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Absolute max/min on an unbounded set

  1. Apr 25, 2008 #1
    1) Find the global max and min values of the function
    f(x,y)=x/[x2+(y-1)2+4] on the first quadrant S={(x,y)|x,y>0}


    Solution: (from textbook example)
    f(x,y)>0 on S and f(0,y)=0, so the minimum is zero.
    Moreover, f(x,y) is less than the smaller of 1/x and 1/(y-1)2, so f vanishes as |(x,y)|->∞. Hence, by theorem (*), f has a maximum on S. This justifies the "existence" of an absolute max on S.
    ...

    theorem (*): Let f be a continuous function on an unbounded closed set S in Rn. If f(x)->0 as |x|->∞ (x E S) and there is a point xo E S where f(xo)>0, then f has an absolute maximum on S.
    ==================================

    I understand why f(x,y) is less than 1/x, but I can't figure out why f(x,y) is less than 1/(y-1)2. Can someone help me, please?

    Secondly, why does f vanish as |(x,y)|->∞? There are many different paths for the length to go to infinity, e.g. x can go to infinity while y is 0 (along x-axis), or y can go to infinity while x is 0 (along y-axis), or x and y may both go to infinity, how can we be sure that in ALL directions to infinity, f will vanish?

    Thanks anyone for explaining!:smile:
     
    Last edited: Apr 25, 2008
  2. jcsd
  3. Apr 25, 2008 #2

    exk

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    f(x,y) is less than 1/(y-1)^2 because no matter what value of x you use the function will evaluate to either 0 for x=0 or something smaller than 1/(y-1)^2. Suppose x=1, you get 1/[5+(y-1)^2], which is smaller than the previous, etc...

    Taking the limit as either variable goes to infinity (keeping the other constant) evaluates to 0. Taking them both as they go to infinity at the same time also goes to 0 (i am not sure if the following explains it for this last case). Apply l'Hopital to evaluate the limit for x , use x=c to evaluate for y, and use l'Hopital with respect to either variable I guess to evaluate it for both at the same time.

    Ok the explanation I just gave is horrible, but it makes sense in my mind.
     
  4. Apr 25, 2008 #3
    But x=1 is just a very special case, how can we prove rigorously using inequalities step-by-step showing that x/[x2+(y-1)2+4] < 1/(y-1)2?



    I believe that |(x,y)|->∞ implies that x->∞ or y->∞ (or both), am I right?
    Since 0 < f(x,y) < 1/x AND 0 < f(x,y) < 1/(y-1)2
    Either x->∞ or y->∞ (or both). If x->∞, apply squeeze theorem to the first inequality. If y->∞, apply squeeze theorem to the second inequality. In either case, f->0. I think this may justify that f vanishes as |(x,y)|->∞. Please let me know if someone is wrong.


    Thank you!
     
  5. Apr 26, 2008 #4

    exk

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    That's pretty much what I was describing, taking x=1 as a special case was just to show an example, in fact what I said and you reaffirmed is that the statements are true for any x or y.

    I forget the exact wording for the squeeze theorem, but if it's anything along the lines of "bigger function converges then so does smaller" then you have your proof.

    What you really need to show rigorously is that f(x,y) is in fact bounded by those 2 functions choosing y=1 for proving 1/x and x=0 for the other one could be used to construct 2 proofs for both variables using mathematical induction. (it doesn't matter that you don't use negative integers in the proof because for the "y" case you get a negative #<positive # due to the squared term on the rhs and in the "x" case you get the squared term on the lhs and that takes care of the negative y values. by symmetry you prove it for all reals, but that doesn't even matter since S is strictly in the positive quadrant.)
     
  6. Apr 26, 2008 #5

    Dick

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    You can't prove that because it's not true. Take y=11 and x=2. Then your inequality becomes 2/108<1/100. That's not true. Your book was playing you for an easy solution sucker. Good job for seeing through it. But just because the book solution lied doesn't mean the conclusion isn't true, does it?
     
    Last edited: Apr 26, 2008
  7. Apr 26, 2008 #6
    Oh, I can't believe that my textbook is lying to me and giving me unnecessary headaches struggling to understand why it is true while in fact it is false :rolleyes:

    So the textbook is wrong!

    Then how can we justify the "existence" of an absolute max on S?
     
  8. Apr 26, 2008 #7

    Dick

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    Think about it. There is an x^2+y^2 in the denominator. Doesn't this go to infinity faster than anything in the numerator or the rest of the denominator regardless of direction? You can do a better job than the textbook solution.
     
  9. Apr 26, 2008 #8
    f(x,y)=x/x2+y2-2y+5

    Sorry, I don't understand...why goes to infinity faster regardless of direction? How can I analyze the situation in every direction?
     
  10. Apr 26, 2008 #9

    Dick

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    What's the limit of x/(x^2+y^2)? Of y/(x^2+y^2)? Etc.
     
  11. Apr 26, 2008 #10

    Vid

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    One way to see it is to convert to polar
    |(x,y)| = sqrt(x^2 + y^2) = r
    t is theta
    rsin(t)/(r^2sin(t) + (rcos(t) - 1)^2 + 4)
    rsin(t)/(r^2sin(t) + r^2cos^2(t) - 2rcos(t) + 5)
    lim r->inf = 0

    Since the limit as r goes to inf doesn't contain any reference to theta, the limit vanishes in all directions.
     
  12. Apr 27, 2008 #11
    Do you mean
    lim x/(x^2+y^2)
    x,y->∞
    and
    lim y/(x^2+y^2) ?
    x,y->∞

    If so, then I think both are 0 because the denominators dominate.
     
  13. Apr 27, 2008 #12
    I think polar is a good idea!

    lim r->inf = 0
    Is this becuase the denominator has a higher degree and dominates?
     
  14. Apr 27, 2008 #13

    Dick

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    Yes. If r=sqrt(x^2+y^2), then |x|<=r and |y|<=r. And r/r^2 goes to 0 as r->infinity.
     
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