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Unconstrained optimization

  1. May 6, 2014 #1
    These 2 questions,I have attempted them for hours but still no outcome:confused:
    For the first question:
    part a) i take the ∇f(x) and set it to be zero;
    then find out(x1,x2)=(0,-1) or((-2a-1)/3,(-2a-4)/3);
    but then for part b), after using second-order necessary condition,
    i have no idea to continue;
    part c): i know if i can prove the function is convex or concave, i can surely conclude the local minimum/maximum is a global minimum/maximum; but after taking the gradient for 2 times, i dont know to work for it in further;

    For Question 2,
    part a) i got 5 possible points (x1,x2)=(0.5,0),(1,-1),(1,1),(-1,-sqrt(3)),(-1,sqrt(3));
    but in part b) i can only prove (0.5,0) is a local maximizer while other 4 points dont satisfy the 2nd conditions(because their determinant of Hessian Matrix is smaller than zero;)
    part c) same, no idea how to do it;
    part d) i find the directional vectors are both [0,0]'; does it mean f(1,-1) is a local minimum?

    Thanks for help:frown:
     

    Attached Files:

  2. jcsd
  3. May 6, 2014 #2
    no body can help?:cry::cry:
     
  4. May 6, 2014 #3

    Ray Vickson

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    Looking for stationary points and examining Hessians only works if the function has a (local) max or min at the point of interest. You can only apply these methods to global optima if you know that the functions have global optima, so first you need to examine that issue: does the function have a finite global max? Global min?
     
  5. May 6, 2014 #4
    First of all, really thanks for your reply;
    These 2 questions both are required to find their global optima, it implies they should have a finite global optima; but what i am confused is how to determine whether those local optima are also global optima? And Even though i know the methods to find local optima.( e,g first order necessary condition, second order condition necessary/sufficient condition...), but i still cannot solve these 2 problems
    Very difficult to me. Could you spend some more time to help me solve them :frown:
     
  6. May 6, 2014 #5

    Ray Vickson

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    No, I cannot do any more, because that would involve my solving the problem for you. I will just say that your reasoning for the existence of global max or min is faulty; just because somebody asks you to find it does not mean it exists--it may, or may not!
     
  7. May 6, 2014 #6

    haruspex

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    As Ray says, don't deduce that.
    They are both smooth functions, defined everywhere. If a global extremum is not also a local extremum, where is it?
     
  8. May 7, 2014 #7
    Okay, you are right. Then i also want to ask, for Q2 part d, if i find a steepest descent directional vector at a point, i.e x*, is zero; can i say that the point is a local minimizer? can the directional vector be zero when the point is a saddle point or maximum point?
     
  9. May 7, 2014 #8

    haruspex

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    Umm.. about what exactly?
    When I wrote
    there is an answer to that question.
    Yes, it will be. The grad is zero if and only if the tangent plane is horizontal. To know whether it's a min, a max, a saddle, or even a horizontal valley or ridge, you need to look at higher derivatives.
     
  10. May 7, 2014 #9
    Okay, thanks very much;
    And all answers are solved, thanks for your reminder;
     
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