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Homework Help: Lagrange Multipliers, maybe?

  1. Aug 15, 2010 #1
    1. The problem statement, all variables and given/known data
    See figure

    2. Relevant equations

    3. The attempt at a solution

    Alright we'll this is my first shot at a question like this, so in all honesty I don't know what concepts this question is testing.

    It mentions finding absolute max/min of a function inside a specific region.

    Is this a Lagrange multipliers problem with 3 constraints? Or is it that this is something else? Perhaps there's another more efficient method of solving this?

    Thanks again!

    Attached Files:

  2. jcsd
  3. Aug 15, 2010 #2


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    Gold Member

    Sketch the region. The answer is obvious
  4. Aug 15, 2010 #3
    See figure attached.

    The answer still isn't very obvious for me.

    Attached Files:

    Last edited: Aug 15, 2010
  5. Aug 15, 2010 #4


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    Gold Member

    Strange. Try plugging in the boundary values maybe youll get lucky. this almost feels like a simplex problem
  6. Aug 15, 2010 #5
    I've been doing independent study on all this stuff so things that may seem obvious to you aren't necessairly obvious to me.

    I'm not entirely sure what you mean why boundary values, is it like the point where [tex]y=x[/tex] and [tex]y= \sqrt{1-x^2}[/tex] intersect?

    I can take a guess at the absolute maximum but I don't think it's right...

    Maybe the point [tex]\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)[/tex]?
  7. Aug 15, 2010 #6
    Try analyzing the points where [tex]\nabla f[/tex] is 0.

    Edit: Without the | |.
    Last edited: Aug 15, 2010
  8. Aug 15, 2010 #7

    [tex] \nabla f = \left( 2x + 2y\right)\hat{i} +\left(2x-2y\right)\hat{j} [/tex]

    So I'm trying to get this to equal the zero vector [tex]\vec{0}[/tex] ?

    If so, then x=0 and y=0.

    I'm still kinda lost.
  9. Aug 15, 2010 #8
    This is an inflection point then. What does this tell you about the function at this point?
  10. Aug 15, 2010 #9
    This would be our absolute minimum.

    The part I'm confused about is how you came to recognize this was an inflection point by equating the gradient of the function to zero.

    Can you explain?
  11. Aug 15, 2010 #10
  12. Aug 15, 2010 #11
    The gradient geometrically tells you how a function is instantaneously changing at any given point. By setting it to 0, you are then looking for points where the function is "flat". This happens in only one of three circumstances, the point you found is a minimum, a maximum, or a saddle point.

    In order to determine the type of point, you have to analyze the Hessian matrix.
  13. Aug 15, 2010 #12
    So now that we've found the minimum, how can we establish the absolute maximum within the bounded region?
  14. Aug 15, 2010 #13
    Take a look at the extreme value theorem (just google it). What it tells you is that the max and min of a continuous function must occur either within the region, or on its boundary. For your case you have shown that the gradient is 0 only at one point which does not lie on the interior of the region. By the extreme value theorem this tells you that the max and min values must occur on the boundary of the region.
  15. Aug 15, 2010 #14
    Then the maximum will simply be at the bounday of the region.

    Therefore the absolute maximum is at the point,

    [tex]\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)[/tex]

  16. Aug 15, 2010 #15
    It could be, I haven't done the calculation. To figure it out you would have to plug y = x and x = sqrt(1-y^2) into your equation for f(x,y). At this point in time it makes sense to check out that link I gave you above since it outlines the procedure and gives you an example.
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