Posted by **ali** on Saturday, July 12, 2014 at 12:08am.

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.

- mathe -
**MathMate**, Saturday, July 12, 2014 at 6:21am
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.

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