A right circular cone is inscribed in a sphere of radius 9 cm, with its vertex and the whole circle of its base on the sphere. The maximum possible volume of the cone is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(288\pi\) cm3
Let h be the height of the cone and r its base radius. Take the centre of the sphere at the origin and the vertex at (0, 9) on the axis. The base circle then lies at height 9 − h, so by Pythagoras
\( r^{2} + (9 - h)^{2} = 81 \)
\( r^{2} = 18h - h^{2} \)
So the volume, as a function of h with \(0 \lt h \lt 18\), is
\( \displaystyle V = \frac{1}{3}\pi r^{2}h = \frac{\pi}{3}(18h^{2} - h^{3}) \)
Setting the derivative equal to zero:
\( \displaystyle V' = \frac{\pi}{3}(36h - 3h^{2}) = \pi h(12 - h) = 0,\ h = 12 \)
Since \(V'' = \pi(12 - 2h) = -12\pi \lt 0\) at h = 12, this is the maximum. Then \(r^{2} = 216 - 144 = 72\) and
\( \displaystyle V = \frac{1}{3}\pi \times 72 \times 12 = 288\pi \) cm3
Why the other options are incorrect
Option (1) \(324\pi\) cm3 — Incorrect. The largest value of \(\frac{\pi}{3}(18h^{2} - h^{3})\) for 0 < h < 18 is \(288\pi\), at h = 12.
Option (2) \(864\pi\) cm3 — Incorrect. This is \(\pi r^{2}h = \pi \times 72 \times 12\): the factor \(\frac{1}{3}\) of the cone formula is missing.
Option (3) \(243\pi\) cm3 — Incorrect. This is the cone with height 9 cm and base radius 9 cm (base through the centre): \(\frac{1}{3}\pi \times 81 \times 9\), which is smaller.