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Using partial derivatives to find a maximum

  1. Aug 6, 2006 #1
    Okay, I have a question that I am not good enough at multivariable calculus yet to answer myself.

    Basically, say I have the following inequality:

    a < sin(xy)/y < b

    with a < b. How can I find out what the maximum value of y is on the interval 0 < y < 1 such that the above inequality holds for some x > 0?

    That is, what is the biggest value of y between 0 and 1, such that for some x > 0, sin(xy)/y is between two limits a and b with a < b?

    Is this even possible...?

    Any information, hints, outright answers, scathing insults etc. would be appreciated! Thank you.
     
  2. jcsd
  3. Aug 6, 2006 #2

    Office_Shredder

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    Given complete control of x, xy can be whatever you want. So imagine instead xy = z, we can assume z is completely independent of y (since we can pick x to satisfy what we need). So you really have:

    a<sin(z)/y<b

    Keeping in mind that sin(z) is between 0 and 1, can you work from there?
     
  4. Aug 7, 2006 #3
    Hopefully this will be enough to solve your problem:

    Theorem: If a function has local extrema at a point, and the first partial derivatives of the function exist at that point, then the first derivatives of the function vanish at that point.

    Theorem: For every function f(x, y), define D(a, b) = [tex]\frac{\partial ^2 f}{\partial x^2} \frac{\partial^2 f}{\partial y^2} - \frac{\partial^2 f}{\partial x \partial y} \frac{\partial^2 f}{\partial y \partial x}[/tex] everywhere the first partial derivatives vanish and the second partial derivatives exist.

    Then, if D(a, b) < 0, then (a, b) is a saddle point of f
    If D(a, b) > 0, and the second derivatives are positive, then (a, b) is a local minimum.
    If D(a, b) > 0, and the second derivatives are negative, then (a, b) is a local maximum.
    If D(a, b) = 0, then any of saddle point, local maximum or minimum, or no extrema could occur at that point - no new information is known.
     
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