How to find Dual Problem I have here an example, where I am not able to find the dual problem:
$(P): J_p(x)= \max_{z_1, z_2 } (z_1 + x z_2)$
subj. to: 
$z_1 + z_2 = 1$ 
$z_1\ge 0$
$z_2\ge0$
where $x\in[0,2]$ a parameter.
The Dual of the Problem is:
$(D): \min_{\lambda} -\lambda $
subj. to:
$\lambda \le-1$
$\lambda \le-x$
I really don't understand how I can derive $(D)$, especially how to derive the constraints of the dual problem.
I can write the lagrangian as:
$L(\lambda, z) = z_1 + xz_2 +\lambda_1z_1 + \lambda_2z_2 + \nu(1-z_1-z_2)$
I can find now $z^*$ by solving direclty the primal problem, but this is not the idea about using the dual problem. How can I derive the dual problem $(D)$?
 A: The utility of the dual problem theory lies on the strong duality theorem, the complementary slackness theorem and the dual simplex algorithm, you can read about it for example here (chapter 3, in spanish unfortunately).
But going to your question, the dual problem of a linear programming problem in maximization cannonical form $$\max c^tx, \text{restricted by}\ Ax\leq b,\ x\geq 0$$ is defined as
$$\min b^tu,\ \text{restricted by}\ A^tu\geq c,\ u\geq 0$$
From this definition it can be proved that the duality is involutive, this is, the dual of the dual problem is the original (or primal) problem.
So, to get the dual problem of of an aritrary linear problem, say with $\leq$, $\geq$ and $=$ restrictions we can do the following, the original problem is
$$\max c^tx,\ \text{restricted by}\ A_1x\leq b_1,\ A_2x\geq b_2,\ A_3x=b_3$$
which is equivalent to
$$\max c^tx,\ \text{restricted by}\ A_1x\leq b_1,\ -A_2x\leq -b_2,\ -A_3x\leq -b_3,\ A_3x\leq b_3$$
writen in a more compact way
$$\max c^tx,\ \text{restricted by}\ \begin{pmatrix}A_1\\-A_2\\-A_3\\A_3\end{pmatrix}x\leq \begin{pmatrix}b_1\\-b_2\\-b_3\\b_3\end{pmatrix}$$
so, by definition, the dual problem is
$$\min \begin{pmatrix}b_1^t&-b_2^t&-b_3^t& b_3\end{pmatrix}\begin{pmatrix}u_1\\u_2\\u_3\\u_4\end{pmatrix},\ \text{restricted by}\ \begin{pmatrix}A_1^t& -A_2^t& -A_3^t& A_3^t\end{pmatrix}\begin{pmatrix}u_1\\u_2\\u_3\\u_4\end{pmatrix}\geq c$$
writing the restrictions in another way
$$A_1^tu_1-A_2^tu_2-A_3^tu_3+A_3^tu_4\geq c$$
which is equivalent to
$$A_1^tu_1-A_2^tu_2+A_3^t(u_4-u_3)\geq c$$
defining $w:=u_4-u_3$ we have $$A_1^tu_1-A_2^tu_2+A_3^tw\geq c$$
So as, $u_1,u_2,u_3,u_4\geq 0$ then $u_1,u_2\geq 0$ and $w$ is not restricted, so finally we have the dual problem
$$\min b_1^tu_1-b_2^tu_2+b_3^tw,\ \text{restricted by}\ A_1^tu_1-A_2^tu_2+A_3^tw\geq c$$
From this, your case is just a particular one.
