91Ó°ÊÓ

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

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: \( x + 2y + u = 4 \) \( 3x + 2y + v = 6 \) \( x \geq 0, y \geq 0, u \geq 0, v \geq 0 \)

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: \( w_1 + 3w_2 \leq 2 \) \( 2w_1 + 2w_2 \leq 5 \) \( w_1 \geq 0, w_2 \geq 0 \)

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 \( C = 4 \).