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Local extrema w/2 var

  1. Aug 24, 2005 #1
    we're doing partial derivatives, and i thought i understood them until:

    [tex]f(x,y)=-x^2-y^2-10xy+4y-4x+2 [/tex]

    we are meant to find and classify all local extrema. i got:

    [tex]f_x=-2x-10y-4 [/tex] and [tex] f_y=-2y-10x+4[/tex]

    i've never had two variables at this stage, and i can't solve for any of the points. the text doesn't seem to say anything about this...
  2. jcsd
  3. Aug 24, 2005 #2
    yay! i'm now a level 30 thumb twiddler
  4. Aug 24, 2005 #3


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    Homework Helper

    You solve [tex]\nabla f = \vec0[/tex] (gradient of f) which basically gives you a system of 2 equations here. The solutions are the 'stationary points' which may be extrema, but you have to check that (e.g. with a test)

    So you solve:

    [tex]\left\{ \begin{gathered}
    \frac{{\delta f}}
    {{\delta x}} = 0 \hfill \\
    \frac{{\delta f}}
    {{\delta y}} = 0 \hfill \\
    \end{gathered} \right.[/tex]
  5. Aug 24, 2005 #4
    mm, i get: [tex]10y=-2x-4 [/tex] and [tex]10x=-2y+4 [/tex] for each of which there are infinite solutions for x and y. i thought that if i combined the equations (no reason) it might do it, but i still got [tex]y=x-1 [/tex]
  6. Aug 24, 2005 #5


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    Homework Helper

    Yes, you have to combine them in a system.

    [tex]\left\{ \begin{gathered}
    2x + 10y + 4 = 0 \hfill \\
    10x + 2y - 4 = 0 \hfill \\
    \end{gathered} \right.[/tex]

    Solving it should give [itex]\left( {\frac{1}
    {2}, - \frac{1}
    {2}} \right)[/itex]

    However, although it's a stationary point, it's not an extremum.
  7. Aug 24, 2005 #6
    aha! thanks, i get it now
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