I found only one critical point using Lagrange multipliers. Must it be a minimizer?

Question

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,

Answer 1

The Lagrange multiplier method doesn't tell you what kind of critical point you've found. Usually, we would work around this by saying that the domain is compact and the objective function is smooth, so the minimum is attained at either a critical point or on the boundary. But in this case the domain is not compact, because it's not closed, and your objective function can't be smoothly extended to the closure of the domain, either.

So you need to do something slightly different, but in the same spirit. The idea is to notice that near $x=0,y=0$ or $z=0$, your objective function becomes very large. So the minimum cannot be in some small neighborhood of these points. So you can modify $x,y,z>0$ into $x,y,z \geq \delta$ where $\delta$ is some positive number. Then your domain becomes compact and you can use the argument I suggested above. If somehow the minimum is attained at $x=\delta,y=\delta$, or $z=\delta$, then just make $\delta$ smaller. Eventually the minimum cannot be there, since your objective function blows up to $+\infty$ near the boundary.

Answer 2

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,