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Find All local maxima and minima and all saddle points of the function

  1. Mar 29, 2012 #1
    1. The problem statement, all variables and given/known data
    ## f\left( x,y\right) =x^{2}-4xy+6x-8y+2y^{2}+10 ##

    ## f_{x}=2x-4y+6=0 ##
    ## f_{y}=-4x-8+4y=0 ##

    ## f_{y}=-4\left( x-y+2\right) ##
    -2=x-y, then solving fx and using this equality

    ## f_{x}=0=2\left( x-2y+3\right) =0 ##

    2(-2 -y+3)=0
    2(1-y)=0
    y=1, then pluggin it to values

    -2=x-y
    -2=x-1
    -1=x, So critical points at (-1,1)

    fxx(-1,1)fyy(-1,1)-0^2=8, which is greater than zero and fxx is too. Therefore, there is a global minima at (-1,1)

    There is no saddle point or global maxima?

    2. Relevant equations



    3. The attempt at a solution

    Is this correct?
     
  2. jcsd
  3. Mar 29, 2012 #2

    LCKurtz

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    Ok until right there. How did you get ##f_{xy}=0\ ##?
     
  4. Mar 31, 2012 #3
    You are right fxy= -4.

    So,
    fxx(-1,1)fyy(-1,1)-(-4^2)=
    fxx=2;
    fyy=4;
    8-16=-8
    Answer:
    So, (-1,1) is a saddle point and there is not enough information(Domain) to find local extrema.

    Is this answer correct?
     
  5. Apr 1, 2012 #4

    LCKurtz

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    Yes, it is a saddle point. But I wouldn't say "there is not enough information(Domain) to find local extrema". I would say there are no relative extrema.
     
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