Maybe it's worthwhile to talk through where the dual comes from. This will take a while, but hopefully the dual won't seem so mysterious when we're done.
Suppose we want to use the primal's constraints as a way to find an upper bound on the optimal value of the primal. If we multiply the first constraint by $9$, the second constraint by $1$, and add them together, we get $9(2x_1 - x_2) + 1(x_1 +3 x_2)$ for the left-hand side and $9(1) + 1(9)$ for the right-hand side. Since the first constraint is an equality and the second is an inequality, this implies $$19x_1 - 6x_2 \leq 18.$$
But since $x_1 \geq 0$, it's also true that $5x_1 \leq 19x_1$, and so $$5x_1 - 6x_2 \leq 19x_1 - 6x_2 \leq 18.$$
Therefore, $18$ is an upper-bound on the optimal value of the primal problem.
Surely we can do better than that, though. Instead of just guessing $9$ and $1$ as the multipliers, let's let them be variables. Thus we're looking for multipliers $y_1$ and $y_2$ to force $$5x_1 - 6x_2 \leq y_1(2x_1-x_2) + y_2(x_1 + 3x_2) \leq y_1(1) + y_2(9).$$
Now, in order for this pair of inequalities to hold, what has to be true about $y_1$ and $y_2$? Let's take the two inequalities one at a time.
The first inequality: $5x_1 - 6x_2 \leq y_1(2x_1-x_2) + y_2(x_1 + 3x_2)$
We have to track the coefficients of the $x_1$ and $x_2$ variables separately. First, we need the total $x_1$ coefficient on the right-hand side to be at least $5$. Getting exactly $5$ would be great, but since $x_1 \geq 0$, anything larger than $5$ would also satisfy the inequality for $x_1$. Mathematically speaking, this means that we need $2y_1 + y_2 \geq 5$.
On the other hand, to ensure the inequality for the $x_2$ variable we need the total $x_2$ coefficient on the right-hand side to be exactly $-6$. Since $x_2$ could be positive, we can't go lower than $-6$, and since $x_2$ could be negative, we can't go higher than $-6$ (as the negative value for $x_2$ would flip the direction of the inequality). So for the first inequality to work for the $x_2$ variable, we've got to have $-y_1 + 3y_2 = -6$.
The second inequality: $y_1(2x_1-x_2) + y_2(x_1 + 3x_2) \leq y_1(1) + y_2(9)$
Here we have to track the $y_1$ and $y_2$ variables separately. The $y_1$ variables come from the first constraint, which is an equality constraint. It doesn't matter if $y_1$ is positive or negative, the equality constraint still holds. Thus $y_1$ is unrestricted in sign. However, the $y_2$ variable comes from the second constraint, which is a less-than-or-equal to constraint. If we were to multiply the second constraint by a negative number that would flip its direction and change it to a greater-than-or-equal constraint. To keep with our goal of upper-bounding the primal objective, we can't let that happen. So the $y_2$ variable can't be negative. Thus we must have $y_2 \geq 0$.
Finally, we want to make the right-hand side of the second inequality as small as possible, as we want the tightest upper-bound possible on the primal objective. So we want to minimize $y_1 + 9y_2$.
Putting all of these restrictions on $y_1$ and $y_2$ together we find that the problem of using the primal's constraints to find the best upper-bound on the optimal primal objective entails solving the following linear program:
$$\begin{align*}
\text{Minimize }\:\:\:\:\: y_1 + 9y_2& \\
\text{subject to }\:\:\:\:\: 2y_1 + y_2& \geq 5 \\
-y_1 + 3y_2& = -6\\
y_2 & \geq 0.
\end{align*}$$
And that's the dual.
It's probably worth summarizing the implications of this argument for all possible forms of the primal and dual. The following table is taken from p. 214 of
Introduction to Operations Research, 8th edition, by Hillier and Lieberman. They refer to this as the SOB method, where SOB stands for Sensible, Odd, or Bizarre, depending on how likely one would find that particular constraint or variable restriction in a maximization or minimization problem.
Primal Problem Dual Problem
(or Dual Problem) (or Primal Problem)
Maximization Minimization
Sensible <= constraint paired with nonnegative variable
Odd = constraint paired with unconstrained variable
Bizarre >= constraint paired with nonpositive variable
Sensible nonnegative variable paired with >= constraint
Odd unconstrained variable paired with = constraint
Bizarre nonpositive variable paired with <= constraint
To get an optimal solution, it is necessary to have a basic solution as a starting point. This means, that the number of constraints must be equal to the number of basic variables. Thus every constraint has to have a positive slack variable or a positive artificial variable. If the slack variable is negative or is not needed, then you add an artificial variable.
There are three cases:
$\color{blue}{\leq-\texttt{constraint}}$
If you have a $\leq$-constraint, then you have to add a slack variable for each constraint.
$2y+z \leq 2 \quad \Longrightarrow \quad 2y+z +s_1=2$
$\color{blue}{=\texttt{-constraint}}$
If you have a $=$-constraint, then you do not have to add a slack variable for each constraint. But you have to add an artificial variable for each constraint.
$x+y+z=4 \quad \Longrightarrow \quad x+y+z+a_1=4$
$\color{blue}{\geq-\texttt{constraint}}$
If you have a $\geq$-constraint, then you have to substract a slack variable for each constraint. Additionally you have to add an artificial variable for each constraint.
$x-2y+z \geq 3 \quad \Longrightarrow \quad x-2y+z-s_2+a_2 = 3$
The initial simplex tableau is
$\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & y & z & s_1 & s_2 & a_1 & a_2 & RHS \\ \hline -2 & 1 & -3 & 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 2 & \color{red} 1 & 1 & 0 & 0 & 0 & 2 \\ \hline 1 & 1 & 1 & 0 & 0 & 1 & 0 & 4 \\ \hline 1 & -2 & 1 & 0 & -1 & 0 & 1 & 3 \\ \hline \end{array} $
The coefficients of the objective function have to be multiplied by (-1), because the objective function has to be maximized.
The initial basic solution is $(x, y, z, s_1, s_2, a_1, a_2)=(0, 0, 0, 2, 0, 4, 3)$
The first pivot element is the red One.
Best Answer
(I feel like I'm taking one for the team here. :) )
Starting with your main questions at the end...
Now for the long list of questions at the start...
No, because of the LTE constraint.
Not if you formulate the problem as a maximization problem the way I described above.
That depends on whether you're used to doing the simplex method by hand for maximization problems or for minimization problems. Since you're talking about converting to a maximization problem, I assume that's the one you're more comfortable with.
Yes, but you'll also have four artificial variables for the GTE constraints - at least for the first phase.
For the second phase, yes. For the first phase, you'll have seven, thanks to the four additional artificial variables.
Yes. See my answer to #2 in your main questions.
Yes. See my answer to #2 in your main questions.
If you're converting to the dual, yes. (Again, though, you don't have to convert to a maximization problem by converting to the dual.)
If you formulate the problem the way I describe in #1 to your main questions, the constraints do not change. The only change to the objective is that you're maximizing $-\phi = -2700x -2400y - 2100z$.