All I got is that $$12y_1 + 7y_2 + 10y_3 = 2(0) + 4(10.4) + 3(0) + 1(0.4)$$ and \[y_2 = 0\] because \(x_6\) or \(s_2\) is in basis. How do I find \(y_1\) and \(y_3\) without going through the simplex method? I took the dual and that doesn't seem to help. I know that \(y_1 = z_5  c_5\) and \(y_3 = z_7  c_7\), but how do I find \(z_i  c_i\)? Edit based on answer: Consider the dual: $$\min z' = 12y_1 +7y_2 + 10y_3$$ s.t. $$\begin{bmatrix} 3 & 1 & 2\ 1 & 3 & 1\ 1 & 2 & 3\ 4 & 3 & 1 \end{bmatrix}\begin{bmatrix} y_1\ y_2\ y_3 \end{bmatrix} \ge \begin{bmatrix} 2\ 4\ 3\ 1 \end{bmatrix}$$ Since the dual of the dual is the primal, the dual of the dual variables are the primal variables. Using complementary slackness on the dual, we have: $$x_1 \mathcal{S}_1 = 0$$ $$x_2 \mathcal{S}_2 = 0$$ $$x_3 \mathcal{S}_3 = 0$$ $$x_4 \mathcal{S}_4 = 0$$ $$\because x_2, x_4 > 0, \mathcal{S}_2 = \mathcal{S}_4 = 0$$ Thus, $$\begin{bmatrix} 1 & 3 & 1\ 4 & 3 & 1 \end{bmatrix}\begin{bmatrix} y_1\ y_2\ y_3 \end{bmatrix} = \begin{bmatrix} 0\ 0 \end{bmatrix}$$ $$\to \begin{bmatrix} 1 & 3 & 1\ 4 & 3 & 1 \end{bmatrix}\begin{bmatrix} y_1\ 0\ y_3 \end{bmatrix} = \begin{bmatrix} 0\ 0 \end{bmatrix}$$ $$\to \begin{bmatrix} 1 & 1\ 4 & 1 \end{bmatrix}\begin{bmatrix} y_1\ y_3 \end{bmatrix} = \begin{bmatrix} 0\ 0 \end{bmatrix}$$ $$\to y_1 = 1, y_3 = 3$$ From Chapter 2 here. asked 19 Apr '16, 14:03 BCLC 
Try using complementary slackness. And I don't think posting your homework here is optimal. answered 19 Apr '16, 14:26 Jeff Linderoth This is not homework. This is an exercise in our handouts. Also, Jeff Linderoth, complementary slackness is I guess what allows me to conclude that $y_2 = 0$. But what about $y_1$ and $y_3$? All complementary slackness gives me is that $y_1x_5 = 0 = y_3x_7$. Also, I edited
(19 Apr '16, 14:44)
BCLC
Use the complementary slackness theorem on the dual. Since $x_2$ and $x_4$ are positive, this tells you something about the dual inequalities corresponding to those primal variables.
(20 Apr '16, 10:34)
Jeff Linderoth
I edited. Thanks Jeff Linderoth ^^
(07 May '16, 06:22)
BCLC
