Lagrange's method to find min/max question I have a doubt when using this method to find min/max of a function. When I can find two (or more solutions) of system equation, so I can easily know it has max and min.
But, the problem is: if the system equation I solve just have only one solution. So, I can know it max or min (by replace arbitrary value to function to compare). For example is min. So, Can I say this function doesn't has max ?
Thanks :)
 A: If you are looking a continuous function, say $f(x,y)$ (there could be more variables) over a compact set (closed and bounded), then there will always be a min and a max. But the min and/or max may occur on the boundary of the region, in which case setting partial derivatives equal to $0$ may (and usually will) fail to find it.
This is the analogue in dimensions $2$ and higher of the fact that if we are looking at a continuous function over a closed interval, the min and/or max may occur at an endpoint.  
For example, if we are trying to find the min or max of $x^2+y^2$ on or inside the square with corners $(1,1)$, $(-1,1)$, $(-1,-1)$, and $(1,-1)$, setting partial derivatives equal to $0$ finds the minimum, which occurs at $(0,0)$, but does not find the maximum, which is at the four corners of the region.
If we have a function say $f(x,y)$ with continuous partial derivatives, and we are interested in finding the min and max of $f$ over all of $\mathbb{R}^2$, there may not be such a min or max. If there is a single place at which the partials are both $0$, then at least one of min or max fails to exist. But even if the partials are both $0$ at several places, there may not be a min or max.
For example, let $f(x,y)=x^3-3x^2+y^3-3y^2$. Then the partials are equal to $0$ 
at $(0,0)$, $(0,2)$, $(2,0)$, and $(2,2)$ but $f$ has neither a min nor a max on $\mathbb{R}^2$.
