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Describing global minimum

  1. Mar 11, 2007 #1
    Does this Fxy have a global Min?

    Heres the question:

    Consider the function fxy= x+y+9/x+1/y. Determine all the local max min and saddle points. DOes f have any global maximum points in the region R where x,y>0 Explain algebraically.

    So I found only one critical point, 3,1 and found it to be a local min.
    As for the global part, I said NO, because the function does not have a boundry x and y goto infinity and this function does not the sort of parabolic shape to have a natural global max or min.

    Is that right? How could I be more graceful with my answer?

    Thanks geniuses :wink:
     
  2. jcsd
  3. Mar 11, 2007 #2
    Everyones lookin and noones sayin anything. :D
     
  4. Mar 11, 2007 #3
    Heres the question:

    Consider the function fxy= x+y+9/x+1/y. Determine all the local max min and saddle points. DOes f have any global maximum points in the region R where x,y>0 Explain algebraically.

    So I found only one critical point, 3,1 and found it to be a local min.
    As for the global part, I said NO, because the function does not have a boundry x and y goto infinity and this function does not the sort of parabolic shape to have a natural global max or min.

    Is that right? How could I be more graceful with my answer?
     
  5. Mar 11, 2007 #4

    mjsd

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    if a function is unbounded from below, it has no global minimum, eg. 1/x has no minimum
     
  6. Mar 11, 2007 #5
    I said, the function does not have any bounds so it does not have a global max.
    I graphed the function with my handy dandy computer graphing program and it
    looks like it could have a global max in the positive x/y region although a portion of the function does goto infinity.

    ?
     
    Last edited: Mar 11, 2007
  7. Mar 11, 2007 #6
    Remember to consider BOTH roots of the grad f = 0 equations, because you get quadratics in both df/dx = 0 and df/dy = 0 (partial derivatives obviously), so that gives you 4 local turning points.

    You're right about no global min, you can see that by considering large and small y and x values, it's not bounded above or below.
     
  8. Mar 12, 2007 #7

    HallsofIvy

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    ?? Do you understand what a global maximum is? If any portion of the function goes to infinity, no finite number can be a global maximum!

    (Local maximum, possibly)
     
  9. Mar 12, 2007 #8

    HallsofIvy

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    Note that it was specified that x,y> 0. The only one of the four critical points in the first quadrant is the one she specified, (3, 1).

    By the way, this same question was posted in the "homework" section. Please do not double post! I am going to merge the two threads.
     
  10. Mar 12, 2007 #9
    "If any portion of the function goes to infinity, no finite number can be a global maximum!"

    I dont think thats true. X^2 has a global min, even if it does goto infinity.
     
  11. Mar 13, 2007 #10
    if you know some tricks about inequality... this problem can be done without ANY calculus.

    by AM-GM (arithmetic mean is greater or equal to geometric mean)

    [tex]\frac{a+b}{2}\ge (ab)^{1/2}[/tex]
    for a, b>0
    with equality iff a=b

    then
    [tex]x+y+9/x+1/y \ge 2\sqrt{x\cdot 9/x}+2
    sqrt{y\cdot 1/y}=7[/tex]

    for equality case, x=9/x, y=1/y, x=3, y=1

    the minimum is [tex]4\sqrt{3}[/tex].

    edit: fixed an embarrassing mistake: the previous inequality is correct but the equality case is impossible.
     
    Last edited: Mar 13, 2007
  12. Mar 13, 2007 #11

    HallsofIvy

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    ?? yes, of course, x2 has a global minimum but neither you in your post, nor I in mine said anything about a minimum.
     
  13. Mar 13, 2007 #12

    mjsd

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    mm... going to infinity and a function has a minimum is two separate thing anyway.. you need it to go to negative infinity then you can't have a global min...common sense it seems.. but r u just confused with what the words mean?
     
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