Extreme value Lagrange multiplier (max or min?)

I am to determine the the range of the volume of a tetrahedron enclosed by the coordinate axes and a tangentplane on the ellipsoid $x^2 + 2y^2 + 3z^2 = 1$. The volume of the tetrahedron can be derived to be given by

$$V(x,y,z) = \frac 1{36xyz}.$$

The constraint is $g(x,y,z) = x^2 + 2y^2 + 3z^2 = 1$ and using Lagrange multiplies, we can arrive at the critical points

$$\left( \pm \frac 1{\sqrt 3}, \pm \frac 1{\sqrt 6}, \pm \frac 13 \right)$$

We can calculate

$$V\left(\frac 1{\sqrt 3}, \frac 1{\sqrt 6}, \frac 13\right) = \frac {\sqrt 2}4$$

But it turns out that this is the smallest volume the tetrahedron can assume but Lagrange's multiplier-method does not give us this information. How can we ascertain that this indeed is a minimum and not a maximum point? Also one thing I wonder is whether there is any boundary to study, to me it appears there is no boundary but perhaps I am mistaken?

I'm going to re-cast a couple of your expressions, as it may be helpful in answering your question. It appears that the tetrahedron for which you wish to extremize the volume is bounded by the three coordinate planes and an inclined plane which is tangent to the given ellipsoid. (We will work with the portion of the ellipsoid that is in the first octant, but analogous results can be obtained in all eight octants.)

If we call the coordinate axis intercepts of the tangent plane $\ x_0 \ , \ y_0 \$ , and $\ z_0 \$ , then the volume of the bounded tetrahedron is

$$\ V \ = \ \frac{x_0 \ y_0 \ z_0}{6} \ \ .$$

Evaluating the normal line at the surface of the ellipsoid $\ x^2 \ + \ 2y^2 \ + \ 3z^2 \ = \ 1 \$ at a point $\ (X , Y, Z) \$ , we find the tangent plane there to be

$$2X \ (x - X) \ + \ 4Y \ (y-Y) \ + \ 6Z \ (z-Z) \ = \ 0$$

$$\Rightarrow \ \ 2X \ x \ + \ 4Y \ y \ + \ 6Z \ z \ = \ 2X^2 \ + \ 4Y^2 \ + \ 6Z^2 \ = \ 2 \ \ ,$$

this last result being found by applying the equation of the ellipsoid. We may conclude from this that the coordinate axis intercepts for a particular tangent plane are given by

$$x_0 \ = \ \frac{1}{X} \ \ , \ \ y_0 \ = \ \frac{1}{2Y} \ \ , \ \ z_0 \ = \ \frac{1}{3Z} \ \ .$$

This leads to the result you show for the volume of the tetrahedron,

$$\ V \ = \ \frac{1}{36 \ XYZ} \ \ .$$

Applying the Lagrange-multiplier method (which I won't recapitulate), you found the tangent point in the first octant which extremizes the tetrahedron's volume to be

$$(X, Y, Z) \ = \ \left( \frac 1{\sqrt 3} \ , \ \frac 1{\sqrt 6} \ , \ \frac 13 \right) \ \ ,$$

and the extremal volume of the tetrahedron as you've given it. Here is a graph of the situation (the ellipsoid has been made a trifle "too large" to make the point of tangency visible):

A limitation of the Lagrange-multiplier method, interpreted geometrically, is that it only locates points on the constraint "surface" where a normal vector lies in the same direction as the normal vector of a specific "level surface" of the function to be extremized. The graph below shows the ellipsoid with the level surface

$$\ V(x,y,z) \ = \ \frac{1}{36 \ xyz} \ = \ \frac{\sqrt{2}}{4} \ \ ,$$

represented in three dimensions as $\ z \ = \ \frac{1}{9\sqrt{2} \ xy} \$ ; the tangent point occurs at the same position as given above. "Lagrange" does not tell us directly whether a particular extremum is a local maximum or minimum of the function $\ V \$ . (Again, the ellipsoid is made a bit "too large" to allow the location of the tangent point to appear in the graph.)

But we are able to say that possible tangent points to the surface of the ellipsoid can be taken all the way to any of the coordinate axes. This can make one or more of the coordinates of the tangent point, $\ X \ , \ Y \ , \ Z \ ,$ equal to zero, sending the corresponding axis intercepts toward infinity. Thus, tetrahedra bounded by the coordinate planes and the tangent plane can have volumes without any upper limit. (Under the geometrical interpretation, the tangent point only occurs on "level surfaces" for larger and larger values of $\ V \$ . ) So the volume found from Lagrange multipliers must be the minimum possible.

• Wow, thank you for that. Is it possible to reason as follows: Lagrange multipliers says that if there is an extremum, the gradients must be parallell. We found one extremum by using Lagrange multipliers but we do not know whether it is maximum or minimum. However we can put in other values and see that bigger values are assumed than the found extremum value. Thus we can conclude that it must be minimum, because if there did exist a maximum, the Lagrange multiplier should have found it but in this case it did not, therefore no maximum? – thelionkingrafiki Jun 14 '14 at 15:01
• Lagrange multipliers may well have found another extremum if it were there (the method sometimes need a bit of "help" to obtain information about the function to be extremized). It is often useful to understand the nature of the function in order to know whether it can even have both a minimum and a maximum (and whether there is only one of each, since there could be local, but not absolute, extrema). In this problem, we can show there is no upper limit for the volume of the tetrahedron. – colormegone Jun 14 '14 at 16:35
• Here is another example of an extremization problem with a minimum, but, by the nature of the geometry involved, no maximum value: Here [math.stackexchange.com/questions/817989/… is – colormegone Jun 14 '14 at 16:36