Find func extremum for the given condition

Question

Find so-called conditional extremum attained by $f(x,y,z)=x \cdot y \cdot z;$ under the following condition: $x^2+y^2+z^2=a^2$. It is thought that solution would rely on Lagrange multipliers method.

Can you help me with this one? I have no idea of how to solve it wisely.

Answer 1

Before we begin, note by symmetry that we expect the answers to be symmetric in $x,y,z$ by which I mean that if $(x_0,y_0,z_0)$ is a point where we have an extrema, then so is any permutation of it. This will be a sanity check on our answer.

I have $f(x,y,z) = xyz$ and $g(x,y,z) = x^2 + y^2 + z^2 - a^2 = 0$. The condition is $Df = \lambda Dg$ which is restated as $$(yz, xz, xy) = \lambda (2x, 2y, 2z)$$ I am going to proactively note some technicalities but in reality when solving this problem, you would have to deal with them as and when they pop up. We note two things:

1. What happens when $\lambda = 0$? Well we have $xz = yz = xy = 0$. It is tempting to say that the only point that satisfies this is $(0,0,0)$ but this is not true. $(l,0,0), (0,m,0) \text{ & } (0,0,n)$ all satisfy these equations. But these are the only solutions. Great, now plugging these into the constraint equation gives us $6$ points to keep in mind: $(±a,0,0), (0,±a,0), (0,0,±a)$. At all these points, $f = 0$.
2. Now we can proceed with the assumption that $\lambda \neq 0$. This actually forces all of $x,y,z$ to be non-zero. Let's see why. WLOG, say $z= 0 $. Then we get $(0,0,xy) = \lambda(2x, 2y, 0)$ This gives us that $x= 0, y = 0$, which is not a point that satisfies the constraint.

Great, now we can proceed to solving this quite easily. A common trick with a single constraint Lagrange multiplier is to eliminate $\lambda$ by "dividing" one vector by another. What I mean by this is to write $$\frac{yz}{x} = \frac{xz}{y} = \frac{xy}{z}$$ These fractions are okay because of point $2$ above. WLOG I'm going to choose the first two to work with. $$\frac{yz}{x} = \frac{xz}{y} \implies y^2z = x^2z \implies y^2 = x^2 \implies y = ±x$$ Similarly, we see that $x, y$ and $z$ are all the same up to a sign. This is exactly our symmetry hunch that we had in the beginning. So the constraint becomes $3x^2 = a^2$ or $x = ±\frac{a}{\sqrt{3}}$. So our $8$ solutions are $(±\frac{a}{\sqrt{3}}, ±\frac{a}{\sqrt{3}}, ±\frac{a}{\sqrt{3}})$ where you take all possible combinations of the $±$. Now the last step is to evaluate $f$ at all these points. There are only two values of $f$ possible, which are $\frac{a^3}{\sqrt{27}}$ and $-\frac{a^3}{\sqrt{27}}$ and so these are the constrained extrema.