Show that a right-circular cylinder of greatest volume that can be inscribed in a right-circular cone that has a volume that is 4/9 the volume of the cone.

Answer 1

Let the cone radius = R
cone height = H
Let cylinder height = h
then cylinder radius, r = R(1-h/H)

Volume of cylinder, V
=πr²h
=πR²(1-h/H)²h

For maximum volume,
dV/dh=0
π(1-h/H)^2R^2-(2πh(1-h/H)R^2)/H=0
which simplifies to:
(H^2-4hH+3h^2)=0
which when solved for h gives
h=H/3 or h=H
h=H will give a zero volume (min.) so reject.
for h=H/3,
Volume of cylinder
=πR²(1-(H/3)/H)²(H/3)
=πR²(2/3)²(H/3)
=(2/3)²(πR²H/3)
=4/9 volume of the cone.