I saw this in a solution for one of the questions in my homework:
Q: Find the closest point to the origin from a curve defined by:
$x^2-xy+y^2-z^2=1$ and $x^2+y^2=1.$
This was written in the solution:
https://i.stack.imgur.com/rapOY.png
why $rk\lt 2$ is required to use Lagrange multipliers?
On
If the rank is smaller than 2 then it means the constraints are linearly dependent so I can't use Lagrange multipliers.
On
You have given the answer yourself. Now let's check!
We are given two constraints $$f(x,y,z):=x^2-xy+y^2-z^2-1=0,\qquad g(x,y,z):=x^2+y^2-1=0$$ hopefully defining a curve $C$. One computes $$\nabla f(x,y,z)=(2x-y, 2y-x,-2z),\qquad\nabla g(x,y,z)=(2x,2y,0)\ .$$ These two gradient fields are defined in all of ${\mathbb R}^3$. We have to check whether they happen to be linearly dependent in certain points of $C$. To this end we compute the "test vector" $${\bf t}:=\nabla f\times\nabla g=(4zy,-4zx,2x^2-2y^2)\ .\tag{1}$$ Points $(x,y,z)$ where ${\bf t}=(0,0,0)$ are bad.
The first two components of ${\bf t}$ vanish iff (i) $x=y=0$ or (ii) $z=0$. Here (i) contradicts $g(x,y)=0$, and (ii) together with $0=f(x,y,z)-g(x,y,z)=-z^2-xy$ implies $xy=0$, which together with $x^2=y^2$ necessitates $x=y=0$, contrary to $g(x,y)=0$. It follows that no bad points are lying on $C$, in other words: The two surfaces $f=0$ and $g=0$ intersect transversally at all points.
We therefore can be sure that the Lagrangian method will bring all conditionally stationary points of the objective function $\Phi(x,y,z):=x^2+y^2+z^2$ to the fore. In the case where some bad points are lying on $C$ you can use Lagrange's method nevertheless. But you have to add the bad points on $C$ to the candidate list produced by Lagrange's method in order to catch the extremum for sure.
$$x^2+y^2+z^2=1+z^2\geq1.$$ The equality occurs for $z=0$, $x^2+y^2=1$ and $xy=0$, which gives all these points:
$(1,0,0)$, $(-1,0,0)$, $(0,1,0)$ and $(0,-1,0)$.