Problem 64

A sphere is cut to the shape of a circular cone. How much of the material can be saved? (See Problem 63)

Solution:
Volume of sphere or radius a:
[math]V_s = \frac{4}{3}\pi a^3[/math]

Volume of cone of radius r and altitude h:
[math]V_c = \frac{1}{3}\pi r^2 \, h[/math]

From the solution of Problem 63:
[math]h = \frac{4}{3}a[/math]

[math]r^2 = 2ah – h^2[/math]
[math]r^2 = 2a(\frac{4}{3}a) – (\frac{4}{3}a)^2[/math]
[math]r^2 = \frac{8}{9} a^2[/math]

[math]V_c = \frac{1}{3}\pi (\frac{8}{9} a^2)(\frac{4}{3}a)[/math]
[math]V_c = \frac{32}{81}\pi a^3[/math]

[math]\text{Material saved:} = \dfrac{V_c}{V_s} \times 100\%[/math]

[math]\text{Material saved:} = \dfrac{\frac{32}{81}\pi a^3}{\frac{4}{3}\pi a^3} \times 100\%[/math]

[math]\text{Material saved:} = 29.63\% \, \text{ (about }30\%\text{)}\,\,[/math]            answer

Problem 65

Find the circular cone of minimum volume circumscribed about a sphere of radius a.

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

By similar triangle:
[math]\sin \theta = \dfrac{a}{h - a} = \dfrac{r}{\sqrt{h^2 + r^2}}[/math]

[math]a\sqrt{h^2 + r^2} = r(h - a)[/math]
[math]a^2 (h^2 + r^2) = r^2 (h – a)^2[/math]
[math]a^2 h^2 + a^2 r^2 = h^2 r^2 – 2ahr^2 + a^2 r^2[/math]
[math]a^2 h^2 = (h – 2a)hr^2[/math]

[math]r^2 = \dfrac{a^2 h}{h - 2a}[/math]

Thus,
[math]V = \dfrac{\pi}{3} \left( \dfrac{a^2 h}{h - 2a} \right)h[/math]

[math]V = \dfrac{\pi a^2}{3} \left( \dfrac{h^2}{h - 2a} \right)[/math]

[math]\dfrac{dV}{dh} = \dfrac{\pi a^2}{3} \left[ \dfrac{(h - 2a)(2h) - h^2 (1)}{(h - 2a)^2} \right] = 0[/math]

[math]2(h – 2a)h - h^2 = 0[/math]
[math]2h – 4a - h = 0[/math]
[math]h = 4a[/math]

altitude of the cone = 4 × the radius of the sphere, a            answer

Another Solution:
For a circle inscribed in a triangle, its center is at the point of intersection of the angular bisector of the triangle called the incenter (see figure).

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

From the figure:
[math]\tan \theta = a/r[/math]

[math]\tan 2\theta = h/r[/math]

[math]\dfrac{2\tan \theta}{1 - \tan^2 \theta} = \dfrac{h}{r}[/math]

[math]\dfrac{2(a/r)}{1 - a^2/r^2} = \dfrac{h}{r}[/math]

[math]\dfrac{2a)}{\dfrac{r^2 - a^2}{r^2}} = h[/math]

[math]h = \dfrac{2ar^2)}{r^2 - a^2}[/math]

Thus,
[math]V = \frac{1}{3}\pi r^2 \left( \dfrac{2ar^2)}{r^2 - a^2} \right)[/math]

[math]V = \frac{2}{3}a \pi \left( \dfrac{r^4)}{r^2 - a^2} \right)[/math]

[math]\dfrac{dV}{dr} = \frac{2}{3}a \pi \left[ \dfrac{(r^2 - a^2)(4r^3) - r^4(2r)}{(r^2 - a^2)^2} \right] = 0[/math]

[math]4r^3(r^2 – a^2) – 2r^5 = 0[/math]
[math]2(r^2 – a^2) – r^2 = 0[/math]
[math]2r^2 – 2a^2 - r^2 = 0[/math]
[math]r^2 = 2a^2[/math]

[math]h = \dfrac{2ar^2}{r^2 - a^2}[/math]

[math]h = \dfrac{2a(2a^2)}{2a^2 - a^2}[/math]

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

[math]h = 4a \,\, [/math]            (ok!)