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Problem 11

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

Expert verified
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: $$w_1 + 3w_2 \leq 2$$ $$2w_1 + 2w_2 \leq 5$$ $$w_1 \geq 0, w_2 \geq 0$$
See the step by step solution

## 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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