A Lagrange multiplier

Question

Given the function

$$f(x,y,z)=xyz$$

subject to the constraint

$$g(x,y,z)=\frac1x+\frac1y+\frac1z-1=0,$$

the method Lagrange multipliers shows that the point $(3,3,3)$ is an extremum of $f(x,y)$ satisfying $g(x,y,z)=0$. This point gives $f(3,3,3)=27$.

But notice also that $f(4,4,2)=32$ and $f(1,-1,1)=-1$, both points satisfying the equation $g(4,4,2)=g(1,-1,1)=0$. But does this not tell me that $(3,3,3)$ is not an extremum? What is wrong with the above?

Answer 1

The method of Lagrange multipliers helps you to find critical points of the function along the constraint. That said, a critical point may be a local max/min (or neither). That's all you can conclude. To look for the absolute max/min (if any) you need to check the value of the function at all those critical points plus the limits as $(x,y,z)$ approaches the boundary of the admissible region, then pick the largest/smallest.

Answer 2

(3,3,3) is a local minimum. It would be the global minimum if you further constrained x,y,z to all being greater than 0.