I have the optimal solution of the primal ($x$ values, which, of course, give me the optimal objective value). However, I am at a complete loss as to how to obtain the $y$ values of the dual in the case of this LP.
Maximize $z = x_{1} + 2x_{2} + 5x_{3} + x_{4}$
Subject to:
$x_{1} + 2x_{2} + x_{3} - x_{4} \le 20$
$-x_{1} + x_{2} + x_{3} + x_{4} \le 12$
$2x_{1} + x_{2} + x_{3} - x_{4} \le 30$
$x_{1},x_{2},x_{3},x_{4} \ge 0$
Given that
($x_{1},x_{2},x_{3},x_{4}) = (10, 0, 16, 6)$
I formulated the dual in standard equality as follows:
Minimize $Z = 20y_{1} + 12y_{2} + 30y_{3}$
$y_{1} + -y_{2} + 2y_{3} - \lambda_{1} = 1$
$2y_{1} + y_{2} + y_{3} - \lambda_{2} = 2$
$y_{1} + y_{2} + y_{3} - \lambda_{3} = 5$
$-y_{1} + y_{2} - y_{3} - \lambda_{4} = 1$
I can see that all constraints in the primal are binding and that the only non-binding constraint in the dual is the second one, since the $\lambda$ for that constraint is not equal to zero. I am familiar with solving duals that have eliminated enough constraints and variables in the dual to solve it graphically, but I am completely stuck with this one. I still have 3 $y$ variables to solve for.
You can apply complementary slackness. Since $x_1$, $x_3$, and $x_4$ are nonzero in the optimal primal solution, the first, third, and fourth constraints are tight (or, the way you're writing them up, $\lambda_1 = \lambda_3 = \lambda_4 = 0$) in the optimal dual solution.
This gives you three equations for the three dual variables, and you can solve them to find out the values of $y_1$, $y_2$, and $y_3$.