Given a tetrahedrom with vertices a,b,c,da,b,c,d its volumen is given by
V = 1/6 << (b-a)xx(c-a),d-a>> V=16⟨(b−a)×(c−a),d−a⟩ where xx × represents the cross product and << cdot , cdot >>⟨⋅,⋅⟩ the scalar product.
Calling
a = {0,0,0}a={0,0,0}
b = {cos theta, 0,0}b={cosθ,0,0}
c={0,sin theta,0}c={0,sinθ,0}
d={cos theta, sin theta,cos phi}d={cosθ,sinθ,cosϕ}
V = 1/6 sin theta cos theta cos phi = V(theta,phi)V=16sinθcosθcosϕ=V(θ,ϕ)
with stationary points located at
grad V = 1/6 {Cos phi Cos(2 theta), -Cos theta sin theta Sin phi} = vec 0∇V=16{cosϕcos(2θ),−cosθsinθsinϕ}=→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
