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

**Problem 62**

Inscribe a circular cylinder of maximum convex surface area in a given circular cone.

**Solution:**

$\dfrac{H - h}{d} = \dfrac{H}{D}$

$h = H - \dfrac{Hd}{D}$

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*

**Problem 63**

Find the circular cone of maximum volume inscribed in a sphere of radius a.

**Solution:**

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

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*