Question:
An objective function \(Z\) of primal variables (\(x_1\) and \(x_2\)) is described below:
\[ \begin{gathered} \text{Minimize } Z = 0.07x_1 + 0.05x_2 \\ \text{subject to} \\ 0.1x_1 \ge 0.4 \\ 0.1x_2 \ge 0.6 \\ 0.1x_1 + 0.2x_2 \ge 2.0 \\ 0.2x_1 + 0.1x_2 \ge 1.8 \\ x_1, x_2 \ge 0 \end{gathered} \]
\(W\) is the objective function of the dual of \(Z\). \(k_1\), \(k_2\), \(k_3\), and \(k_4\) represent the corresponding dual variables. Which one of the following options represents the correct form of \(W\)?
An objective function \(Z\) of primal variables (\(x_1\) and \(x_2\)) is described below:
\[ \begin{gathered} \text{Minimize } Z = 0.07x_1 + 0.05x_2 \\ \text{subject to} \\ 0.1x_1 \ge 0.4 \\ 0.1x_2 \ge 0.6 \\ 0.1x_1 + 0.2x_2 \ge 2.0 \\ 0.2x_1 + 0.1x_2 \ge 1.8 \\ x_1, x_2 \ge 0 \end{gathered} \]
\(W\) is the objective function of the dual of \(Z\). \(k_1\), \(k_2\), \(k_3\), and \(k_4\) represent the corresponding dual variables. Which one of the following options represents the correct form of \(W\)?
\[ \begin{gathered} \text{Minimize } Z = 0.07x_1 + 0.05x_2 \\ \text{subject to} \\ 0.1x_1 \ge 0.4 \\ 0.1x_2 \ge 0.6 \\ 0.1x_1 + 0.2x_2 \ge 2.0 \\ 0.2x_1 + 0.1x_2 \ge 1.8 \\ x_1, x_2 \ge 0 \end{gathered} \]
\(W\) is the objective function of the dual of \(Z\). \(k_1\), \(k_2\), \(k_3\), and \(k_4\) represent the corresponding dual variables. Which one of the following options represents the correct form of \(W\)?
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Build the dual by transposing the constraint coefficient matrix, one dual constraint per primal variable.
Updated On: Jul 27, 2026
- \[ \begin{gathered} \text{Maximize } W = 0.4k_1 + 0.6k_2 + 2.0k_3 + 1.8k_4 \\ \text{subject to} \\ 0.1k_1 + 0.1k_3 + 0.2k_4 \le 0.07 \\ 0.1k_2 + 0.2k_3 + 0.1k_4 \le 0.05 \\ k_1, k_2, k_3, k_4 \ge 0 \end{gathered} \]
- \[ \begin{gathered} \text{Maximize } W = 0.4k_1 + 0.6k_2 + 2.0k_3 + 1.8k_4 \\ \text{subject to} \\ 0.1k_1 + 0.1k_2 + 0.2k_4 \le 0.07 \\ 0.1k_2 + 0.2k_3 + 0.1k_4 \le 0.05 \\ k_1, k_2, k_3, k_4 \ge 0 \end{gathered} \]
- \[ \begin{gathered} \text{Maximize } W = 0.4k_1 + 0.6k_2 + 2.0k_3 + 1.8k_4 \\ \text{subject to} \\ 0.1k_1 + 0.1k_2 + 0.2k_4 \le 0.05 \\ 0.1k_1 + 0.2k_3 + 0.1k_4 \le 0.07 \\ k_1, k_2, k_3, k_4 \ge 0 \end{gathered} \]
- \[ \begin{gathered} \text{Maximize } W = 0.4k_1 + 0.6k_2 + 2.0k_3 + 1.8k_4 \\ \text{subject to} \\ 0.1k_1 + 0.1k_3 + 0.2k_4 \le 0.05 \\ 0.1k_2 + 0.2k_3 + 0.1k_4 \le 0.07 \\ k_1, k_2, k_3, k_4 \ge 0 \end{gathered} \]
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The Correct Option is A
Solution and Explanation
Step 1: Set up the primal in a coefficient table.
Write the four constraints as rows and the two variables as columns: row 1 has coefficients (0.1, 0) for \(x_1, x_2\) with RHS 0.4; row 2 has (0, 0.1) with RHS 0.6; row 3 has (0.1, 0.2) with RHS 2.0; row 4 has (0.2, 0.1) with RHS 1.8. The primal is a minimization with greater-or-equal constraints and non-negative variables, which is already in standard dual form.
Step 2: Turn the RHS values into the dual objective.
Each primal constraint gets one dual variable, so the dual objective uses the primal RHS values as its coefficients: \(W = 0.4k_1 + 0.6k_2 + 2.0k_3 + 1.8k_4\), and since the primal was a minimization, the dual becomes a maximization.
Step 3: Turn the primal costs into the dual RHS.
Each primal variable gets one dual constraint, and its RHS is that variable's cost coefficient in \(Z\): \(x_1\) gives RHS 0.07, and \(x_2\) gives RHS 0.05.
Step 4: Fill in the dual constraint coefficients by reading down the columns.
For the \(x_1\) column, read (0.1, 0, 0.1, 0.2) down rows 1 to 4, giving \(0.1k_1 + 0.1k_3 + 0.2k_4 \le 0.07\) (the zero coefficient for \(k_2\) is simply dropped). For the \(x_2\) column, read (0, 0.1, 0.2, 0.1), giving \(0.1k_2 + 0.2k_3 + 0.1k_4 \le 0.05\). Since the primal had greater-or-equal constraints and non-negative variables, the dual inequality direction flips to less-or-equal.
Final Answer:
Option (A) has exactly these two constraints with the correct RHS values 0.07 and 0.05, while the others swap a coefficient between \(k_2\) and \(k_3\) or swap the two RHS values. \[ \boxed{\text{Option A}} \]
Write the four constraints as rows and the two variables as columns: row 1 has coefficients (0.1, 0) for \(x_1, x_2\) with RHS 0.4; row 2 has (0, 0.1) with RHS 0.6; row 3 has (0.1, 0.2) with RHS 2.0; row 4 has (0.2, 0.1) with RHS 1.8. The primal is a minimization with greater-or-equal constraints and non-negative variables, which is already in standard dual form.
Step 2: Turn the RHS values into the dual objective.
Each primal constraint gets one dual variable, so the dual objective uses the primal RHS values as its coefficients: \(W = 0.4k_1 + 0.6k_2 + 2.0k_3 + 1.8k_4\), and since the primal was a minimization, the dual becomes a maximization.
Step 3: Turn the primal costs into the dual RHS.
Each primal variable gets one dual constraint, and its RHS is that variable's cost coefficient in \(Z\): \(x_1\) gives RHS 0.07, and \(x_2\) gives RHS 0.05.
Step 4: Fill in the dual constraint coefficients by reading down the columns.
For the \(x_1\) column, read (0.1, 0, 0.1, 0.2) down rows 1 to 4, giving \(0.1k_1 + 0.1k_3 + 0.2k_4 \le 0.07\) (the zero coefficient for \(k_2\) is simply dropped). For the \(x_2\) column, read (0, 0.1, 0.2, 0.1), giving \(0.1k_2 + 0.2k_3 + 0.1k_4 \le 0.05\). Since the primal had greater-or-equal constraints and non-negative variables, the dual inequality direction flips to less-or-equal.
Final Answer:
Option (A) has exactly these two constraints with the correct RHS values 0.07 and 0.05, while the others swap a coefficient between \(k_2\) and \(k_3\) or swap the two RHS values. \[ \boxed{\text{Option A}} \]
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