Discussion
Consider the LP problem :
Max \(Z = 4x_{1} + 2x_{2}\)
Subject to the constraints :
\(-x_{1} - x_{2} \le -3\)
\(-x_{1} + x_{2} \ge -2\) and
\(x_{1}, x_{2} \ge 0.\)
Then the dual of the LP problem is :
Min. \(W = ay_{1} + by_{2}\)
\(-py_{1} + qy_{2} \ge 4\)
\(-y_{1} - y_{2} \ge 2\) and \(y_{1}, y_{2} \ge 0\)
where the values of \(a, b, p, q\) are :
Answer: C
The values are \(a = -3, b = 2, p = 1, q = 1\).
For a maximisation problem, write every constraint as \(\le\). The second constraint \(-x_{1} + x_{2} \ge -2\) becomes \(x_{1} - x_{2} \le 2\). The primal is then
\[ \begin{aligned} -x_{1} - x_{2} &\le -3 \\ x_{1} - x_{2} &\le 2 \end{aligned} \]
The dual minimises \(W = -3y_{1} + 2y_{2}\) (the right-hand sides), with one constraint for each primal variable using the columns of coefficients: for \(x_{1}\), \(-y_{1} + y_{2} \ge 4\), and for \(x_{2}\), \(-y_{1} - y_{2} \ge 2\). Matching \(-py_{1} + qy_{2} \ge 4\) gives \(p = 1, q = 1\), and \(a = -3, b = 2\). The other options get a sign wrong in the objective or in the first constraint.
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