# Optimization problem: minimization

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1. Dec 3, 2014

### rayge

1. The problem statement, all variables and given/known data
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 $z = 0$.

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. Dec 3, 2014

### RUber

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)?

3. Dec 3, 2014

### 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.

4. Dec 3, 2014

### rayge

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.

5. Dec 3, 2014

### RUber

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).

6. Dec 4, 2014

### Ray Vickson

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.