62 - 63 Maxima and minima: cylinder inscribed in a cone and cone inscribed in a sphere

$h = H - \dfrac{Hd}{D}$
 

Convex surface area of the cylinder:
$A = \pi d \, h$

$A = \pi d \left( H - \dfrac{Hd}{D} \right)$

$A = \pi H d - \dfrac{\pi H}{D}d^2$
 

The cone is given, thus H and D are constant
$\dfrac{dA}{dd} = \pi H - \dfrac{2\pi H}{D}d = 0$

$\pi H = \dfrac{2\pi H}{D}d$

$d = \frac{1}{2} D$
 

Diameter of cylinder = radius of cone           answer

From the figure:
$r^2 = a^2 - (h - a)^2$

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

$r^2 = 2ah - h^2$
 

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

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

The sphere is given, thus radius a is constant.
$\dfrac{dV}{dh} = \frac{1}{3} \pi \, ( \, 4ah - 3h^2 \,) = 0$

$4ah = 3h^2$

$h = \frac{4}{3} a$
 

Altitude of cone = 4/3 of radius of sphere           answer