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Global extreme values

  1. Jul 1, 2008 #1
    determine the global extreme values of the function on the given domain:

    f(x,y)=x^2+2y^2 , 0<=x,y<=1





    3. The attempt at a solution

    - I know you need to evaluate the critical points first and then use the boundary to find the maximum and minimum values, but I can't seem to get the right answer. I think I'm just misunderstanding the way you go about a problem like this. Please help.
     
  2. jcsd
  3. Jul 1, 2008 #2

    Defennder

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    Finding critical points is one way to solve it, but it's a little tedious and unnecessary. Just start by examining the function [tex]f(x,y) = x^2 + 2y^2 [/tex].

    Note that the function is greater or equal to 0. That should tell you something about the extreme values. And note that while x is bounded below, it's not bounded above, what does that say about the function?
     
  4. Jul 1, 2008 #3
    - would the function always being great than or equal to zero mean that the global minimum would be 0?

    with x not being bounded above, that would mean that it contains points arbitrarily far from the origin, but I'm not sure what that would mean for finding the maximum.
     
  5. Jul 1, 2008 #4

    Defennder

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    Yes, for the first part.

    For the second, does the question say anything about the function necessarily having a maximum?
     
  6. Jul 1, 2008 #5
    - no, but the answer in the back of the book has a maximum and a minimum (0). I just don't see how they got the answer they got for the maximum.
     
  7. Jul 1, 2008 #6

    Dick

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    If the function has a max, it must be on the boundary of 0<=x,y<=1, yes?
     
  8. Jul 2, 2008 #7
    - right. so how would you set up the problem to find that max? the problem before it, I wasn't sure if I did it right, but I got the right answer. I did this problem the same way and got the wrong answer, so I just think I got lucky doing the problem before incorrectly. So I guess I'm not really sure how to set a problem like this up.
     
  9. Jul 2, 2008 #8

    HallsofIvy

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    What does your book say that maximum is?
     
  10. Jul 2, 2008 #9

    Dick

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    The boundary consists of four line segments. One of them is x=0, 0<=y<=1. What's the maximum along that segment? What about the other three?
     
  11. Jul 2, 2008 #10
    - the book has the max at 3, but I got 2 for my max.
     
  12. Jul 2, 2008 #11

    Dick

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    Try ALL of the boundary segments. How about x=1, 0<=y<=1?
     
    Last edited: Jul 2, 2008
  13. Jul 2, 2008 #12
    - this is what I did to try and find my maximum value:

    y=0, 0<=x<=1 >>> f(x,0)=x^2 >>> f(1,0) = 1

    y=1, 0<=x<=1 >>> f(x,1)=x^2+2 >>> f(0,0)=2

    x=0, 0<=y<=1 >>> f(0,y)=2y^2 >>> f(0,1)=2

    x=1, 0<=y<=1 >>> f(1,y)=1+2y^2 >>> f(1,0)=1

    I have no clue if thats the correct way of setting these problems up because the book isn't very clear on it, but thats how I got the answer for the first problem and it was right.
     
  14. Jul 2, 2008 #13

    Dick

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    Somehow, you are completely missing the possibility that x=1 and y=1. f(1,1)=3.
     
  15. Jul 2, 2008 #14

    HallsofIvy

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    Basically, then, your error was not looking at the "boundaries of the boundaries"- that is, the endpoints of the line segments.
     
  16. Jul 2, 2008 #15
    - alright, I see now. I was going a lot off the one example in the book and they didn't use that possibility (just the 4 I used). Can somebody explain the proper way to set these problems up?
     
  17. Jul 2, 2008 #16
    So, we need to look at total of 12 cases (some will be duplicates)?

    4 cases he described and 2 more for each of those cases
    1)y=1, 0<=x<=1 >>> f(x,1)=x^2+2 >>> f(0,0)=2
    2) y =1, x = 0
    3) y = 1,x = 1

    In case of circular dimension/any dimension that can be described by a function, we do two times?
    >find min/max of area
    >find min/max along the circular line (using deltas/something)
     
  18. Jul 2, 2008 #17

    Defennder

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    Oh wow, I thought he meant x>=0 and y<=1 rather than both x,y in [0,1].
     
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