Is there a way to find optimal solution of an LPP problem with an optimal solution for dual which is found using graphical method?

Question

For the following LPP problem, $$ \text{Minimise} \quad 12x+8y+8z \\ \text{s.t} \quad 2x+2y+2z ≥ 1 \\ 3x+ y- z ≥ 1\\ \text{Where} \quad x, y, z ≥ 0 $$ I converted it to dual with variables $r$ & $s$ & using graphical method found that optimal solution is $r=3/2$ & $s=2$ with maximum value of dual being 5. I know minimum value of the given LPP will also be 5. But how can I use this solution to find optimal values $x,y,z$?

Answer 1

The tool for solving the problem with having optimal solution of dual problem is Complementary Slackness theorem. Summary of this theorem :

if $\lambda_i$ is slack variable for constraint of dual problem corresponding to $x_i$ (variable of primal problem), then $\lambda_i x_i=0$.

so you have this problem : $$ LP_1:\left\{ \begin{array}{l} min \quad 12x+8y+8z \\ 2x+2y+2z \geq 1 \\ 3x+y−z \geq 1 \\ x,y,z \geq 0 \end{array} \right. \longleftrightarrow LP_1^*:\left\{ \begin{array}{l} max \quad r+s \\ 2r+3s \leq 12 \\ 2r+s \leq 8 \\ 2r-s \leq 8 \\ r,s \geq 0 \end{array} \right. $$ with solving the dual problem $LP_1^*$, we find that $r^*=3$ , $s^*=2$ (you have a typo in value of $r$). if you put these values to dual problem constraints you can see that first and second inequalities get their maximum values and become equality at optimal point. but not the third one $2\times3-2 = 1 <8$.
using complementary slackness theorem we can conclude variable of $LP_1$ corresponding to third constraints of $LP_1^*$ which is $z$ should be zero. so $z^*=0$.
and again using these theorem because $r* \neq 0$ and $s^* \neq 0$ , then both constraints of $LP_1$ are equality at optimal point. knowing these facts we get : $$ \left\{ \begin{array}{l} 2x^*+2y^*+2z^* = 1 \\ 3x^*+y^*−z^* = 1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 2x^*+2y^* = 1 \\ 3x^*+y^* = 1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x^* = \frac{1}{4} \\ y^* = \frac{1}{4} \end{array} \right. $$ hence optimal solution of $LP_1$ is $(x^*,y^*,z^*) = (\frac{1}{4},\frac{1}{4},0)$.