finding the dual I am supposed to find the dual of 
max $c^Tx$
subject to $a \le Ax \le b$
$l \le x \le u$.
In order to find the dual I think I have to write it in standard form, the standard form is:
max $Ax'$
subject to:
$A'x'\le b'$
$x' \ge 0$.
The way I try to transform it is like this:
max $c^Tx$
subject to $Ax \le b$
$-Ax \le-a$
$x \le u$
$-x \le -l$
Any tips on how to proceed?, thanks.
 A: Rewrite the problem into its canonical form.
\begin{align}
\max z = c^T x & \\
\text{s.t. } \begin{bmatrix} A \\ -A \\ I \\ -I \end{bmatrix} x &\le \begin{bmatrix} b \\ -a \\ u \\ -l \end{bmatrix}
\end{align}
Now, the standard form will be very clear.
\begin{align}
\max z = c^T x & \\
\text{s.t. } Ax + s_1 &= b \\
-Ax + s_2 &= -a \\
x + s_3 &= u \\
-x + s_4 &= -l \\
s_i &\ge 0 \,\forall i = 1,\dots,4
\end{align}
Depending on the sign of $a,b,l,u$, you may need to multiply both sides of a constraint by $-1$ so that the RHS of these four equalities is positive.  


*

*Write the dual from the standard form is very hard and inefficient.

*Memorizing the table is hard.

*Use SOB method for converting primals to duals.
Therefore, the dual is the following LPP.
\begin{align}
\max z = \begin{bmatrix} b \\ -a \\ u \\ -l \end{bmatrix}^T \begin{bmatrix} t_1 \\ t_2 \\ t_3 \\ t_4 \end{bmatrix} & \\
\text{s.t. } \begin{bmatrix} A \\ -A \\ I \\ -I \end{bmatrix}^T \begin{bmatrix} t_1 \\ t_2 \\ t_3 \\ t_4 \end{bmatrix} &= c \\
t_i &\ge 0 \,\forall i = 1,\dots,4
\end{align}
Simply the above LPP.
\begin{align}
\max z = b^T t_1 - a^T t_2 + u^T t_3 - l^T t_4 & \\
\text{s.t. } A^T t_1 - A^T t_2 + t_3 - t_4 &= c \\
t_i &\ge 0 \,\forall i = 1,\dots,4
\end{align}
