# 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 :)

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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 $\mathbb{R}^2$.

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can you give me an example, please. My teacher says the thing opposite :( that we can sure that no max for that function (in above example) – hqt Jun 8 '12 at 6:47
@AndréNicolas: That's not strictly true; since there are inequality constraints, in this case you would need to look at 9 possibilities for a solution. You have chosen the one with no constraints active. Looking at all the other possibilities will indeed throw up the corner points (in fact all corner points, and the mid-points of the edges too). The Lagrange conditions are necessary. – copper.hat Jun 8 '12 at 7:37
Sorry. Can I ask you, when I read on Wiki, constraint is an equation. and when I ask my teacher, my teacher say when constraint is inequality, we use another method. – hqt Jun 8 '12 at 7:54
There are many different sorts of constraints. Equality constraints and inequality constraints are two types... – copper.hat Jun 8 '12 at 8:12
Nitpick: the maximum of $x^2+y^2$ on $[-1,1]\times[-1,1]$ also occurs at $(-1,1)$ and $(1,-1)$. – Rahul Jun 8 '12 at 17:43