Q.Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height $h$ and semi vertical angle $\alpha$ is one-third that of the cone and the greatest volume of cylinder is $\frac{4}{27}\pi h^3 \tan^2 \alpha$.

Concept: Optimization Word Problem — maximizing the volume of a cylinder inscribed in a cone.

Let the cone have height $h$ and base radius $R = h \tan \alpha$. Inscribe a cylinder of radius $r$ and height $H$. By similar triangles, the cylinder’s top touches the cone’s slant surface, so:

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

Let $x = H/h$. Then $V = \pi h^3 \tan^2\alpha \cdot (1 - x)^2 x$. Differentiate with respect to $x$:

$\frac{dV}{dx} = \pi h^3 \tan^2\alpha \left[ (1 - x)^2 - 2x(1 - x) \right] = \pi h^3 \tan^2\alpha (1 - x)(1 - 3x).$ …

We have a right circular cone of height $h$ and semi-vertical angle $\alpha$. That means the radius of the cone’s base is $R = h \tan \alpha$. Inside this cone, we inscribe a cylinder of radius $r$ and height $x$, with its axis along the cone’s axis. The cylinder touches the cone’s lateral surface all around — so the top face of the cylinder is a circle that just fits inside the cone at a certain height.

The key geometric insight: from a side view, the cone is a triangle, and the cylinder is a rectangle inscribed in it. The top corners of the rectangle lie on the sloping sides of the triangle. This gives a direct linear relation between $r$ and $x$ via similar triangles.

Let’s work it through.

1. Set up the geometry. Draw the cone with vertex at the top and base at the bottom. Place the vertex at the origin of a coordinate system for convenience. The cone’s axis is vertical. At a distance $y$ measured downward from the vertex, the radius of the cone’s cross-section is $y \tan \alpha$. The cylinder of height $x$ sits inside: its top face is at some distance from the vertex, and its bottom face rests on the cone’s base (or somewhere inside — but for maximum volume, the cylinder will touch the cone’s lateral surface along its entire height, so its top face is at a distance $d$ from the vertex, and its bottom face is at distance $d + x$). However, a cleaner approach: let the cylinder’s height be $x$, measured from the base upward. Then the distance from the vertex to the top of the cylinder is $h - x$. At that height, the cone’s radius is $(h - x) \tan \alpha$. This must equal the cylinder’s radius $r$, because the cylinder’s top edge touches the cone. So we have:

2. Write the volume of the cylinder.

   Volume $V = \pi r^2 x = \pi \big[(h - x) \tan \alpha\big]^2 x = \pi \tan^2 \alpha \cdot x (h - x)^2.$

3. Maximise $V$ with respect to $x$.

   Since $\pi \tan^2 \alpha$ is a positive constant, we maximise $f(x) = x (h - x)^2$ for $0 < x < h$.

   Differentiate:

Factor $(h - x)$:

4. Set $f'(x) = 0$. Either $h - x = 0$ (which gives $x = h$, a degenerate cylinder of zero radius) or $h - 3x = 0$, so $x = h/3$. The second derivative test or sign analysis confirms this gives a maximum: for $x < h/3$, $f'(x) > 0$; for $x > h/3$, $f'(x) < 0$. …

Method: Optimizing a Cylinder Inscribed in a Cone (Similar Triangles)

This method finds the maximum volume of a cylinder inscribed in a cone by using similar triangles — rather than a sphere's Pythagorean constraint — to link the cylinder's radius and height, then reducing to a single-variable calculus problem.

Steps

Step 1: Set up the cone's dimensions and the similar-triangle relation

For a cone of height $h$ and semi-vertical angle $\alpha$, the base radius is $R = h\tan\alpha$. Inscribe a cylinder of radius $r$ and height $x$ (measured from the cone's base) with its top rim touching the cone's slant surface. Measuring from the vertex, the cylinder's top is at a distance $h-x$, and the cone's radius there is $(h-x)\tan\alpha$ — this must equal the cylinder's radius:

This is the similar-triangles constraint — always sketch the side view (a triangle with a rectangle inside it) to read this relation off directly.

Step 2: Write the volume in one variable

Step 3: Differentiate and factor

Setting $V'(x)=0$ gives $x=h$ (degenerate, zero radius) or $x=\frac{h}{3}$.

Step 4: Confirm the maximum via a sign check …

Common Mistakes

Mistake 1: Measuring the cylinder's radius from the vertex instead of from the base

A student sets $r=x\tan\alpha$ (using $x$, the cylinder's own height measured from the base) instead of $r=(h-x)\tan\alpha$. Why it's wrong: the cone's radius at a given point depends on the distance from the apex, and the cylinder's top is at distance $h-x$ from the apex, not $x$ — using the wrong distance flips which factor decreases and changes the final answer. Correct approach: always sketch the side view and measure explicitly from the vertex down to the cylinder's top rim.

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