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Global extreme values

  1. Jul 29, 2008 #1
    find global min/max: f(x,y)= x^3-2y ; 0<=x, y<=1





    3. The attempt at a solution

    - I know that the critical points don't work since there are none, but what I'm not understanding is how to draw the square from 0<=x, y<=1. The solutions has the points (0,0), (1,0), (1,1), (0,1), but I'm not sure how you determine that. I guess what is getting me is why x is limited at 1 for the square. I understand how to work the problem except for this part. Please help.
     
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  3. Jul 29, 2008 #2

    HallsofIvy

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    If there are no critical point in the interior of the square, there cannot be a global max or min there. So the global max or min must be on the boundary. It will be best to handle that in four parts:
    1) x= 0. f(0, y)= -2y for 0<= y<= 1.
    2) x= 1. f(1, y)= 3- 2y for 0<= y<= 1.
    3) y= 0. f(x, 0)= 3x2 for 0<= x<= 1
    4) y= 1. f(x, 1)= 3x2- 2 for 0<= x<= 1
    You will need to check each point at which the derivative is 0 in those intevals.

    And, of course, a max or min might occur at an endpoint of any one of those intervals. That's what gives you the vertices as possible max or min points.
     
  4. Jul 29, 2008 #3
    - I understand how to work the problem, what I'm unsure about is how you chose the points. if x is >=0, then why is 1 chosen? Couldn't x be any point from 0->infinity?
     
  5. Jul 29, 2008 #4

    HallsofIvy

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    What is the boundary of the region you are given?
     
  6. Jul 29, 2008 #5
    x=0 and y=1, correct?
     
  7. Jul 29, 2008 #6

    Defennder

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    Which part of the solution are you addressing? There are 4 parts.
     
  8. Jul 30, 2008 #7

    HallsofIvy

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    No, that is not correct. Try again. The boundary of a square is four line segments.
     
  9. Jul 30, 2008 #8
    This is the part I'm having trouble figuring out. All we're given is that 0<=x and y<=1. That only gives 2 boundaries. So I'm trying to figure out how they got the other two. I would think what was given is not a closed region (only two boundaries). Maybe I'm just thinking too much about this problem and not seeing something. with the information given, I want to know how you determine the points (0,0), (1,0), (1,1) and (0,1) (those are the 4 points used in the solution to make up the square).
     
  10. Jul 30, 2008 #9

    HallsofIvy

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    No, no, no!! "Only 2 boundaries" wouldn't give a closed and bounded region and there would be not be a "global maximum". "[itex]0\le x, y \le 1[/itex]" is standard shorthand for "[itex]0\le x\le 1[/itex] and [itex]0\le y\le 1[/itex]".
     
  11. Jul 30, 2008 #10
    - Okay, thats all I needed to know. Nothing in the book (not that I've seen) said anything about that being shorthand notation. Thats the only thing I was trying to figure out, was how they were getting the boundaries...the problem itself is easy. Are there any other short hand notations that are used when defining boundaries? That seems like a pretty important thing to note if its going to be used in a book (I guess it should be pointed out that my school changed books, so I had a different book that never used short hand notation for my first 2 calculus classes).
     
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