Question:
The dual of a LPP is
Minimize $w = 4w_1 + 6w_2 + 5w_3 - w_4$
The dual of a LPP is
Minimize $w = 4w_1 + 6w_2 + 5w_3 - w_4$
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The dual of a minimization problem is a maximization problem, and the objective coefficients and constraints switch roles between the primal and the dual.
Updated On: Sep 4, 2025
- Maximize $z = -3x_1 + 2x_2$
- Maximize $z = x_1 + x_3$
- Maximize $z = x_3 - x_4$
- Maximize $z = 3x_1 - 2x_2$
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The Correct Option is D
Solution and Explanation
The dual of a linear programming problem (LPP) can be formulated from the given primal. For a given primal problem, we can construct the dual by considering the constraints and the objective function of the primal problem.
- The primal objective function coefficients become the right-hand side (RHS) values of the dual constraints.
- The dual variables correspond to the primal constraints. In this case, the primal is a minimization problem with constraints involving $w_1$, $w_2$, $w_3$, and $w_4$. The dual formulation maximizes the expression \(z = 3x_1 - 2x_2\), making (D) the correct answer.
- The primal objective function coefficients become the right-hand side (RHS) values of the dual constraints.
- The dual variables correspond to the primal constraints. In this case, the primal is a minimization problem with constraints involving $w_1$, $w_2$, $w_3$, and $w_4$. The dual formulation maximizes the expression \(z = 3x_1 - 2x_2\), making (D) the correct answer.
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