**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:**

$V_s = \frac{4}{3}\pi a^3$

Volume of cone of radius r and altitude h:

$V_c = \frac{1}{3}\pi r^2 \, h$

$h = \frac{4}{3}a$

$r^2 = 2ah - h^2$

$r^2 = 2a(\frac{4}{3}a) - (\frac{4}{3}a)^2$

$r^2 = \frac{8}{9} a^2$

Thus,

$V_c = \frac{1}{3}\pi (\frac{8}{9} a^2)(\frac{4}{3}a)$

$V_c = \frac{32}{81}\pi a^3$

$\text{Material saved:} = \dfrac{V_c}{V_s} \times 100\%$

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

$\text{Material saved:} = 29.63\% \, \text{ (about }30\%\text{)}$ *answer*

**Problem 65**

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

**Solution:**

$V = \frac{1}{3}\pi r^2 h$

By similar triangle:

$\sin \theta = \dfrac{a}{h - a} = \dfrac{r}{\sqrt{h^2 + r^2}}$

$a\sqrt{h^2 + r^2} = r(h - a)$

$a^2 (h^2 + r^2) = r^2 (h - a)^2$

$a^2 h^2 + a^2 r^2 = h^2 r^2 - 2ahr^2 + a^2 r^2$

$a^2 h^2 = (h - 2a)hr^2$

$r^2 = \dfrac{a^2 h}{h - 2a}$

Thus,

$V = \dfrac{\pi}{3} \left( \dfrac{a^2 h}{h - 2a} \right)h$

$V = \dfrac{\pi a^2}{3} \left( \dfrac{h^2}{h - 2a} \right)$

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

$2(h - 2a)h - h^2 = 0$

$2h - 4a - h = 0$

$h = 4a$

Altitude of the cone = 4 × the radius of the sphere, a *answer*

**Another Solution:**

For the problem:

$V = \frac{1}{3}\pi r^2 h$

From the figure:

$\tan \theta = a/r$

$\tan 2\theta = h/r$

$\dfrac{2\tan \theta}{1 - \tan^2 \theta} = \dfrac{h}{r}$

$\dfrac{2(a/r)}{1 - a^2/r^2} = \dfrac{h}{r}$

$\dfrac{2a)}{\dfrac{r^2 - a^2}{r^2}} = h$

$h = \dfrac{2ar^2)}{r^2 - a^2}$

Thus,

$V = \frac{1}{3}\pi r^2 \left( \dfrac{2ar^2)}{r^2 - a^2} \right)$

$V = \frac{2}{3}a \pi \left( \dfrac{r^4)}{r^2 - a^2} \right)$

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

$4r^3(r^2 - a^2) - 2r^5 = 0$

$2(r^2 - a^2) - r^2 = 0$

$2r^2 - 2a^2 - r^2 = 0$

$r^2 = 2a^2$

$h = \dfrac{2ar^2}{r^2 - a^2}$

$h = \dfrac{2a(2a^2)}{2a^2 - a^2}$

$h = \dfrac{4a^3}{a^2}$

$h = 4a$ (*okay!*)

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