I am attempting to find the dual of the following linear program
Max $2x_1+x_2$, subject to:
$12x_1+3x_2\leq60$,
$3x_1-x_2\geq -7$,
$x_2\leq 10$,
$x_1\geq 0$,
$x_2$ free.
To do this I took the transpose of
$\begin{bmatrix} 12&3&60\\ 3&-1&-7\\ 0&1&10\\ 2&1 \end{bmatrix}$
and got
$\begin{bmatrix} 12&3&0&2\\ 3&-1&1&1\\ 60&-7&10 \end{bmatrix}$
which gives me the linear program
min $60v_1-7v_2+10v_3$ subject to:
$12v_1+3v_2\geq 2$,
$3v_1-v_2+v_3\leq1$
Now I am unsure as to the nonegitivity constraints because I have one more variable $v_3$ and because $x_2$ was a free variable. Any help would be appreciated.
The constraint of a primal problem actually corresponds to the variable to the dual problem.
Your primal problem is a maximization problem.
The first constraint of your primal problem is $$12x_1 + 3x_2 \leq 60$$
Let's associate a variable $v_1$ to it. When we go from a primal maximization constraint to a dual minimization variable, $\leq$ in the constraint, corresponds to $\geq$ in the variable. Hence $\color{blue}{v_1 \geq 0}$.
I will leave the sign of $v_2$ and $v_3$ as an exercise.
When we go from a primal maximization variable to a dual minimization constraint, the sign is preserved.
Hence $x_1 \geq 0$ corresponds to $$12v_1+3v_2 \geq 2.$$
$x_2$ being free corresponds to $$3v_1-v_2+v_3\color{red}=1$$