Given a tetrahedrom with vertices #a,b,c,d# its volumen is given by

#V = 1/6 << (b-a)xx(c-a),d-a>> # where # xx # represents the cross product and #<< cdot , cdot >># the scalar product.

Calling

#a = {0,0,0}#

#b = {cos theta, 0,0}#

#c={0,sin theta,0}#

#d={cos theta, sin theta,cos phi}#

#V = 1/6 sin theta cos theta cos phi = V(theta,phi)#

with stationary points located at

#grad V = 1/6 {Cos phi Cos(2 theta), -Cos theta sin theta Sin phi} = vec 0#

giving the pairs

#{
(theta = 0, phi= -pi/2),
(theta = 0,phi = pi/2),
(theta= -(3 pi)/4, phi= 0),
(theta= -(3 pi)/4, phi= pi),
(theta = -pi/2, phi= -pi/2),
(theta = -pi/2, phi= pi/2),
(theta= -pi/4, phi= 0),
(theta= -pi/4,phi=pi),
(theta= pi/4,phi= 0),
(theta = pi/4, phi=pi),
(theta = pi/2, phi= -pi/2),
(theta = pi/2, phi= pi/2),
(theta = (3 pi)/4, phi= 0),
(theta= (3 pi)/4, phi= pi),
(theta =pi, phi= -pi/2),
(theta = pi, phi=pi/2))#

The stationary point qualification is done with the characteristic polynomial roots, after the hessian matrix calculation for each point.

The hessian matrix is given by

#grad (grad V(theta,phi)) =1/6(
(-2 Cos phi Sin(2theta), -Cos(2theta) Sin phi),
(-Cos(2theta) Sin phi, -Cos phi Cos theta Sin theta)
) #

The condition for stationary point to be a maximum is that

its characteristic polynomial must have two negative roots. This is guaranteed for the solutions

#((theta= -(3 pi)/4, phi= 0),
(theta= -pi/4,phi=pi),
(theta= pi/4,phi= 0),
(theta= (3 pi)/4, phi= pi))#

for which the volumen is

#V = 1/12#