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

From the figure:
[math]z^2 = (x/2)^2 + (x/2)^2[/math]
[math]z^2 = x^2 / 2[/math]

[math]z^2 + (h – a)^2 = a^2[/math]
[math]x^2 / 2 + (h – a)^2 = a^2[/math]
[math]x^2 = 2a^2 – 2(h – a)^2[/math]
[math]x^2 = 2a^2 – 2h^2 + 4ah – 2a^2[/math]
[math]x^2 = 4ah – 2h^2[/math]

[math]V = \frac{1}{3}(4ah – 2h^2)h[/math]
[math]V = \frac{1}{3}(4ah^2 – 2h^3)[/math]

[math]\dfrac{dV}{dh} = \frac{1}{3}(8ah – 6h^2) = 0[/math]

[math]6h^2 = 8ah[/math]
[math]h = \frac{4}{3} a[/math]

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:
[math]h^2 + r^2 = L^2[/math]

The length of pole is given, thus L is constant
[math]2h + 2r \dfrac{dr}{dh} = 0[/math]

[math]\dfrac{dr}{dh} = -\dfrac{h}{r}[/math]

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

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

[math]r + 2h \dfrac{dr}{dh} = 0[/math]

[math]r + 2h \left( -\dfrac{h}{r} \right) = 0[/math]

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

r^2 = 2h
[math]r = \sqrt{2} \, h[/math]

[math]\text{radius } = \sqrt{2} \times \text{ altitude } \,\, [/math]            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:
[math]A = \pi rL \,\, [/math] where [math]L = \sqrt{h^2 + r^2}[/math]
[math]A = \pi r\sqrt{h^2 + r^2} \,\, [/math]

[math]\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[/math]

[math]hr \dfrac{dh}{dr} + r^2 + (h^2 + r^2) = 0[/math]

[math]\dfrac{dh}{dr} = - \dfrac{2r^2 + h^2}{rh}[/math]

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

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

[math]r \dfrac{dh}{dr} + 2h = 0[/math]

[math]r\left( - \dfrac{2r^2 + h^2}{rh} \right) + 2h = 0[/math]

[math]–2r^2 – h^2 + 2h^2 = 0[/math]
[math]h^2 = 2r^2[/math]
[math]h = \sqrt{2}\,r[/math]

[math]\text{height } = \sqrt{2} \times \text{ radius } \,\,[/math]            answer