Problem 62

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

Solution:
By similar triangle:
[math]\dfrac{H - h}{d} = \dfrac{H}{D}[/math]

[math]h = H - \dfrac{Hd}{D}[/math]

Convex surface area of the cylinder:
[math]A = \pi d \, h[/math]

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

[math]A = \pi H d - \dfrac{\pi H}{D}d^2[/math]

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

[math]\pi H = \dfrac{2\pi H}{D}d[/math]

[math]d = \frac{1}{2} D[/math]

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:
[math]V = \frac{1}{3} \pi r^2 h[/math]

From the figure:
[math]r^2 = a^2 – (h – a)^2[/math]
[math]r^2 = a^2 – (h^2 – 2ah + a^2)[/math]
[math]r^2 = 2ah – h^2[/math]

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

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

[math]4ah = 3h^2[/math]
[math]h = \frac{4}{3} a[/math]

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