A right circular cylinder is inscribed in a right circular cone of ba…

The correct answer is Option (2) → \(48\pi\) cm³

Let the cylinder have radius r and height h. Its top rim touches the slant side of the cone. By similar triangles, the cone's radius at height h is r:

    \( \displaystyle \frac{r}{6} = \frac{9 - h}{9} \)

    \( \displaystyle h = 9 - \frac{3r}{2} \)

The volume of the cylinder, for 0 < r < 6, is

    \( \displaystyle V = \pi r^{2}\left(9 - \frac{3r}{2}\right) = \pi\left(9r^{2} - \frac{3}{2}r^{3}\right) \)

Setting the derivative equal to zero:

    \( \displaystyle V' = \pi\left(18r - \frac{9}{2}r^{2}\right) = \frac{9\pi}{2}\,r(4 - r) = 0 \)

    \( r = 4 \)

Since \(V'' = \pi(18 - 9r) = -18\pi \lt 0\) at r = 4, this is the maximum. Then h = 9 − 6 = 3, and

    \( V = \pi \times 16 \times 3 = 48\pi \) cm³

Why the other options are incorrect

• Option (1) \(36\pi\) cm³ — Incorrect. This is less than the volume \(48\pi\) obtained at r = 4, so it is not the maximum.

• Option (3) \(54\pi\) cm³ — Incorrect. The largest possible volume is \(48\pi\), so \(54\pi\) cannot be reached.

• Option (4) \(108\pi\) cm³ — Incorrect. This is the volume \(\frac{1}{3}\pi \times 36 \times 9\) of the cone itself.