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Calculating Stationary Points of a Function

  1. Mar 10, 2010 #1
    1. The problem statement, all variables and given/known data

    Finding the stationary point(s) of the function:

    [tex]f(x,y) = xy - \frac{y^{3}}{3}[/tex]

    .. on the line defined by [itex]x+y = -1[/itex].

    For each point, state whether it is a minimum or maximum.

    2. Relevant equations

    .. within the problem statement and solutions.

    3. The attempt at a solution

    This is what I have so far:

    [tex]f(x,y) = xy - \frac{y^{3}}{3}[/tex]

    [tex]g(x,y) = x+y-1 = 0[/tex]

    Therefore need to extemise:

    [tex]F(x,y,\lambda) = f + \lambda g = xy - \frac{y^{3}}{3} + \lambda(x+y-1)[/tex]

    So calculating the partial derivatives:

    [tex]\frac{\partial F}{\partial x} = y + \lambda = 0[/tex]

    [tex]\frac{\partial F}{\partial y} = x - 3\left(\frac{y^{2}}{3}\right) + \lambda = x - y^{2} + \lambda = 0[/tex]

    [tex]\frac{\partial F}{\partial \lambda} = x + y - 1 = 0[/tex]

    Then need to look for all consistent solutions:

    [tex]1. y = \lambda[/tex]

    .. but now I'm stuck on what to do now, seemto have done something wrong because I can't get more consistent soluations and then nice simultaneous equations to equate :confused:
     
  2. jcsd
  3. Mar 10, 2010 #2

    tiny-tim

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    Hi Hart! :smile:

    (have a lambda: and a curly d: ∂ :wink:)

    (Actually, it's y = -λ, isn't it? :wink:)

    ok, now substitute that in the other two equations, and you'll get a quadratic equation in λ. :smile:
     
  4. Mar 10, 2010 #3
    OK.. So from the first partial derivative equation I get that:

    [tex]y = -\lambda[/tex]

    .. then put this into the second partial derivative equation, which gives:

    [tex]x - \lambda^{2} + \lambda = 0[/tex]

    .. hence:

    [tex]x = \lambda^{2} - \lambda[/tex]

    .. then put this into the third partial derivative equation, which gives:

    [tex]\lambda^{2} - \lambda - \lambda - 1 = \lambda^{2} - 2\lambda - 1 = 0[/tex]

    .. then solve quadratic equation to get values of [itex]\lambda[/itex]:

    [tex]\lambda = \frac{2 \pm \sqrt{(4) + (4)}}{2} = 1 \pm \sqrt{2}[/tex]

    .. hence:

    [tex]\lambda = 1 + \sqrt{2} , 1 - \sqrt{2}[/tex]

    .. yes??!

    .. note: but:

    [tex]\lambda = 1 \pm \sqrt{2} \neq - y[/tex]

    :confused:
     
  5. Mar 10, 2010 #4

    tiny-tim

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    What's wrong with that? :confused:

    y = -1 - √2, x = 2 + √2;

    y = -1 + √2, x = 2 - √2.
     
  6. Mar 10, 2010 #5
    um.. nothing! I don't know why I thought there was an issue there earlier! :redface: .. maybe just going a bit mad! :wink:

    Right, so since now have:

    [tex]y = -1 - \sqrt{2}, x = 2 + \sqrt{2}[/tex]

    [tex]y = -1 + \sqrt{2}, x = 2 - \sqrt{2}[/tex]

    .. then there are just 2 stationary points of the function, which are those 2 above?

    Also, to find if the points are maxima or minima, just need to put each set of [itex](x,y)[/itex] values into this equation:

    [tex]f(x,y) = xy - \frac{y^{3}}{3}[/tex]

    .. and if [itex]f(x,y) > 0[/itex] then maxima point, and if [itex]f(x,y) < 0[/itex] then minima point?
     
    Last edited: Mar 10, 2010
  7. Mar 10, 2010 #6

    tiny-tim

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    (why do you keep going into LaTeX?!)

    No, that only tells you the value of f(x,y) …

    you'll need more than that to check whether the value is a local maximum or minimum or stationary point.
     
  8. Mar 10, 2010 #7
    .. calculate second derivatives? negative = max point, positive = min point?
     
  9. Mar 11, 2010 #8

    tiny-tim

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    Hi Hart! :smile:

    (just got up :zzz: …)
    Yes, except you'll need the second derivatives "dot" the line. :wink:
     
  10. Mar 12, 2010 #9
    erm.. so for the first set of values I have:

    [tex]f(x,y) = xy - \left(\frac{y^{3}}{3}\right) = \left(\left(2+\sqrt{2}\right)\left(-1-\sqrt{2}\right)\right) - \left(\frac{\left(-1 - \sqrt{2}\right)}{3}\right) = \left(-4 - 3\sqrt{2}\right) - \left(-7 + 5\sqrt{2}\right) = 3 + 2\sqrt{2} [/tex]

    which is positive, so this is a minium stationary point.

    .. correct?!
     
  11. Mar 12, 2010 #10

    tiny-tim

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    what happened to the second derivatives? :confused:
     
  12. Mar 12, 2010 #11
    OH.. I just forgot them! :redface:

    ok, so:

    [tex]f(x,y) = xy - \left(\frac{y^{3}}{3}\right)[/tex]

    [tex]\frac{\partial f(x,y)}{\partial xy} = 1 - y^{2}[/tex]

    [tex]\frac{\partial f(x,y)}{\partial xy} = 2y[/tex]

    .. correct? (I'm not too good on partial derivations :frown:)

    So then:

    [tex]2y = 2\left(-1 - \sqrt{2}\right) = -2 - 2\sqrt{2}[/tex]

    .. which is positive.

    [tex]2y = 2\left(-1 + \sqrt{2}\right) = -2 + 2\sqrt{2}[/tex]

    .. which is negative.

    Hopefully somewhat getting there! :smile:
     
  13. Mar 12, 2010 #12

    tiny-tim

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    hmm … this has become needlessly complicated.

    It would have been easier at the beginning, instead of using the λ method (i know it has a name, but i've forgotten it :redface:), to simply use the directional derivative.

    Since g(x,y) in this case is a straight line, in the direction (1,-1), all you needed to do was to find the directional derivative, ∇gf, in that direction, ie ∂f/dx - ∂f/∂y, which was y + y2 - x, and put that = 0 (subject to x + y = 1).

    Substituting x = 1 - y gives ∇gf = 2y + y2 - 1 = 0 …

    this is the same final equation as before, but without using λ ! :wink:

    Now, to distinguish between maxima and minima, you need to know whether ∇gf is increasing or decreasing, so just consider d/dy (∇gf). :smile:
     
  14. Mar 12, 2010 #13
    .. I was doing in the [itex]\lambda[/itex] way (um.. lagrange multiplier?!) as this is how I was taught a similar problem and hence should really use this way.

    Looking back through my notes on this, once the values of x and y have been found (as they have been) then just put these into the original equation for f and then can deduce if the stationary point is a min or a max.

    Basically, for the first point:

    [tex]x = 2 + \sqrt{2}, y = -1 - \sqrt{2}[/tex]

    .. input the values into:

    [tex]f(x,y) = xy - \frac{y^{3}}{3}[/tex]

    and consider whether is positive or negative, and hence max or min point.

    Then repeat for second point.
     
  15. Mar 12, 2010 #14

    tiny-tim

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    (oh yes, Lagrange multiplier! :rolleyes:)

    But the problem with that method is that f(x,y) isn't necessarily positive at a maximum or negative at a minimum.

    You can change it slightly, and say that if there are only two turning-points, at A and B, then either f(A) > f(B) or f(B) > f(A) …

    in either case, the first one must be a maximum, and the second a minimum! :smile:

    (of course, that doesn't rule out points of inflexion, so perhaps for completeness the values towards infinity in both directions should also be considered)
     
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