A right circular cylinder is placed inside a right circular cone of base radius 6 cm and height 12 cm so that the cylinder's base lies on the cone's…

Let the cylinder have radius r and height h. By similar triangles, at height h the cone's radius is 6(1 − h/12), and this must equal r: r = 6 − h/2 ⇒ h = 12 − 2r.

V = πr²h = πr²(12 − 2r) = π(12r² − 2r³).

dV/dr = π(24r − 6r²) = 0 ⇒ r = 4 (the other root r = 0 is the minimum). Then h = 12 − 8 = 4.

V = π × 16 × 4 = 64π cm³.

Check nearby: r = 3 → 9 × 6π = 54π; r = 5 → 25 × 2π = 50π – both smaller ✓.

Tip: The optimal inscribed cylinder in a cone always has radius = 2/3 of the cone's radius and height = 1/3 of the cone's height.