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Applied Mathematics: For the Managerial, Life, and Social Sciences

Soo T. Tan

Chapter 6: Chapter 6

Problem 11

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Construct the dual problem associated with the primal problem. Solve the primal problem. $$ \begin{array}{rr} \text { Minimize } & C=2 x+5 y \\ \text { subject to } & x+2 y \geq 4 \\ & 3 x+2 y \geq 6 \\ & x \geq 0, y \geq 0 \end{array} $$

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The optimal solution of the primal problem is $ x = 2, y = 0 $ with a minimal value of $ C = 4 $. The dual problem is given by: Objective: Maximize: $ W = 4w_1 + 6w_2 $ Subject to: $ w_1 + 3w_2 \leq 2 $ $ 2w_1 + 2w_2 \leq 5 $ $ w_1 \geq 0, w_2 \geq 0 $

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Step 1: Identify the primal problem's objective and constraints

The primal problem is given by: Objective: Minimize: $ C = 2x + 5y $ Constraints: $ x + 2y \geq 4 $ $ 3x + 2y \geq 6 $ $ x \geq 0, y \geq 0 $

Step 2: Write the primal problem in standard form

To write the primal problem in standard form, we need to rewrite the inequalities as equalities. We can introduce slack variables $u$ and $v$ to make the inequalities as equalities. $ x + 2y + u = 4 $ $ 3x + 2y + v = 6 $ So the new primal problem becomes: Minimize: $ C = 2x + 5y $ Subject to: $ x + 2y + u = 4 $ $ 3x + 2y + v = 6 $ $ x \geq 0, y \geq 0, u \geq 0, v \geq 0 $

Step 3: Construct the dual problem

The dual problem is a linear programming problem that can be constructed from the primal. We will have two variables, say $w_1$ and $w_2$, representing the constraints. To form the dual, we will reverse the objective and constraints, meaning minimization will become maximization, and vice versa. Objective: Maximize: $ W = 4w_1 + 6w_2 $ Subject to: $ w_1 + 3w_2 \leq 2 $ $ 2w_1 + 2w_2 \leq 5 $ $ w_1 \geq 0, w_2 \geq 0 $

Step 4: Solve the primal problem

To solve the primal problem: Minimize: $ C = 2x + 5y $ Subject to: $ x + 2y + u = 4 $ $ 3x + 2y + v = 6 $ $ x \geq 0, y \geq 0, u \geq 0, v \geq 0 $ We can use the Simplex method or graphical method. Here, we'll solve using the graphical method. First, we draw the constraint lines for the primal: 1. For the constraint $x + 2y + u = 4$, when $x = 0$, $y = 2$, and when $y = 0$, $x = 4$. 2. For the constraint $3x + 2y + v = 6$, when $x = 0$, $y = 3$, and when $y = 0$, $x = 2$. Plotting these lines and finding the feasible region, we find the intersection points of the constraint lines with the axes and each other to obtain the following feasible region: - The origin $ (0,0) $ - The intersection of the first constraint with the y-axis: $ (0,2) $ - The intersection of the second constraint with the x-axis: $ (2,0) $ - The intersection of the constraints: $ (1,1.5) $ Now we evaluate the objective function at these points: 1. $ C(0,0) = 2(0) + 5(0) = 0 $ 2. $ C(0,2) = 2(0) + 5(2) = 10 $ 3. $ C(2,0) = 2(2) + 5(0) = 4 $ 4. $ C(1,1.5) = 2(1) + 5(1.5) = 9.5 $ The minimum of these values occurs at point $ (2,0) $ with an objective value of $ C = 4 $. The optimal solution of the primal problem is $ x = 2, y = 0 $ with a minimal value of $ C = 4 $.

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Chapter 6: Chapter 6

Construct the dual problem associated with the primal problem. Solve the primal problem. $$ \begin{array}{rr} \text { Minimize } & C=2 x+5 y \\ \text { subject to } & x+2 y \geq 4 \\ & 3 x+2 y \geq 6 \\ & x \geq 0, y \geq 0 \end{array} $$

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Step 1: Identify the primal problem's objective and constraints

Step 2: Write the primal problem in standard form

Step 3: Construct the dual problem

Step 4: Solve the primal problem

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More chapters from the book ‘Applied Mathematics: For the Managerial, Life, and Social Sciences’

Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

Chapter 12

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