Solving linear programming problem with given information I want to find the optimal solution to a following linear programming problem:
$$8x_1+120x_2+114x_3\to \min$$
$$x_1+7x_2+3x_3\geq 4,$$
$$x_1+5x_2+5x_3\geq 5,$$
$$x_1+3x_2+10x_3\geq 9,$$
$$x_1+2x_2+15x_3\geq11,$$
$$x_1\geq0, \ \ \ x_2\geq 0, \ \ \ x_3\geq0$$
and solution for its dual problem. I already found the dual problem:
$$4y_1+5y_2+9y_3+11y_4\to \max$$
$$y_1+y_2+y_3+y_4\leq 8,$$
$$7y_1+5y_2+3y_3+2y_4\leq 120,$$
$$3y_1+5y_2+10y_3+15y_4\leq 114,$$
$$y_1\geq 0, \ \ \ y_2\geq 0 , \ \ \ y_3\geq 0 , \ \ \ y_4\geq 0.$$
I also know that in the primal problem solution $x^*_1>0, \ x^*_3>0$ for $x^*=(x_1^*,x_2^*,x_3^*)$ and $y_1^*=y_2^*=0$ for $y^*=(y_1^*,y_2^*,y_3^*,y_4^*)$. How should I use that information to find other coordinates in the optimal solution?
 A: Let $e_1,e_2,e_3,e_4$ be the slack variables of the primal, and $f_1,f_2,f_3$ the slack variables of the dual.
By the complementary slackness conditions:


*

*$x_1^*>0 \quad \Rightarrow \quad f_1=0$

*$x_3^*>0 \quad \Rightarrow \quad f_3=0$


And since $y_1^*=y_2^*=0$, it follows that $y_3^*$ and $y_4^*$ satisfy
\begin{cases}
y_3^*+y_4^*=8 \\
10y_3^*+15y_4^*=114
\end{cases}
Solving for $y_3^*$ and $y_4^*$ yields
\begin{cases}
y_3^*= \frac{6}{5} \\
y_4^*= \frac{34}{5}
\end{cases}
Plugging these values in the second constraint, we get
$$
0+0+3\frac{6}{5}+2\frac{34}{5}+f_2=120\quad \Rightarrow \quad f_2 =\frac{514}{5}
$$
Again, by complementary slackness:


*

*$y_3^*>0 \quad \Rightarrow \quad e_3=0$

*$y_4^*>0 \quad \Rightarrow \quad e_4=0$

*$f_2>0 \quad \Rightarrow \quad x_2^*=0$


Therefore $x_1^*$ and $x_3^*$ satisfy 
\begin{cases}
x_1^*+0+10x_3^*=9 \\
x_1^*+0+15x_3^*=11
\end{cases}
which gives us
\begin{cases}
x_1^*= 5 \\
x_3^*= \frac{2}{5}
\end{cases}
Plugging these values in the two first constraints, we get $e_1=\frac{11}{5}$, $e_2=2$ and finally $y_1^*=y_2^*=0$ by complementary slackness again.
In summary:
$$
\boxed{
(x_1^*,x_2^*,x_3^*)=(5,0,\frac{2}{5})\quad \mbox{and}\quad (y_1^*,y_2^*,y_3^*,y_4^*)=(0,0,\frac{6}{5},\frac{34}{5})
}$$
