Discussion
Let \(y = (\alpha, -1)^{T}\), \(\alpha \in \mathbb{R}\) be a feasible solution for the dual problem of the linear programming problem
Maximize: \(5x_{1} + 12x_{2}\)
subject to: \(x_{1} + 2x_{2} + x_{3} \leq 10\)
\(2x_{1} - x_{2} + 3x_{3} = 8\)
\(x_{1}, x_{2}, x_{3} \geq 0\).
Which of the following statements is TRUE?
Answer: D
The answer is \(\alpha \geq 7\).
The primal is a maximisation with one 'at most' constraint and one equality, so its dual is: minimise \(10y_{1} + 8y_{2}\) with \(y_{1} \geq 0\), \(y_{2}\) free, and one 'at least' constraint per primal variable, from the columns of the coefficients:
\[ \begin{aligned} y_{1} + 2y_{2} &\geq 5 \\ 2y_{1} - y_{2} &\geq 12 \\ y_{1} + 3y_{2} &\geq 0 \end{aligned} \]
\[ \begin{aligned} y_{2} = -1: \ \alpha - 2 &\geq 5 \Rightarrow \alpha \geq 7 \\ 2\alpha + 1 &\geq 12 \Rightarrow \alpha \geq 5.5 \\ \alpha - 3 &\geq 0 \Rightarrow \alpha \geq 3 \end{aligned} \]
All three must hold, and the strictest is \(\alpha \geq 7\). The ranges starting at 3 or 5.5 satisfy only the weaker constraints, and \(\alpha \lt 3\) fails all three.
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