Kuhn Tucker Conditions with fewer non-negativity constraints than number of variables I have a following type of problem:

$Maximize\,\, F(s,x,y,z)$
$s,x,y,z$    

s.t. (i) $g(x,y,z) \le I$
(ii) $x \ge 0$
(iii) $y \ge 0$
(iv) $s &gt; 0$

That is there is no non negativity constraint on variable $z$. I have read about the Kuhn-Tucker method in the books but in all those cases the problem is fourmulated such that the non-negativity constraints apply to all the choice variables. How would my K-T optimality conditions change if I have no non-negativity constraint on $z$?

My sense is that other than the regular K-T conditions for $x$ and $y$ and Lagrange multiplier, $\lambda$, we would have (a) $\frac{\partial \Large{L}}{\partial z} = 0$ (i.e. no inequality here) and NO such condition as (b) $z\,.\frac{\partial \Large{L}}{\partial z} = 0$ (which is obviously true given (a)). My entire confusion is because I have not seen such a formulation in books where K-T conditions are discussed and just want to be sure about my steps.  

What would the conditions be in case of $s$, where the inequality is strict?  

Here I would think that with strict inequality the problem is not well defined. The constraint set is not compact (not closed).

Thanks a lot in advance.
Non-negativity constraints have nothing special and are not critical for the general validity of the Karush-Kuhn-Tucker approach. First, realize that we could have $x \geq a &gt;0$ and then we could write $x-a \geq 0$ and view this "non-negativity" constraint as just one more inequality constraint on the solution.

When a decision variable is weakly bounded, the new phenomenon that emerges is that we may have the partial derivative non-zero (strictly negative, for the maximization case), if at the solution this decision variable lies on the boundary.

If it is unbounded, than we must have the first-order derivative equal to zero. As another question pointed out, this may create issues in certain contexts, but it does not invalidate the approach.

As for your second question, indeed strict inequalities may create problems, but don't jump immediately to a "not well-defined" heavy statement. The problem will arise only if (stars denote the optimal solution)

$$\frac {\partial F(s^*,x^*,y^*,z^*)}{\partial s^*} &lt;0$$

which would "push" the solution vector towards the unreachable zero-lower boundary for $s$. To avoid that, we require that at the solution the first-order derivative w.r.t to $s$ equals zero, and see whether we can indeed obtain a solution vector. If we can, it means that the $s&gt;0$ constraint does not even "tend to" be binding, so you can ignore it.