Let us consider the following primal problem : Maximize \(z = 2x_{1} + x_{2}\) Subject to \(x_{1} - x_{2} \leq 4\), \(x_{1} - x_{2} \leq 2\), \(x_{1}, x_{2} \geq 0\) What will be its dual problem ?
Dual infeasible
Dual infeasible option is correct.
The primal problem is \(\max z = 2x_{1} + x_{2}\) subject to \(x_{1} - x_{2} \leq 4\), \(x_{1} - x_{2} \leq 2\), \(x_{1}, x_{2} \geq 0\). Only the difference \(x_{1} - x_{2}\) is bounded, so setting \(x_{2} = x_{1} - 1\) keeps both constraints tight while pushing \(z = 2x_{1} + (x_{1} - 1) = 3x_{1} - 1 \to \infty\).
By the strong duality theorem, a primal LP that is unbounded forces its dual to be infeasible — if the dual had any feasible solution, weak duality would bound the primal objective from above, contradicting unboundedness. Hence the dual of this problem is infeasible.
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