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Optimization problem: minimization

  1. Dec 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Minimize the function [itex]f(x,y) = \sqrt{x^2 + y^2}[/itex] subject to [itex]x + y \leq 0[/itex]. Show that the function [itex]MP(z)[/itex] is not differentiable at [itex]z = 0[/itex].

    2. Relevant equations

    3. The attempt at a solution
    I haven't gotten anywhere because I don't understand why the solution isn't trivial, i.e. (0,0). Any suggestions as to where to start are welcome. Thanks!
  2. jcsd
  3. Dec 3, 2014 #2


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    Your trivial solution gives f(x,y)=0. Are there any others? If there are, then they are all minimizers. If not, then I think you are done.
    What about the differentiable piece? What is meant by MP(z)?
  4. Dec 3, 2014 #3


    Staff: Mentor

    Your constraint is the line y = -x together with the half-plane below it.

    What is MP(z)? That information should be in the problem statement.
  5. Dec 3, 2014 #4
    sorry about that. MP(z) is the infimum of the set of all values of [itex]f(x,y)[/itex] satisfying [itex]x + y \leq z[/itex], i.e. the minimum value of f(x,y) if it exists, or the greatest lower bound of it. The domain of MP(z) is the set of all z such that there are points (x,y) satisfying [itex]x + y \leq z[/itex].

    [itex]MP(z) = inf\{f(x) | x \epsilon C, g(x) \leq z\}[/itex]
    where C is a convex set, in this case R^2

    I haven't worked on showing the function isn't differentiable at z=0 yet, I'll probably have questions about that too. I'm mostly confused about why there is anything to solve in the first part of the problem.
  6. Dec 3, 2014 #5


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    The solution to the first part is straightforward. Assume there is a minimum that is not (0,0) and contradict your assumption by showing that f(x,y)>0=f(0,0).
  7. Dec 4, 2014 #6

    Ray Vickson

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    The optimal solution IS trivial---you can see that geometrically--- and it is at (0,0), as you have stated.

    If you have studied the Karush-Kuhn-Tucker conditions, you will see that (0,0) satisfies the necessary conditions for a minimum, but in the equivalent convex problem ##\min \; x^2 + y^2 \;\; \text{subject to } \; x + y \leq 0##. Since this is a convex programming problem, any local minimum is a global minimum, so the origin is provably the only solution---again, though, that is about as obvious as you can get.
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