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#
