1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding local max, min and saddle points

  1. Oct 3, 2011 #1
    1. The problem statement, all variables and given/known data

    f(x,y)=(1+xy)(x+y)



    2. Relevant equations



    3. The attempt at a solution

    I started out by expanding and got:

    [itex]x+y+x^2y+xy^2[/itex]

    Then I found all my partial derivatives and second derivatives:

    [itex]f_{x}=1+2xy+y^2, f_{y}=1+2xy+x^2, f_{xx}=2y, f_{yy}=2x, f_{xy}=2(x+y)[/itex]

    I know that both first partial derivatives must equal zero so I get:

    [itex]f_{x}=1+2xy+y^2=0[/itex] and [itex]f_{y}=1+2xy+x^2=0[/itex]

    This is the part I am stuck at; I can't find the critical points. I notice that there is symmetry so I tried subtracting the equations but I got y=x and got:

    [itex]f_{x}=1+2x(x)+(x)^2=1+2x^2+x^2=0=1+3x^2=0---->x^2=-\frac{1}{3}[/itex]

    I also tried setting [itex]f_{x}[/itex] and [itex]f_{y}[/itex] equal to each other but that didn't seem to work.

    Thanks in advance for the help
     
    Last edited by a moderator: Oct 3, 2011
  2. jcsd
  3. Oct 3, 2011 #2
    I think I may have got it. When I got that [itex]x^2=y^2[/itex] I didn't account for that y could equal (-x). I did that and it got the right answer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding local max, min and saddle points
Loading...