Problem 66

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

Solution:
Volume of pyramid:
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:
From the figure:
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
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:
Area of covering material:
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