Optimization problem: minimization

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SUMMARY

The optimization problem involves minimizing the function f(x,y) = √(x² + y²) under the constraint x + y ≤ 0. The trivial solution at (0,0) is confirmed as the only minimizer, as any other point would yield f(x,y) > 0. Additionally, the function MP(z), defined as the infimum of f(x,y) for points satisfying the constraint, is shown to be non-differentiable at z = 0. The discussion emphasizes the application of the Karush-Kuhn-Tucker conditions to validate the global minimum at the origin.

PREREQUISITES
  • Understanding of convex optimization principles
  • Familiarity with the Karush-Kuhn-Tucker conditions
  • Knowledge of differentiability in mathematical functions
  • Basic concepts of infimum and convex sets in R²
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  • Study the Karush-Kuhn-Tucker conditions in detail
  • Learn about differentiability and its implications in optimization
  • Explore the concept of infimum in the context of convex sets
  • Investigate geometric interpretations of optimization problems
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rayge
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Homework Statement


Minimize the function f(x,y) = \sqrt{x^2 + y^2} subject to x + y \leq 0. Show that the function MP(z) is not differentiable at z = 0.

Homework Equations

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!
 
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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)?
 
rayge said:

Homework Statement


Minimize the function f(x,y) = \sqrt{x^2 + y^2} subject to x + y \leq 0. Show that the function MP(z) is not differentiable at z = 0.

Homework Equations

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!
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.
 
sorry about that. MP(z) is the infimum of the set of all values of f(x,y) satisfying x + y \leq z, 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 x + y \leq z.

Formally:
MP(z) = inf\{f(x) | x \epsilon C, g(x) \leq z\}
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.
 
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).
 
rayge said:

Homework Statement


Minimize the function f(x,y) = \sqrt{x^2 + y^2} subject to x + y \leq 0. Show that the function MP(z) is not differentiable at z = 0.

Homework Equations

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!
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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