Maximum Volume of Cylinder Inside Cone

This question was seen in the GCSE 1985 paper as well as more recently the New South Wales HSC 2015 Mathematics paper. That kinda proves the recycling of questions over time and across different education systems. This is a question I like due to the purely algebraic nature of calculus used.

Question

The diagram shows a cylinder of radius \( x \) and height \( y \) inscribed in a cone of radius \( R \) and height \( H \), where \( R \) and \( H \) are constants.

  1. Show that the volume, \( V\) of the cylinder can be written as:

    \[ V = \frac{H}{R}\pi x^2(R-x)\]

  2. By considering the inscribed cylinder of maximum volume, show that the volume of any inscribed cylinder does not exceed \( \frac{4}{9}\) of the volume of the cone.

Solution

Part 1 Solution

1.

By looking at the triangle with sides \( H\) and \(R\), and \(y\) and \(R-x\), we can easily see that they are similar as they are equiangular.

Corresponding sides in similar triangles are in the same ratio, hence:

\[ \frac{H}{R} = \frac{y}{R-x}\]

Rearranging this:

\[y = \frac{H}{R}(R-x)\]

Now the volume of a cylinder is \( V = \pi r^2 h\) and the radius of the cylinder in the question is \(x\), the height is \(y\):

\[\begin{align*}V &= \pi x^2 y \\ V & = \pi x^2 (\frac{H}{R}(R-x))\\ V & = \frac{H}{R}\pi x^2 (R-x) \end{align*}\]

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Part 2 Solution

2.

I don’t want to use product rule as that can potentially make a mess of things, so it’s best to expand the expression for the volume of the cylinder:

\[\begin{align*}V &= H\pi x^2 – \frac{H}{R}\pi x^3\\  \frac{dV}{dx} &= 2H\pi x – \frac{3H\pi}{R} x^2\\ \frac{d^2V}{dx^2}& = 2H\pi – \frac{6H\pi }{R}x\end{align*}\]

For maximum volume of the cylinder, \(\frac{dV}{dx} = 0\), which leads to:

\[\begin{align*}2H\pi x – \frac{3H\pi}{R} x^2 & =0\\2x – \frac{3x^2}{R} & = 0\\2-\frac{3x}{R} &= 0\\x &= \frac{2R}{3} \end{align*}\]

In the second last line above, we can divide by \(x\) because \(x=0\) is not a possible radius of the cylinder.

Substitute \( x = \frac{2R}{3}\) into the second derivative to confirm that it is indeed a maximum:

\[\begin{align*}\frac{d^2V}{dx^2} & = 2H\pi-\frac{6H\pi}{ R}\cdot\frac{2 R}{3}\\ & = 2H\pi – 4H \pi\\ & = 2H\pi \\ &< 0\end{align*}\]

This implies that the graph of \(V\) at \(x=\frac{2R}{3}\) is concave down, and is hence a local maximum. Since it is also the only stationary point in the domain, it is also the absolute maximum.

Therefore:

\[\begin{align*}\mbox{Max }V & = H\pi \left(\frac{2R}{3}\right)^2-\frac{H}{R}\pi\left(\frac{2R}{3}\right)^3\\&=H\pi\frac{4R^2}{9}-\frac{H}{R}\pi \cdot \frac{8R^3}{27}\\& =H\pi\frac{4R^2}{9}-H\pi\frac{8R^2}{27}\end{align*}\]

Now we factorise the volume of the cone out of this expression, i.e. \(\frac{1}{3}\pi R^2H\):

\[\begin{align*}&=\frac{1}{3}\pi R^2 H \left( \frac{4}{3} – \frac{8}{9}\right)\\ &=\frac{1}{3}\pi R^2 H \left( \frac{12}{9} – \frac{8}{9}\right)\\ & = \frac{1}{3}\pi R^2 H \left(\frac{4}{9}\right)\end{align*}\]

Since the maximum volume of the cylinder is \(\frac{4}{9}\) of the volume of the cone, then the volume of any inscribed cylinder does not exceed that amount.

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