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Multivar Max Value

  1. Oct 3, 2006 #1
    OK, I have a question about this problem. I'm solving for the absolute maximum value for a function of two variables. I THINK i know what i'm doing, but feel free to rip into me and tell me that I'm clueless :frown:

    The function is
    [tex]f(x , y) = \frac{-2x}{x^2+y^2+1}[/tex]

    on the boundary of [tex] \left\{\begin{array}{r} 0 \leq x \leq 2\\ \|y\| \leq x \end{array}\right.[/tex]

    Anywho, I've computed the partials easily enough
    [tex]F_x = \frac{2(x^2-y^2-1)}{(x^2+y^2+1)^2}[/tex]
    [tex]F_y = \frac{4xy}{(x^2+y^2+1)^2}[/tex]

    So, i set each of those equal to zero to determine when there will be a critical point. So they will occur when:

    [tex] \left\{\begin{array}{r} x^2-y^2-1=0\\xy=0 \end{array}\right.[/tex]

    So right of the bat I know that (0,0) and (1,0) are critical points. But what i'm not sure of is whether or not that's all of them. I'm pretty sure that it is, but I'm having a bit of trouble convincing myself why non-integer values of x between 0 and 2 won't work.
    Last edited: Oct 3, 2006
  2. jcsd
  3. Oct 3, 2006 #2


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    Think of it in terms of the sets of points where those equations are true. x^2-y^2=1 is true on a hyperbola intersecting the x-axis at +/-1. xy=0 is true when x or y is 0, so along both the x and y axes. So what are the points in both these sets, and in the domain of interest? ( (0,0) isn't a critical point by the way).
  4. Oct 3, 2006 #3
    i'm afraid that didn't help me too much. I'm not entirely sure i understand what you mean by "So what are the points in both these sets, and in the domain of interest?"

    I'm not sure how to get from the end of my first post to having the crit points determined

    edit: wait, i see now. It's the points along the x axis where the hyperbola intersects. So (0,1) and (0,-1) right? because the hyperbola eq'n doesn't interset the y axis at all.
    Last edited: Oct 3, 2006
  5. Oct 3, 2006 #4


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    Right. And you don't care about (0,-1), right?
  6. Oct 3, 2006 #5
    right :)

    thanks a lot for your help StatusX!
  7. Oct 3, 2006 #6
    wait, those should both be (1,0) and (-1,0) right? Not (0,1)...
  8. Oct 3, 2006 #7


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    Yea, x=1, y=0.
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