Determine the volume of the largest cylinder that can be cut from a given log

A log is the shape of the frustum of a right circular cone with height 12 feet, base with diameter $(4+h)$ feet, and top diameter of 4 feet. Find the volume of the largest right circular cylinder that can be cut from the log if the axis of the cylinder coincides with the axis of the log.

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Let $r$ denote the radius of the right circular cylinder, and $a$ its height. Then, the volume is given by

$V = \pi r^2 a.$

First, we find an equation for $a$ in terms of $r$ . A graph depicting the edge of the frustum will help:

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So, in the graph, $a$ lies on the line connecting the origin to the top edge of the frustum. The equation of this line is $y = \frac{24}{h} x$ since it has slope $\frac{24}{h}$ (since the height of the frustum is 12 feet and the horizontal distance between the lower edge and upper edge is $\frac{h}{2}$ ). Therefore,

$a = \frac{24}{h} \left( 2 +\frac{h}{2} - r \right).$

Hence, we have the volume $V$ of the cylinder we are cutting out given by

$V = \pi r^2 a = \pi r^2 \left( \frac{48}{h} - \frac{24r}{h} + 12 \right).$

Since we want to maximize this volume, we take the derivative with respect to $r$ and look for critical points,

$\frac{dV}{dr} = \frac{96\pi r}{h} - \frac{72 \pi r^2}{h} + 24 \pi r.$

Setting this equal to 0 we have,

$\begin{align*} &&\frac{96 \pi r}{h} - \frac{72 \pi r^2}{h} + 24 \pi r &= 0 \\[9pt] \implies && 96 \pi r - 72 \pi r^2 + 24 \pi r h &= 0 \\[9pt] \implies && 4 - 3r + h &= 0 \\[9pt] \implies && r &= \frac{4+h}{3}. \end{align*}$

We also have an additional constraint from the problem which is that we should require $r \geq 2$ since the maximum height of the cylinder we are cutting is 12 feet and

$a = \frac{24}{h} \left( 2 +\frac{h}{2} - r \right) \leq 12 \quad \implies \quad r \geq 2.$

Since $r = \frac{4+h}{3}$ this implies that this value is valid for $h \geq 2$ . So if $h \geq 2$ we have

$a = \frac{24}{h} \left( 2 + \frac{h}{2} - r \right) = \frac{24}{h} \left( \frac{4+h}{6} \right).$

Plugging this into the formula for the volume of the cylinder we obtain a maximum for the volume of

$\begin{align*} V &= \pi r^2 a = \pi \left( \frac{(4+3)}{3} \right)^2 \left( \frac{24}{h} \right) \left( \frac{4+h}{6} \right) \\[9pt] &= \frac{4 \pi}{9h} (4+h)^3. \end{align*}$