The method called Lagrange Multipliers is used to find critical points of $f(x_1,x_2,\ldots,x_n)$, when $f$ is constrained to the level set $S = \{ x\in \mathbb{R}^n \, | \, g(x_1,x_2,\ldots,x_n)=0 \}$.
These critical points may be used to locate local minima and maxima.
When $S$ is a compact set, we know that a global max and min of $f$ exist on $S$.
Do we also have to check what happens on the boundary of $S$, in case a boundary exists ?
Best Answer
Another example to supplement that of Sharkos: $f(x,y)=x+y$ and $g(x,y)=\sqrt{x}+\sqrt{y}-1.$ Then the only critical point found by LaGrange is at $(1/4,1/4)$ where $f=1/2$, but the max of $f=1$ occurs at both points $(1,0),(0,1)$. The constraint region here is the inwardly bent curve $y=(1-\sqrt{x})^2$ for $0 \le x \le 1$ and has $(1,0),(0,1)$ as its boundary points.
ADDED: Teddy has in a comment noted that when $\nabla g$ is not defined, one considers that to be a critical point. So here is another example, similar to the above, which avoids that issue. Let $f(x,y)=x+y$ and $g(x,y)=x^{3/2}+y^{3/2}-1$. Note that the implicit radicals in the constraint impose that $x,y \ge 0$, and that this means we have $$\nabla g = ((3/2)x^{1/2},(3/2)y^{1/2}),$$ which exists and is nonzero at all points on the constraint curve (it's the zero vector only at the origin, clearly not on the constraint curve).
In this case the constraint curve is the graph of $y=(1-x^{3/2})^{2/3}$ on the interval $[0,1]$, having endpoints at $(1,0),(0,1)$, and the graph bends outward away from the origin. There is now only one critical point at $(x,y)=(a,a)$ where $a=(1/4)^{1/3}$. At that point, $f=2^{1/3}\approx. 1.259$, while at the "boundary of constraint region" points $(1,0),(0,1)$ we have $f=1$. So if we seek the minimum we would have to look at the boundary points of the constraint region, at non-critical points.