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Global extrema, Calc III

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    I really, really don't understand. My work shows one thing, and a simple look at the function shows another.

    f(x,y) = .25x^2 + 5y^7 + 6y^2 - 3x

    Find global max and min if they exist.

    2. Relevant equations



    3. The attempt at a solution

    Took partial derivatives, solved each one for 0.

    .5x - 3 = 0 => x = 6
    35y^6 + 12y = 0 => y = 0

    So a critical point is at (6,0), period.

    Discriminant comes out to be

    [.5][210y^5 + 12] - 0

    Which comes out to be 6, since y = 0.

    Therefore the discriminant is > 0, and the partial wrt x is > 0, so a global min occurs at (6,0). Period.



    But wait! I submit this answer, and it basically says "nah, forget all that math crap. As y grows without bound, so does the function, so no global max or min."

    ???

    Shouldn't the normal process lead me to that conclusion SOME HOW?
     
  2. jcsd
  3. Feb 28, 2012 #2

    Dick

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    The discriminant doesn't tell you anything about whether it is a global max or min. It tells you whether is a local max or min.
     
  4. Feb 28, 2012 #3
    I see...

    but, why? and furthermore, local to what region?
     
  5. Feb 28, 2012 #4

    Dick

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    Local to the neighborhood of the critical point. And why? Because derivatives tell about the properties of a function near the point where you evaluate them, they don't tell you about what happens far away.
     
  6. Feb 28, 2012 #5
    That makes sense. What determines the boundaries of the neighborhood of the critical point?
     
  7. Feb 28, 2012 #6

    Dick

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    It depends on the class of function you are dealing with. In this case it would be the higher derivatives. But that's not even important. The big point is just that the discriminant only tells you about local properties. Let's just say, "some neighborhood".
     
  8. Feb 28, 2012 #7
    Got it, thanks a lot!
     
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