When is the volume of this cylinder inside a cone a maximum?

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The volume of a cylinder is given by (area of the base)\(\times\)(height).

Since the radius of the cylinder is \(s\), the area of its base is \(\pi s^2\).

We now need to find the height of the cylinder which we shall call \(a\). For this, consider the following cross section of the diagram above.

From the two similar triangles at the left hand side we can see that \[\begin{align*} \frac{h}{r}&=\frac{a}{2r-s} \\ \iff\quad a&=\frac{h}{r}(2r-s) \end{align*}\]

Thus the volume of the cylinder is \[\frac{\pi s^2 \times h(2r - s)}{r}\] as required.

We have \[\begin{align*} V &= \frac{\pi s^2 h (2r - s)}{r} \\ &= (2 \pi h)s^2 - \frac{\pi h}{r}s^3 \\ \frac{dV}{ds} &= (2 \pi h) 2s - \frac{\pi h}{r}3s^2\\ &= \pi h s \left(4-\frac{3s}{r}\right). \end{align*}\]

Putting \(\dfrac{dV}{ds} = 0\), we have \[s = 0 \quad \text{ or } \quad 4-\frac{3s}{r} = 0.\]

The second case is clearly the one we require, giving \(s = \dfrac{4}{3}r\). This yields \[ V_{max} = \frac{32\pi r^2 h}{27}. \]