with respect to $x$:

Set $\frac{dV}{dx} = 0$. Since $x > 0$, we have

This is the only critical point in the interior of the domain.

4. Find the corresponding height

From the constraint:

So the cylinder’s height is exactly one-third of the cone’s height.

5. Verify it’s a maximum

Check the second derivative or simply note that $V(x)$ is a cubic with a positive coefficient on $x^2$ and negative on $x^3$, so it rises then falls. The single critical point must be a maximum. Alternatively:

At $x = \frac{2r}{3}$, $\frac{d^2V}{dx^2} = \pi h (2 - 4) = -2\pi h < 0$, confirming a maximum.

6. Compute the maximum volume …