TITLE: I found only one critical point using Lagrange multipliers. Must it be a minimizer?
QUESTION [9 upvotes]: I am trying to minimize 
$$V(x,y,z) = \frac {a^2b^2c^2}{6xyz}$$
subject to 
$$\frac {x^2}{a^2} + \frac{y^2}{b^2} + \frac {z^2}{c^2} = 1$$
and for $x,y,z>0$.
I found one critical point; evaluating it gives a function value of 
$$V= \frac {\sqrt{3}}{2}|abc|$$
This agrees with the solution that I am looking at, but why does it have to be a minimum?
Thanks,

REPLY [4 votes]: When you approach the boundary (that is, when $x \searrow 0$, $y \searrow 0$ or $z \searrow 0$), the function $V$ goes to infinity. Thus you can assume that 

$V$ admits no maximum on your domain considered
If $V$ admits a minimum, it won't be within a $\varepsilon$-range of the boundary. 

So you can add the conditions $x \ge \varepsilon, y \ge \varepsilon$ and $z \ge \varepsilon$ for $\varepsilon > 0$ small enough. This makes your new domain of definition compact, thus the function attains a minimum and a maximum on this new domain. However, since our function goes to infinity as $x \searrow 0$ (resp. $y$ and $z$), we can assume that the maximum is attained if and only if either $x,y$ or $z = \varepsilon$ ; in other words, that the values on the boundary are large, hence never minimal. So your minimum cannot be in the boundary, thus must be in the interior. If you found only one critical point in the interior, it has to be a minimum since the minimum exists and is a critical point.
Hope that helps,