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I am slightly confused with what I am asked to do when trying to prove that a boundary point satisfies the first order necessary conditions for it to be a minimum, ie that the dot product of the gradient of the function at the point and any feasible direction at the point is greater or equal to 0.

The question is, how do you find the feasible directions of a boundary point?

The definition is: $\exists \alpha_0$ such that $\forall \alpha \in (0,\alpha]$ $x+\alpha*d$ is in the feasible set.

What my lecturer does is setting $\alpha_0$ to 1 and exclude the forall condition.

I think this is wrong, or I do not understand it.

I give you an example which I would like to solve, and that was solved by my lecturer using the "$\alpha=1$ method."

$f(x) = x_2, with f: R^2-> R$ with the constraint that $x_1^2 + x_2 \ge0$.

Is the point $[0,1]^T$ satisfying the first order conditions? Can we find some conditions that the vectors which are feasible directions have to satisfy?

Can you please show me your working? And with usage of $\alpha$.

Thanks!

elaRosca
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1 Answers1

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I realised that the general problem of finding all feasible directions at a point, does not have a simple analytic solution, that you can just plug in in your problem and solve it.

However, in my case I had to find the feasible directions of the boundary points of the region $x_1^2+x_2^2>=1$. Hence the boundary points are the points on the unit circle.

In order to find the feasible directions, I proved that any feasible direction has to move the point $(x_1, x_2)$ above the tangent at $(x_1, x_2)$ to the circle, for any positive $\alpha$ (where $\alpha$ comes from the definition of feasible direction).

Hence, using the equation of the tangent at a circle, we get that $x_1 * (x_1 + \alpha * d_1) + x_2 * (x_2 + \alpha * d_2) >=1 $

Hence $x_1 * d_1 + x_2 * d_2 >=0 $ is a condition required for a feasible direction.

elaRosca
  • 1,093