Lagrange Multiplier. Justifying point is a minimum

Question

A problem in my problem book

I've found the constants using Lagrange multipliers to be,

a=c=√3
b=2√3

Which I'm pretty sure is correct however, why is it so, that these values are minimums?

Answer 1

In general, the Lagrange multipliers method gives only a necessary condition for some point to be an extremum of some function $F(x)$ subject to the condition $G(x)=c$, where $F,G$ are differentiable functions from $\mathbb{R}^n$ to $\mathbb{R}$, and $c$ some constant. The necessary condition is that at an extremum point, the gradient of $F$ needs to be proportional to the gradient of $G$. Now, a lot of things can happen: the point found may be a maximum, a minimum, or neither. Some further analysis of $F$ is required in order to determine which is it.

In your particular case your parametrized ellipsoid was obtained from the Euclidean unit ball by applying a linear transformation whose representing matrix with respect to the cannonical basis is a diagonal matrix with $(a,b,c)$ on the diagonal. The volume of the ellipsoid is therefore the volume of the unit ball in $\mathbb{R^3}$ multiplied by the absolute value of the determinant of the matrix carrying it to the ellipsoid, which is just $|abc|$. So in your case the function you want to minimize is $F(a,b,c)=abc$, and the constraint is given by the condition that the ellipsoid passes through the given point, i.e.

$$G(a,b,c)={1\over a^2}+{4\over b^2}+{1\over c^2}-1=0.$$

So now you found that $a=c=\sqrt{3}, b=2\sqrt{3}$ is a candidate point for a constrained extremum. At this point of the anaylsis, you still do not know whether it is a maximum, a minimum, or neither. In fact, it is my honest opinion that the usual methods of differential calculus do not lend themselves in a generous fashion to this particular problem. Therefore, I propose an alternative route.

A point $(a,b,c)$ that minimizes $abc$ subject to the above constraint will also minimize $a^2b^2c^2$ subject to the same constraint, and clearly will also minimize $a^2b^2c^2/4$, hence will maximize the quantity $\left({1\over a^2}{4\over b^2}{1\over c^2}\right)^{1/3}$, subject to the same constraint. By the AG-GM inequality we have: $$\left({1\over a^2}{4\over b^2}{1\over c^2}\right)^{1/3}\leq {{1\over a^2}+{4\over b^2}+{1\over c^2}\over 3}={1\over 3}$$ where we have used our constraint. We know that there is equality in the AM-GM inequality if and only if all the terms are equal, which is precisely what happens when $a=c=\sqrt{3}$ and $b=2\sqrt{3}$. So for these numbers the quantity $\left({1\over a^2}{4\over b^2}{1\over c^2}\right)^{1/3}$ is maximized with respect to the constraint. It follows that for these numbers the volume of the ellipsoid is minimal.

So aposteriori, we did not have to use Lagrange's multipliers.