Lagrange Multipliers Question - some extremum points are missed (not detected) by the method

Question

I was reading about Lagrange multipliers. My book says that if $f(x,y,z)$ has in the point $(x_0, y_0, z_0)$ an extremum under the condition that $g(x,y,z) = k$, and if $\nabla g \ne (0,0,0)$, then there exists a number $\lambda$ such that

$$\nabla f (x_0, y_0, z_0) = \lambda \nabla g (x_0, y_0, z_0) \tag{1}$$
and
$$g(x_0, y_0, z_0) = k \tag{2}$$

But then I was working through this simple example.

Find the extremums of $f(x,y,z) = xyz$ under the condition
$g(x,y,z) = x+y+z = 1$ where $x,y,z \ge 0$

I noticed that here the Lagrange method (i.e. solving the system $(1), (2)$) misses to detect/find the extremum points $P (x,y,z)$, such that e.g. $x = 0$ and $y+z = 1$ and $y,z$ are positive.

It only detects/finds the four points $(1/3, 1/3, 1/3), (0,0,1), (0,1,0), (0,0,1)$.

So what am I missing here?

I guess in my book the theorem is stated somewhat informally, and that's why the Lagrange method is missing to detect these points.

So why are these points $P$ missed? I mean, they are obviously extremum (minimum) points of $f$ but they do not satisfy the system $(1), (2)$ for any $\lambda$.

Answer 1

The method of Lagrange multipliers works when we are after the locally extreme points of a class $C^1$ function $f$ on a region of the type $g(x_1,x_2,\ldots,x_n)=k$, for a class $C^1$ function $g$. That's not the case here, since your region is a strict subset of $\{(x,y,z)\in\Bbb R^3\mid g(x,y,z)=1\}$ rather then the whole set. So, there can be (and, in this case, there are) locally extreme points at the boundary of your region.

Answer 2

This is an additional remark to the correct reply from José:

it is possible (though not always feasible) to introduce new variables and in the process shift the problem to a higher dimensional one that has no boundary.

Indeed by introducing $x_0,y_0,z_0$ and add the conditions $$ x=x_0^2\\ y=y_0^2\\ z=z_0^2\\ $$ We can define the Lagrangian by $$ L=xyz+\lambda(x+y+z-1)+\mu_x(x-x_0^2)+\mu_y(y-y_0^2)+\mu_z(z-z_0^2) $$ Now, if we look at the critical points $\nabla L=0$ in the 10-dimensional space we find all critical points, even those that were originally on the boundary.

Answer 3

Too long for a comment.

For instance. Using the Lagrangian depicted in the answer from @maxmilgram we can find the following stationary points

$$ \left[ \begin{array}{ccccccccccc} f& x & y & z & \lambda & \mu_x&\mu_y&\mu_z & x_0^2& y_0^2& z_0^2\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & y & 1-y & 0 & (y-1) y & 0 & 0 & 0 & y & 1-y \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & x & 0 & 1-x & 0 & 0 & (x-1) x & 0 & x & 0 & 1-x \\ 0 & x & 1-x & 0 & 0 & 0 & 0 & (x-1) x & x & 1-x & 0 \\ \frac{1}{27} & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & -\frac{1}{9} & 0 & 0 & 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \end{array} \right] $$

where $f = x y z$. Note that in the third row we have conditions: $y \ge 0$ and $1-y\ge 0$ etc.