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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*}

If 0 \leq h < 2 then we have r = 2 and so a = \frac{24}{h} \cdot \frac{h}{2} = 12. Therefore,

    \[ V = \pi r^2 a = 48 \pi. \]

Putting this together into one equation we have

    \[ V = \begin{dcases} 48 \pi & \text{if } 0 \leq h < 2 \\ \frac{4 \pi}{9h} (4+h)^3 &\text{if } h \geq 2. \end{dcases} \]

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