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 above LP problem is :
Min. : \(W = py_{1} + qy_{2}\)
Subject to the constraints :
\(ry_{1} + sy_{2} \ge 4\)
\(-y_{1} - y_{2} \ge 2\)
\(y_{1}, y_{2} \ge 0,\)
where the values of \(p, q, r, s\) are :
Answer: D
The values are \(p = -3, q = 2, r = -1, s = 1\).
For a maximization primal, first write every constraint as \(\le\). The second constraint \(-x_{1} + x_{2} \ge -2\) becomes \(x_{1} - x_{2} \le 2\), so the primal is
\[ \begin{aligned} -x_{1} - x_{2} &\le -3 \\ x_{1} - x_{2} &\le 2 \end{aligned} \]
The dual minimizes \(-3y_{1} + 2y_{2}\) (right-hand sides), so \(p = -3\), \(q = 2\). Its constraints come from the columns: the \(x_{1}\) column \((-1, 1)\) gives \(-y_{1} + y_{2} \ge 4\), so \(r = -1\), \(s = 1\); the \(x_{2}\) column \((-1, -1)\) gives \(-y_{1} - y_{2} \ge 2\), exactly as printed.
The other options get the signs wrong, typically by converting the second constraint incorrectly or by flipping the first row's signs.
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