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Homework Help: Stationary Points

  1. May 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the four stationary points of the function:
    u(x, y) = 4x^3 − 18(x^2)y + 24x(y^2) − 120y

    Determine whether they are maxima, minima or saddle points.

    2. Relevant equations

    To find stationary points use:

    E= (d^2u/dxy)^2 - [(d^2u/dx^2) * (d^2u/dy^2)]

    E>0 saddle
    E < 0 Maximum -> d^2u/dx^2 < 0
    Minimum -> d^2u/dx^2 > 0

    3. The attempt at a solution

    du/dx = 12x^2 - 36xy +24y^2 (1)
    du/dy = -18x^2 + 48xy-120 (2)

    Stationary points mean both du/dx and du/dy are equal to 0. Here I should find simultaneous solution of equations (1) and (2). This is where I get stuck and I am not sure how to find them.

    I have done the next part though:

    E= (-36x + 48y)^2 - (1152x^2 - 1728xy)

    But I need the stationary points to find the min/max/saddle points.
     
  2. jcsd
  3. May 2, 2010 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Try factoring one of those equations.
     
  4. May 2, 2010 #3
    Ok i've tried:

    du/dx = 12x^2 - 36xy +24y^2 (1)
    du/dy = -18x^2 + 48xy-120 (2)

    (1) x^2 - 3xy + 2y^2 = 0
    (2)-3x^2 + 8xy -20 = 0

    Eliminate x^2 ; xy +6y^2 - 20= 0 ; x = 20/y - 6y
    Or
    Eliminate xy ; x^2 - 16y^2 + 60 = 0 (not sure how to factorise)

    Do I sub that back into the previous equtions?

    I am still not sure where to go form here.
     
  5. May 2, 2010 #4

    Mark44

    Staff: Mentor

    The first equation can be written in factored form as
    (x - y)(x - 2y) = 0

    BTW, you should post problems like this in the Calculus & Beyond section, not this section.
     
    Last edited by a moderator: May 2, 2010
  6. May 3, 2010 #5
    Thank you! I think I know what I need to do now with the other equation now.

    Noted, i'll post in the other section.
     
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