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Max/min with partial derivatives

  1. Jul 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that f(x,y) = -(x^2 - 1)^2 - (yx^2-x-1)^2 has only two critical points, and both are maxima.

    3. The attempt at a solution
    Set partial derivatives (wrt x and y) to zero to find critical pts.

    f_x = -2(x^2 - 1)(2x) - 2(yx^2 - x - 1)(2xy - 1) = 0
    f_y = -2(yx^2 - x - 1)(x^2) = 0


    -2(x^2 - 1)(2x) - 2(yx^2 - x - 1)(2xy - 1) = -2(yx^2 - x - 1)(x^2)
    (x^2 - 1)(2x) + (yx^2 - x - 1)(2xy - 1) = (yx^2 - x - 1)(x^2)
    2x^3 - 2x + 2(x^3)(y^2) - 2(x^2)y - 2xy - yx^2 + x + 1 = yx^4 - x^3 - x^2
    3x^3 + x^2 - x + 1 = yx^4 -2(x^3)(y^2) + 2(x^2)y + 2xy
    3x^3 + x^2 - x + 1 = y(x^4 -2(x^3)(y) + 2(x^2) + 2x)


    And that's about as far as I got.
     
  2. jcsd
  3. Jul 6, 2009 #2
    look at the f_y equation. What can you deduce about its solutions straightaway (it's already factorized!) ? Check if each solution works by comparing with f_x equation. I see only two workable ones. I suppose I'll leave it to you to check for maxima.
     
  4. Jul 6, 2009 #3
    well, from f_y, I can deduce that x=0 or yx^2 - x - 1 = 0.

    If x=0, from f_x, y= -2.

    If yx^2 - x - 1 = 0,
    y = (x+1)/(x^2)

    Subbing into f_x,

    -2(x^2 - 1)(2x) - 2((x+1)-x-1)(2x(x+1)/(x^2) - 1) = 0
    -2(x^2 - 1)(2x) = 0
    x(x^2 - 1) = 0
    x = 0 (as above), x = 1 or x = -1

    If x = 1, y = 2
    If x = -1, y = 0

    Using the second derivatives test,
    D = (f_xx)(f_yy) - (f_xy)^2
    where

    f_xx = -2[6x^2 - 2 + 6(x^2)(y^2) - 4xy - 2y]
    f_yy = -2[x^4]
    f_xy = -2[4yx^3 - 3x^2 - 2x]

    At (0,-2), f_xx = -2(-2+4) = -4; D = (-4)(0) - 0 = 0
    At (1,2), f_xx = -2(6-2+24-8-4) = -32; D = (-32)(-2) - (3)^2 > 0
    At (-1,0), f_xx = -2(6-2) = -8, D = (8)(-2) - (-5)^2 < 0

    So, I essentially have a saddle point at (-1,0), a maximum at (1,2) and an unknown at (0,-2). The problem with this is that:

    1) there are three critical points, somehow, and not two.
    2) only one is a maximum.
     
  5. Jul 6, 2009 #4
    "If x=0, from f_x, y= -2."

    Are you sure?

    f_x = -2(x^2 - 1)(2x) - 2(yx^2 - x - 1)(2xy - 1)
    = -2(0-1)(0) - 2(-1)(-1) = -2

    its f_x, NOT y.

    Thus, what does this mean?
     
  6. Jul 6, 2009 #5
    *backtracks*

    ooooooooooookay. so.

    when x=0, f_x = -2, but f_x = 0, so x=0 is not a solution.

    which brings me back to the points (-1,0) and (1,2). looking at my calculations for (-1,0), turns out that it's horrendously wrong.


    At (-1,0), f_xx = -2(6-2) = -8, D = (-8)(-2) - (0-3+2)^2 = 16 - 1 > 0

    Which gives a maximum. Hooray. Thanks for the help.
     
  7. Jul 6, 2009 #6
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