How do you find the maximum volume V(theta, phi) of the tetrahedron, with vertices at (0, 0, 0), (cos theta, 0, 0), (0, sin theta, 0) and (cos theta, sin theta, cos phi)?

Aug 20, 2016

$\frac{1}{12}$

Explanation:

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

$V = \frac{1}{6} \left\langle\left(b - a\right) \times \left(c - a\right) , d - a\right\rangle$ where $\times$ represents the cross product and $\left\langle\cdot , \cdot\right\rangle$ the scalar product.

Calling

$a = \left\{0 , 0 , 0\right\}$
$b = \left\{\cos \theta , 0 , 0\right\}$
$c = \left\{0 , \sin \theta , 0\right\}$
$d = \left\{\cos \theta , \sin \theta , \cos \phi\right\}$

$V = \frac{1}{6} \sin \theta \cos \theta \cos \phi = V \left(\theta , \phi\right)$

with stationary points located at

$\nabla V = \frac{1}{6} \left\{C o s \phi C o s \left(2 \theta\right) , - C o s \theta \sin \theta S \in \phi\right\} = \vec{0}$

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

$V = \frac{1}{12}$