# Homework Help: Global Extreme Values (what?)

1. Oct 17, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data
This problem relates to nothing we talked about in class...
Determine the global extreme values of f(x,y) = x^2 + 2y^2 on the unit square D={(x,y):0<x,y<1}

So first of all I don't see how D={(x,y):0<x,y<1} defines a square.

df/dx = 2x
df/dy = 4y

anyone wanna guide this lost sheep?

2. Oct 17, 2013

### brmath

The square: 0 < x < 1 defines a vertical strip in the plane where y can be anything and x is in that interval. They also say 0 < y < 1. That defines a horizontal strip in the plane where x can be anything and y is in that interval. But they say both x and y are limited. So you have to look at where those two strips cross. And it is a square.

A "global extremum" means a point ($x_0, y_0$) in D where the $f( x_0, y_0) \ge f(x,y)$ for every (x,y) in the domain of f (that is, in your square D); or $f( x_0, y_0) \le f(x,y)$ in that domain .

In this case there are two ways to go about finding the answer. One is "by inspection" and the other is to apply the mathematical techniques for finding extrema.

Let's look at inspection: $f(x,y) = x^2 + 2y^2$. What is the smallest value that can take on? What happens as x and y get larger -- does the function increase or decrease? So what is the largest value it can take on (be very careful here, because the boundaries of D are not included.)

Mathematical techniques (used for more complicated functions where inspection is impossible): the theory says that if there is an extremum it is either where both $f_x \text { and } f_y = 0$; or it is on the boundary of the domain (which in this case you do not have).

Now if $f_x \text { and } f_y = 0$ are both 0, there are 3 cases: it is a local minimum; it is a local maximum, or it isn't either. That depends on the determinant of
$\begin{pmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy}\\ \end{pmatrix}$evaluated at the point where $f_x and f_y$ = 0.

Was this information given to you? Can you figure out where df/dx and df/dy are 0?

If you get that far, look around in your textbook, or online. There are plenty of sources for saying what the determinant tells you about whether there is a min, a max, or neither.

And finally, just because you find a local minimum or maximum doesn't mean it is global. There might be another minimum point in the domain which is lower than the first minimum you find.

That may be a bit daunting to you, but I promise the f you were given is very easy to work with.

3. Oct 17, 2013

### Ray Vickson

Your region is an open square: the corners are (0,0), (0,1), (1,0) and (1,1). However, these corners, and their joining edges, are not actually in the square; that is, the boundary of the square is not itself in the square. Therefore, the problem---exactly as written---has no solution; that is, there is no max and no min! There are suprema and infima, but because your feasible set is OPEN, the suprema and infima are not attained in this case. If you changed the set to {(x,y): 0 ≤ x,y ≤ 1} then you would have a well-posed problem. (This is very similar to asking what is the minimum value of x, subject to x > 0. Well, there is no minimum, because x = 0 has not been allowed.)

Considering the closed-rectangle case, the max and min problems are very different: the min problem is a so-called convex programming problem, and so any local constrained min is automatically a global min (global inside the rectangle, that is). However, the max problem is non-convex optimization problem, whose nature is "combinatoric" in general. In this specific case it is easy enough, but in more general problems of this type (say with a much more complicated constraint set) the problem would---in the worst case---be solvable (as far as we currently know) only by enumerating all the possibilities and just finding the best one numerically. (This gets into the are of so-called NP-hard problems; the max problem is a special case of NP-hard.)

4. Oct 17, 2013

### brmath

Ray, the problem is perfectly well posed. There is an absolute minimum at (x,y) = 0 and no maximum. Did you read my answer?

5. Oct 18, 2013

### Ray Vickson

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No, as written, the problem is NOT well-posed; did you read my answer? Basically, the question is asking for the largest or smallest ellipse centered at (0,0) that intersects the square; since the square is OPEN there are no maxima and minima. If the inequalities were non-strict there would be obvious solutions: the smallest "ellipse" would collapse to the point (x,y) = (0,0)---but that point is not in the specified square. Similarly, the largest ellipse would be the one passing through (1,1), but again, (1,1) is excluded from the specified open square. If the problem had, instead, specified NON-STRICT inequalities, the solutions would have been at (0,0) and (1,1). Both of these are Kuhn-Tucker points: (0,0) satisfies $f_x \geq 0$ and $f_y \geq 0$, which are the constrained versions of the optimality conditions at the lower bounds. The point (1,1) satisfies $f_x > 0$ and $f_y > 0$ which are the necessary conditions for a maximum at the upper bounds.

It may be that the OP was asked a well-posed question (having non-strict inequalities), but was careless in re-writing the problem here, perhaps not realizing just how important that issue actually turns out to be.

6. Oct 18, 2013

### brmath

Ray, what do you think a "global extremum is"?

7. Oct 18, 2013

### Ray Vickson

It is an "attained" infimum or supremum. In optimization books and papers it is usually said that the problem has no max and/or min if the supremum (or infimum) is not attained, as is the case in the problem posed in this thread. Many of the on-line sources (Wikipedia, etc.) are sloppy on this point, but one source has it right: see answer O in
http://math.stackexchange.com/questions/456546/what-is-the-definition-of-an-extremum-of-a-function

Again, in optimization one often speaks of local and/or global max or min over constrained regions, so a function like x^2 has a global max in the interval 0 <= x <= 1, even though, of course, it has no global max on the whole line. However, sticking with this example, the function x^2 on the open interval 0 < x < 1 does not have a maximum---it does, of course, have a supremum.

In this context, a point $x_0$ is a global max of $f(x)$ in the interval I if $x_0 \in I$ and $f(x) \leq f(x_0) \; \forall x \in I$. A local max is one in which $f(x_0) \leq f(x) \; \forall x \in I \cap N,$ for some neighborhood $N \subset \mathbb{R}$. This, of course, allows an endpoint of I to be a local or global max---provided that the endpoint is actually in I.

8. Oct 18, 2013

### brmath

Exactly. That is what was asked for, nothing more or less. This function has no attained supremum on the (open) square in question, and therefore has no max in the domain. However, the function certainly has an attained infimum at 0 and therefore has a minimum, which is global within that square, and in fact in the entire plane. There is nothing "sloppy" about any of this.

I did not look up anything on the internet in responding to this post. I have significant experience with max and min in multi dimensions and recently guided a senior college student through an honors paper which included determination of extrema as one of several chapters.

9. Oct 18, 2013

### Dick

Quarreling about exact definitions seldom leads anywhere. Definitions are just definitions. Ray stated his definition, you didn't. I don't see in what sense this function attains an infimum at 0 and doesn't attain a supremum. 'Global' to me means global in the domain of the function, not global outside of that domain. What's your definition?

10. Oct 18, 2013

### brmath

You know, I was thinking the square was centered at 0, which is silly because I knew perfectly well it was not, and in fact so defined it in my first post. That said, no the function has no extrema in the square. If it had been centered at 0, as I deludedly was thinking, it would have a minimum there

I stated my definition in the first post I sent:
As far as I can tell that is equivalent to what Ray said.

Thank you for joining this discussion.

11. Oct 18, 2013

### Dick

Yes, it is. So we all agree!

12. Oct 19, 2013

### Ray Vickson

Good. Now I can relax.