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Critical points (x-1)^4 (x-y)^4

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Find critical points. [itex]f(x,y) = (x-1)^4 + (x-y)^4[/itex]

    3. The attempt at a solution

    [itex]\frac{\partial f}{\partial x} = 4(x-1)^3 + 4(x-y)^3 [/itex]

    [itex]\frac{\partial f}{\partial y} = - 4(x-y)^3 [/itex]

    [itex]\frac{\partial f}{\partial x} = 0 [/itex],

    [itex]\frac{\partial f}{\partial y} = 0 [/itex],

    [itex]- 4(x-y)^3 = 0 [/itex], x=y

    [itex]4(y-1)^3 + 4(y-y)^3 = 0[/itex],

    We have a critical point at: y=1,x=1

    So as the both part of the function ( [itex](x-1)^4[/itex] and [itex](x-y)^4[/itex] ) will always be grater or equeal 0, and the critical point gives us 0, then I guess it's a min, no?

    So if I'm not wrong, then why is inconclusive the second derivative text AC-B^2=0

    A=Fxx, C=Fyy, B=Fxy

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 20, 2013 #2

    Ray Vickson

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    Homework Helper

    The second derivatives are all zero at the stationary point, so second-order conditions give no information. However, the function f(x,y) is strictly convex, so any stationary point must be a global min.

    BTW: having second-order things = 0 does not guarantee minimality: the *sufficient* second-order conditions need a strict inequality ">0". Another way to see this is to note that g(x,y) = -f(x,y) also has all second derivatives = 0, but for it the stationary point is a maximum!
  4. Feb 20, 2013 #3
    Understood! Thanks :)
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