Find the absolute maximum and minimum values of $f(x,y,z)=2x+y$ subject to the constraint $x+y+z=1$.
This comes from Marsden's Vector Calculus book 3.4 #14.
I tried to use Lagrange multipliers. So I had $g(x,y,z)=x+y+z-1$.
Taking $ \operatorname{grad}f= L\operatorname{grad} g$, I got the four equations $$2=L \\ 1=L\\ 0=L\\ \text{and } x+y+z=1.$$ So no solution? Is there no max and min subject to this constraint?
I also thought of it as $z=1-x-y$. Then $x$ and $y$ can be anything we want and $z$ will make up for it, so there's no max or min because we can keep putting in larger or smaller numbers, as long as $z=1-x-y$.
I also know that when f is constrained to a closed and bounded region, then we're guaranteed to find a max/min on that closed and bounded region. However, I think that this region is not closed and bounded (because like I said $x,y,z$ can be anything)...so we're not guaranteed a max/min, but I don't know if that automatically means there is no max/min.
Looks good. And indeed, the fact that the region is not compact makes it so that your function can be unbounded, even when subjected to the constraint. In fact in this case it's easy to see how to do it: take $x=M,y=0,z=1-M$, where $M$ is a big number (either positive or negative).
Note that the region not being compact does not give a guarantee; for instance you could have $f(x,y)=x^2+y^2$ confined to a line, and it will still have a minimum.