actually USING the lagrangian dual to solve a problem I don't get the maximisation bit in the dual problem...
Let's consider the standard form LP:

minimise $c^\top x$ 
subject to $Ax = b, x \geq 0$

This has lagrangian 
$$\mathcal{L} = c^\top x + \nu ^\top (Ax-b) - \lambda^\top x$$ $$= -\nu^\top b + (c^\top + \nu^\top A - \lambda^\top) x = -\nu^\top b + (c + A^\top\nu -\lambda)^\top x$$
so the dual function is 
$$g(\lambda,\nu) = \inf_x\ \mathcal{L}=\begin{cases}-\nu^\top b, & \text{if } c + A^\top\nu -\lambda=0\\-\infty, & \text{otherwise}\end{cases}$$
and consequently, the dual we're trying to solve is

maximise $-\nu^\top b$
subject to $c + A^\top\nu -\lambda\geq 0$

This will yield some optimal $\lambda_*,\nu_*$ for which $g(\lambda_*,\nu_*) \geq c^\top x_*$ for optimal $x_*$.
Alright, I stared at this long enough to convince myself it (kind of) makes sense, so let's try that with, say,
$c = \begin{bmatrix}3\\9\\1\end{bmatrix}$, $A = \begin{bmatrix}2 & 1 & 0\\1 & 0 & 2\end{bmatrix}$, $b = \begin{bmatrix}8\\8\end{bmatrix}$
or 

minimise $3x+9y + z$
subject to:
$2x +  y = 8$
$x + 2z = 8$
$x,y,z \geq 0$

I do believe the result should be $x = 4, y = 0, z = 2$ with a minimum objective of $14$.
So ...
$-\nu^\top b = -8\nu_1 - 8\nu_2$, which we want to maximise
$c + A^\top \nu - \lambda = 0 \equiv \begin{cases}
3 + 2\nu_1 + \nu_2 -\lambda_1 = 0\\
9 + \nu_1 - \lambda_2 = 0\\
1 + 2\nu_2 - \lambda_3 = 0\end{cases}$
That's three equations for five variables and I expect the remaining two equations to be produced by the fact we're maximising (why else would we have bothered to determine the dual?)...
Except I have no idea how to continue from here.
Any help?
 A: I think the problem you are facing is because you're adding the sign constraints into the lagrangian. For a linear program in standard form: $$\begin{align} 
   \text{min} \quad c^Tx  &\phantom{=}  \\ 
   \text{such that}\quad  \mathrm{A}x &= b \\
   x &\geq 0\end{align}$$
, the usual way to form the lagrangian for the above is:
$$\begin{align} 
\min_{x\geq0} \quad c^Tx &\phantom{=} + p^T(b-\mathrm{A})x \\
\end{align}$$
If we now form the cost function for the lagrangian $g(p)$:
$$\begin{align}
     g(p) &= \min_{x\geq0} \big(c^Tx + p^T(b - \mathrm{A}x) \big)\\ 
     &= p^Tb  \ + \ \min_{x\geq0} (c^T - p^T\mathrm{A})x
   \end{align}$$
Now this has the dual: 

$$\begin{align} 
   \max \quad p^Tb  &\phantom{=}  \\ 
   \text{such that}\quad  p^T\mathrm{A} &\leq c^T \\
   p &\lessgtr  0\end{align}$$ 

Now, in case of your example, the dual would be: 

$$\begin{align} 
   \max \quad 8p_1 +8p_2 &\phantom{=}  \\ 
   \text{such that}\quad 2p_1 + 2p_2 &\leq 3 \\
    p_1 &\leq 9 \\
    2p_2 &\leq 1\\
   p &\lessgtr  0\end{align}$$ 

This could easily be solved by the graphical method to find the optimal value and $p$. 
