# 66 - 68 Maxima and minima: Pyramid inscribed in a sphere and Indian tepee

**Problem 66**

Find the largest right pyramid with a square base that can be inscribed in a sphere of radius a.

**Solution:**

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

From the figure:

$z^2 = (x/2)^2 + (x/2)^2$

$z^2 = x^2 / 2$

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

$x^2 / 2 + (h - a)^2 = a^2$

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

$x^2 = 2a^2 - 2h^2 + 4ah - 2a^2$

$x^2 = 4ah - 2h^2$

$V = \frac{1}{3}(4ah - 2h^2)h$

$V = \frac{1}{3}(4ah^2 - 2h^3)$

$\dfrac{dV}{dh} = \frac{1}{3}(8ah - 6h^2) = 0$

$6h^2 = 8ah$

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

Altitude of pyramid = 4/3 × radius of sphere, a *answer*

**Problem 67**

An Indian tepee is made by stretching skins or birch bark over a group of poles tied together at the top. If poles of given length are to be used, what shape gives maximum volume?

**Solution:**

$h^2 + r^2 = L^2$

The length of pole is given, thus L is constant

$2h + 2r \dfrac{dr}{dh} = 0$

$\dfrac{dr}{dh} = -\dfrac{h}{r}$

Volume of tepee:

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

$\dfrac{dV}{dh} = \frac{1}{3} \pi \left( r^2 + 2rh \dfrac{dr}{dh} \right) = 0$

$r + 2h \dfrac{dr}{dh} = 0$

$r + 2h \left( -\dfrac{h}{r} \right) = 0$

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

$r^2 = 2h^2$

$r = \sqrt{2} \, h$

$\text{radius } = \sqrt{2} \times \text{ altitude}$ *answer*

**Problem 68**

Solve Problem 67 above if poles of any length can be found, but only limited amount of covering material is available.

**Solution:**

$A = \pi rL \,\, $ where $L = \sqrt{h^2 + r^2}$

$A = \pi r\sqrt{h^2 + r^2} \,\, $

$\dfrac{dA}{dr} = \pi r \left( \dfrac{2h\dfrac{dh}{dr} + 2r}{2\sqrt{h^2 + r^2}} \right) + \pi \sqrt{h^2 + r^2} = 0$

$hr \dfrac{dh}{dr} + r^2 + (h^2 + r^2) = 0$

$\dfrac{dh}{dr} = - \dfrac{2r^2 + h^2}{rh}$

Volume of tepee:

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

$\dfrac{dV}{dr} = \frac{1}{3} \pi \left( r^2 \dfrac{dh}{dr} + 2rh \right) = 0$

$r \dfrac{dh}{dr} + 2h = 0$

$r\left( - \dfrac{2r^2 + h^2}{rh} \right) + 2h = 0$

$-2r^2 - h^2 + 2h^2 = 0$

$h^2 = 2r^2$

$h = \sqrt{2}\,r$

$\text{height } = \sqrt{2} \times \text{ radius}$ *answer*