Problem 62

Inscribe a circular cylinder of maximum convex surface area in a given circular cone.

Solution:
By similar triangle:
\dfrac{H - h}{d} = \dfrac{H}{D}

h = H - \dfrac{Hd}{D}

Convex surface area of the cylinder:
A = \pi d \, h

A = \pi d \left( H - \dfrac{Hd}{D} \right)

A = \pi H d - \dfrac{\pi H}{D}d^2

The cone is given, thus H and D are constant
\dfrac{dA}{dd} = \pi H - \dfrac{2\pi H}{D}d = 0

\pi H = \dfrac{2\pi H}{D}d

d = \frac{1}{2} D

diameter of cylinder = radius of cone            answer

Problem 63

Find the circular cone of maximum volume inscribed in a sphere of radius a.

Solution:
Volume of the cone:
V = \frac{1}{3} \pi r^2 h

From the figure:
r^2 = a^2 – (h – a)^2
r^2 = a^2 – (h^2 – 2ah + a^2)
r^2 = 2ah – h^2

V = \frac{1}{3} \pi \, ( \, 2ah – h^2 \,) \, h
V = \frac{1}{3} \pi \, ( \, 2ah^2 – h^3 \,)

The sphere is given, thus radius a is constant.
\dfrac{dV}{dh} = \frac{1}{3} \pi \, ( \, 4ah – 3h^2 \,) = 0

4ah = 3h^2
h = \frac{4}{3} a

altitude of cone = 4/3 of radius of sphere            answer