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Stationary points

  1. Feb 8, 2008 #1
    1. The problem statement, all variables and given/known data
    determine the stationary points of the function f(x,y) = (x^2 + 1/2(xy) + y^2)e^(x+y)


    2. Relevant equations



    3. The attempt at a solution
    first i got

    df/dx= (x^2+(1/2)xy+y^2)e^(x+y)+(2x+(1/2)y)e^(x+y)

    df/dy= (x^2+(1/2)xy+y^2)e^(x+y)+((1/2)x+2y)e^(x+y)

    i then let them both = 0

    and i get

    (x^2+2x+(1/2)xy+(1/2)y+y^2)=0

    and

    (x^2+(1/2)x+(1/2)xy+2y+y^2)=0

    i've tried different to solve them simultaniously and then sub back in. I've tried to factorise them which i think i should do but cant seem to get it could anyone give me some pointers please?
     
  2. jcsd
  3. Feb 8, 2008 #2

    EnumaElish

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    Using the quadratic formula you can solve df/dx = 0 as y [itex]\in[/itex] {1/4 (-1 - Sqrt[1 - 32 x - 16 x^2 - 8 xy]), 1/4 (-1 + Sqrt[1 - 32 x - 16 x^2 - 8 xy])}, then sub into df/dy = 0.
     
    Last edited: Feb 8, 2008
  4. Feb 8, 2008 #3

    HallsofIvy

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    Subtracting the second from the first gives you (3/2)x- (3/2)y= 0 or y= x. Put that back into either equation (because of the symmetry) gives you [itex]x^2+ 2x+ (1/2)x^2+ (1/2)x+ x^2= (5/2)x^2+ (5/2)x= 0[/itex] or x(x+ 1)= 0.

     
  5. Feb 8, 2008 #4

    HallsofIvy

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    Subtracting the second from the first gives you (3/2)x- (3/2)y= 0 or y= x. Put that back into either equation (because of the symmetry) gives you [itex]x^2+ 2x+ (1/2)x^2+ (1/2)x+ x^2= (5/2)x^2+ (5/2)x= 0 or x(x+ 1)= 0[/itex].

     
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