Largest Cone In Sphere

Using the formula formula for circumradius of an isosceles triangle of base $a$ and height $h$ :

$R = \frac{1}{8}(\frac{a^2}{h} + h)$

Taking a section of the sphere along its central axis, we will have a triangle whose base is the cone base diameter and whose height is the cone height, which have to be circumscribed by a circle of radius 1. Thus, denoting cone radius as $r$ and height as $h$ :

$1 = \frac{1}{8}(\frac{4r^2}{h} + h)$

$r^2 = 2h - h^2$

So cone volume is:

$v = \frac{\pi}{3}r^2 h = \frac{\pi}{3} (2h^2 - h^3)$

Differetiating $v$ with respect to $h$ and making it equal to $0$ :

$\frac{\pi}{3}(4h - 3h^2) = 0$

$h = \frac{4}{3}$

Plugging this in $v$ :

$v = \frac{\pi}{3}(2\cdot\frac{16}{9} - \frac{64}{27})$

$v = \pi\cdot\frac{32}{81}$

Hence, $A = 32$ , $B = 81$ and $A+B = \fbox{113}$

Let's examine this problem via a plane section of the unit sphere (using the xy-plane and the circle x^2 + y^2 = 1):

Let the slanted edges of the cone in question be modeled as the lines y = (+/-)m*x + 1 (with the apex situated at the point (0,1). Substitution of this expression into that of our unit circle gives:

x^2 + [1 (+/-) mx]^2 = 1;

or x^2 + (1 (+/-)2mx + m^2 * x^2) = 1;

or (m^2 + 1)*x^2 (+/-)2mx = 0;

or x = 0 and x = (+/-) 2m/(m^2 + 1). Solving for the corresponding ordinate gives y = -sqrt(1 - x^2) = -sqrt[1 - (2m/(m^2+1))^2] = -[(m^2 - 1)/(m^2 + 1)]. So the endpoints of the cone's base are (x,y) = (-2m/(m^2+1), -(m^2-1)/(m^2+1)) and (2m/(m^2+1), -(m^2-1)/(m^2+1)). The radius of the cone equals 2m/(m^2 + 1) and its height equals 1 + (m^2 - 1)/(m^2 + 1) = (2m^2)/(m^2 + 1). So, the volume can now be expressed as a function of the slope m:

V(m) = (pi/3) * [2m/(m^2 + 1)]^2 * (2m^2)/(m^2 + 1) = (pi/3) * (8m^4)/(m^2 + 1)^3 (i). Differentiating (i) and setting it equal to zero produces:

dV/dm = (pi/3) * [(32m^3)(m^2 + 1)^3 - (8m^4) * 3(m^2+1)^2 * (2m)] / (m^2 + 1)^6 = 0;

or (pi/3) * [(32m^3)*(m^2 + 1) - (48m^5)] / (m^2 + 1)^4 = 0;

or 32m^5 + 32m^3 - 48m^5 = 0;

or 32m^3 - 16m^5 = 0;

or (16m^3)*(m^2 - 2) = 0;

or m = 0, (+/-)sqrt(2). Thus, the minimum volume is attained when m = 0, or V(0) = 0. The maximum volume must therefore be attained at either m = sqrt(2) or -sqrt(2), which both critical values result in V_max = (pi/3) * (8)(4)/(2+1)^3 = (32/81)*pi.