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Finding max and min of a function of several variables

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Find max/min/saddle point of the function :

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

    = x^2*y + y^2*x + x + y

    F_x = 2xy + y^2 + 1
    F_xx = 2y

    F_y = x^2 + 2yx + 1
    F_yy = 2x

    Need help on finding the critical points :

    So I have to solve this equations :

    F_y = x^2 + 2yx + 1
    F_x = y^2 + 2yx + 1
     
  2. jcsd
  3. Oct 18, 2009 #2

    Dick

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    Subtract the two equations and get x^2-y^2=0. That's (x+y)(x-y)=0. Suggests you split the solution into two cases. Can you go from there?
     
  4. Oct 18, 2009 #3
    I think its supposed to be (y+x)(y-x). What should I do next?

    That function will be 0 when x = y or when x = -y


    Now do I plug it in into F_x or F_y ?
     
  5. Oct 18, 2009 #4

    Dick

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    (x+y)(x-y)=0 and (y+x)(y-x)=0 are both the same condition. Assume x=y and try to solve F_x=0 and F_y=0, any solutions? Now assume x=(-y) and do the same exercise.
     
  6. Oct 18, 2009 #5
    So we have :

    F_y = x^2 + 2yx + 1
    F_x = y^2 + 2yx + 1

    and our test case is x = y , -y

    case : x = y

    F_x = y^2 + 2yx + 1
    = y^2 + 2y^2 + 1 = 0

    3y^2 = - 1

    x = y has no real answer


    case x = -y

    F_x = y^2 + 2yx + 1

    = y^2 - 2y^2 + 1

    -y^2 = -1

    y = 1, -1

    so when x = -y , y has to equal -1 or 1 for the function F(x,y) to have a horizontal plane.


    Using the 2nd derivative test

    D = F_xx*F_yy -[F_xy]^2

    = (2y)(2x) - (2x + 2y)^2

    = 4xy - 4x^2 + 8xy + 4y^2

    = xy - x^2 + 2xy + y^2

    = y^2 - x^2 + 3xy

    so inputting values are (-y, - 1) and (-y , 1)

    Case P(-y,-1)

    D = y^2 - x^2 + 3xy

    = (-1)^2 - (-y)^2 + 3(-y)(-1)
    = 1 - y^2 + 3y

    0 = y^2 - 3y - 1

    Then what, use calculator to find its root?
     
  7. Oct 18, 2009 #6

    Dick

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    Aren't your critical points (1,-1) and (-1,1)? Why are you treating y like some unknown value? I thought you solved for it.
     
  8. Oct 18, 2009 #7
    Oh so in this snippet :

    case x = -y

    F_x = y^2 + 2yx + 1

    = y^2 - 2y^2 + 1

    -y^2 = -1

    y = 1, -1


    I just input y back into the equation x = -y ?
     
  9. Oct 18, 2009 #8

    Dick

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    Sure you do. x=(-y). You assumed that. And with good reason.
     
  10. Oct 18, 2009 #9
    Ok thanks then.

    I can solve it from here .

    So to recap (because it helps me learn better ) I had to do the following.

    For the function F(x,y) = ... , find F_x, F_xx, F_xy, F_y , F_yy

    Then I had to solve the system of equation of F_x, and F_y, in which I
    found the equation x^2 - y^2, which has two cases where it will have a critical point, at x = y and at x = -y.

    Inputting this values into F_x or F_y gives me a non real answer for x = y and for x = -y,
    it gives me y = -1, 1. So I input those values into the equation x = -y, (question : technically could I have also inputed x = y , -y into the equation F_x or F_y or x^2 - y^2 ? ). From that I have two point , (1,-1) and (-1,1).

    With these points I can use the second derivative test to check whether its a saddle point minimum point or a maximum point.

    That should be correct , right ?
     
  11. Oct 18, 2009 #10

    Dick

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    Sure that's correct. I'm a little unsure what you are unsure about. To be a critical point you need F_x=F_y=0. If F_x-F_y=(x+y)(x-y)=0, then you can certainly conclude that for any critical point either x=y or x=(-y), right?
     
  12. Oct 18, 2009 #11
    Yep, thanks for your help. I really appreciate it. I have an exam tomorrow at 9am.
     
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