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

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

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

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

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

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

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

$r^2 = 2h^2$

$r = \sqrt{2} \, h$
 

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

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