I am new to Optimization so I think the following question may be very easy, but I'm not sure how to solve it.
The dual of an LP is an LP. If we solve the dual LP, we can get the optimal value for the primal problem. But:
How do we get the optimal decision variables for the primal?
Does it make a difference if we only relax some of the constraints?
My Work for Finding Dual of LP
\begin{align*} &\text{min } &c^Tx \\ &\text{s.t. } &Ax = b \\ &&Bx \le d \\ \end{align*}
We form the Lagrangian ($\mu \ge 0$): \begin{align*} L(x, \lambda, \mu) &= c^Tx + \lambda^T (Ax - b) + \mu^T (Bx - d) \\ &= (c + A^T \lambda + B^T \mu)^Tx - \lambda^Tb - \mu^Td \end{align*}
Thus we see that the dual function is $$d(\lambda, \mu) = \inf_x L(x, \lambda, \mu) = \begin{cases} - \lambda^Tb - \mu^Td \ \ \ &\text{if } \ \ \ c + A^T \lambda + B^T \mu = 0 \\ - \infty & \text{otherwise} \end{cases}$$ and therefore the dual problem is \begin{align*} &\text{max } &-\lambda^Tb - \mu^Td \\ &\text{s.t. } &c + A^T \lambda + B^T \mu = 0\\ \end{align*}