Discussion
The height of a right circular cone of maximum volume that can be enclosed within a hollow sphere of radius \(R\) is
Answer: C
The height is \(\frac{4}{3}R\).
Put the cone's apex on the sphere and its base circle on the sphere too. If the height is h, the base lies a distance h - R from the centre, so the base radius r satisfies \(r^{2} = R^{2} - (h - R)^{2} = 2Rh - h^{2}\).
\[ \begin{aligned} V &= \frac{\pi}{3}(2Rh^{2} - h^{3}) \\ \frac{dV}{dh} &= \frac{\pi}{3}(4Rh - 3h^{2}) = 0 \\ h &= \frac{4R}{3} \end{aligned} \]
The second derivative, \(4R - 6h\), is negative there, so this is a maximum. h = R puts the base through the centre (a cone of volume \(\pi R^{3}/3\), smaller than the optimum \(32\pi R^{3}/81\)), and 3R/2 and 5R/4 are other heights with smaller volume.
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