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 $\lambda$.
Best Answer
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.