I am trying to employ a particular technique required to determine a dual problem of the following linear programming:
$$ \text{min } x_1 -3x_2-x_3 $$ $$ \text{subject to}\begin{cases} 3x_1 - x_2 + 2x_3 \geq 1\\ -2x_1 + 4x_2 +0x_3 \leq 12\\-4x_1 + 3x_2 +3x_3 = 14 \end{cases} $$ The first thing I tried was to reformulate in terms of dot products, giving us the equivalent problem $$ \text{min } c_0^Tx $$ $$ \text{subject to}\begin{cases} c_1^Tx \leq -1\\ c_2^Tx \leq 12\\c_3^Tx \leq 14\\c_4^Tx \leq -14 \end{cases} $$ where the equality has been turned into two inequalites and we have that $$ c_0 = (1,-3,-1), c_1 =(-3,1,-2),c_2 = (-2,4,0), c_3 = (-4,3,3), c_4 = (4,-3,-3) $$ Thus, we can further simplify to
$$ \text{min } c_0^Tx $$ $$ \text{subject to }Ax\leq b $$ Where we have put the $c_i$s into $A$ as rows, and $b^T = (-1,12,14,-14)$.
Thus, we can write our Lagrange function $$ L(x,\lambda) = \langle c_0, x \rangle + \langle \lambda, Ax-b \rangle $$ and our dual Lagrangian function as \begin{align*} h(\lambda) &= \text{inf}_{x\in\mathbb{R}^n} \{\langle c_0, x \rangle + \langle \lambda, Ax-b \rangle\}\\ &= -\text{sup}_{x\in\mathbb{R}^n} \{-\langle c_0, x \rangle - \langle A^T\lambda, x \rangle\} - \langle b, \lambda \rangle \end{align*} We notice that this is the definition of the convex conjugate, so \begin{align*} h(\lambda) &= -f^{\ast}(-A^T\lambda)-\langle b, \lambda\rangle\\ &=\begin{cases} \langle b, \lambda \rangle,& \text{if } -A^T\lambda = c_0\\ -\infty,& \text{if }-A^T\lambda \neq c_0 \end{cases} - \langle b, \lambda\rangle\\ &=\begin{cases} 0,& \text{if } -A^T\lambda = c_0\\ -\infty,& \text{if } -A^T\lambda \neq c_0 \end{cases} \end{align*}
Which just results in the dual problem of maximising zero, which is strange, and I am not sure I have the right answer, as I plugged this problem into a dual problem calculator and it didn't give me this result.
This problem is unconstrained, you can show that if you use the last constraint $−4x_1+3x_2+3x_3=14$ and solve for $x_3$, the problem is reduced to
$$ \text{min } \frac{1}{3}(-14 - x_1 - 6x_2) $$ $$ \text{subject to}\begin{cases} 17x_1 - 9x_2 &\geq& -25\\ x_1 - 2x_2 &\geq& -6 \end{cases} $$
If you move along the line $x_2 = 0$, the constraint on $x_1$ becomes $25+17x_1 \geq 0 $ and the objective function will have the form
$$ \frac{1}{3}(-14 - x_1) $$
which continuously decreases for increasing values of $x_1$, so the problem is unbound!