Mindyard | An objective function Z of primal variable ( x 1 , x 2 ) is described. K 1 , K 2 , K 3 , K 4 are the dual variables. Min  Z = 0.07 x 1 + 0.05 x 2 Subject to: 0.1 x 1 ≥ 0.4 0.1 x 2 ≥ 0.6 0.1 x 1 + 0.2 x 2 ≥ 2.0 0.2 x 1 + 0.1 x 2 ≥ 1.8 x 1 , x 2 ≥ 0 W is the objective function of dual of Z . Which option is correct? Max W = 0.4 K 1 + 0.6 K 2 + 2.0 K 3 + 1.8 K 4 0.1 K 1 + 0.1 K 3 + 0.2 K 4 ≤ 0.07 0.1 K 2 + 0.2 K 3 + 0.1 K 4 ≤ 0.05 K 1 , K 2 , K 3 , K 4     , Max W = 0.4 K 1 + 0.6 K 2 + 2.0 K 3 + 1.8 K 4 0.1 K 1 + 0.1 K 3 + 0.2 K 4 ≥ 0.07 0.1 K 2 + 0.2 K 3 + 0.1 K 4 ≥ 0.05 K 1 , K 2 , K 3 , K 4 , Max W = 0.4 K 1 + 0.6 K 2 + 2.0 K 3 + 1.8 K 4 0.1 K 1 + 0.1 K 3 + 0.2 K 4 ≥ 0.05 0.1 K 2 + 0.2 K 3 + 0.1 K 4 ≥ 0.07 K 1 K 2 K 3 K 4 , Max W =0.4K1+0.6K2+2.0K3+1.8K40.1K2+0.1K3+0.2K4≥0.070.1K1+0.2K2+0.1K4≥0.05K1K2K3K4

An objective function $Z$ of primal variable $(x_{1}, x_{2})$ is described. $K_{1}, K_{2}, K_{3}, K_{4}$ are the dual variables.

$Min Z = 0.07 x_{1} + 0.05 x_{2}$

Subject to:

$0.1 x_{1} \geq 0.4$

$0.1 x_{2} \geq 0.6$

$0.1 x_{1} + 0.2 x_{2} \geq 2.0$

$0.2 x_{1} + 0.1 x_{2} \geq 1.8$

$x_{1}, x_{2} \geq 0$

$W$ is the objective function of dual of $Z$ . Which option is correct?

Options

$Max W = 0.4 K_{1} + 0.6 K_{2} + 2.0 K_{3} + 1.8 K_{4}$

$0.1 K_{1} + 0.1 K_{3} + 0.2 K_{4} \leq 0.07$

$0.1 K_{2} + 0.2 K_{3} + 0.1 K_{4} \leq 0.05$

$K_{1}, K_{2}, K_{3}, K_{4}$

$Max W = 0.4 K_{1} + 0.6 K_{2} + 2.0 K_{3} + 1.8 K_{4}$

$0.1 K_{1} + 0.1 K_{3} + 0.2 K_{4} \geq 0.07$

$0.1 K_{2} + 0.2 K_{3} + 0.1 K_{4} \geq 0.05$

$K_{1}, K_{2}, K_{3}, K_{4}$

$Max W = 0.4 K_{1} + 0.6 K_{2} + 2.0 K_{3} + 1.8 K_{4}$

$0.1 K_{1} + 0.1 K_{3} + 0.2 K_{4} \geq 0.05$

$0.1 K_{2} + 0.2 K_{3} + 0.1 K_{4} \geq 0.07$

$K_{1} K_{2} K_{3} K_{4}$

$Max W = 0.4 K_{1} + 0.6 K_{2} + 2.0 K_{3} + 1.8 K_{4}$

$0.1 K_{2} + 0.1 K_{3} + 0.2 K_{4} \geq 0.07$

$0.1 K_{1} + 0.2 K_{2} + 0.1 K_{4} \geq 0.05$

$K_{1} K_{2} K_{3} K_{4}$

Correct Answer :

Option A

$Max W = 0.4 K_{1} + 0.6 K_{2} + 2.0 K_{3} + 1.8 K_{4}$

$0.1 K_{1} + 0.1 K_{3} + 0.2 K_{4} \leq 0.07$

$0.1 K_{2} + 0.2 K_{3} + 0.1 K_{4} \leq 0.05$

$K_{1}, K_{2}, K_{3}, K_{4}$

Solution :

To convert a primal Linear Programming Problem (LPP) to its dual form, we follow standard duality rules step-by-step.

Step 1: Identify Primal Elements
The given primal problem is a minimization problem:
Primal Objective Function:

$Min Z = 0.07 x_{1} + 0.05 x_{2}$

Subject to the constraints:
1.

0.1 x_{1} + 0.0 x_{2} \geq 0.4

(associated dual variable:

K_{1}

)
2.

0.0 x_{1} + 0.1 x_{2} \geq 0.6

(associated dual variable:

K_{2}

)
3.

0.1 x_{1} + 0.2 x_{2} \geq 2.0

(associated dual variable:

K_{3}

)
4.

0.2 x_{1} + 0.1 x_{2} \geq 1.8

(associated dual variable:

K_{4}

)
with non-negativity condition

x_{1}, x_{2} \geq 0

Step 2: Dual Objective Function
Since the primal is a Minimization problem, the dual will be a Maximization problem.
The coefficients of the dual objective function $W$ come from the right-hand side (RHS) constants of the primal constraints:

$Max W = 0.4 K_{1} + 0.6 K_{2} + 2.0 K_{3} + 1.8 K_{4}$

Step 3: Dual Constraints
The transpose of the primal coefficient matrix defines the dual constraints. Since all primal constraints are of the type $\geq$ , all dual constraints will be of the type $\leq$ , and their RHS values are the coefficients of the primal objective function.

For primal variable $x_{1}$ (column coefficients: 0.1, 0.0, 0.1, 0.2 and primal cost coefficient 0.07):

$0.1 K_{1} + 0.1 K_{3} + 0.2 K_{4} \leq 0.07$

For primal variable $x_{2}$ (column coefficients: 0.0, 0.1, 0.2, 0.1 and primal cost coefficient 0.05):

$0.1 K_{2} + 0.2 K_{3} + 0.1 K_{4} \leq 0.05$

Step 4: Non-negativity Restrictions
All dual variables are non-negative:

$K_{1}, K_{2}, K_{3}, K_{4} \geq 0$

Thus, the correct dual formulation matches Option 1:

$Max W = 0.4 K_{1} + 0.6 K_{2} + 2.0 K_{3} + 1.8 K_{4}$