When forming the Lagrangian of an optimization problem, why don't we include all constraints in the Lagrangian? For example, the optimization problem $$ \begin{align} \max_{x}&\quad f(x)\\ \nonumber \text{subject to } \quad & g_i(x) \le0,\,i=1,\ldots,m\\ \quad & h_j(x)=0,\,j=1,\ldots,l\\ \quad & x\in X \end{align} $$ has a Lagrangian of $ \mathcal{L}(x,\mu,\lambda) = f(x) + \sum_{i=1}^m\mu_ig_i(x) + \sum_{j=1}^l\lambda_jh_j(x) $. Look at slide 17 of http://www.ise.ncsu.edu/fangroup/or706.dir/Lecture10.pdf.
Why didn't we dualize $x\in X$?
This raises the question of what sort of constraints are allowed to be lumped into $x\in X$? Does this depend on whether the constraint is complicating (that is, non-separable) or not?
I am especially confused since I've seen optimization problems that have constraints of the form $x\ge0$ and yet they are dualized into the Lagrangian. The constraint $x\ge0$ is not complicating (it is clearly separable), so why can't it be lumped into $x\in X$ and left out of the Lagrangian?