Math Problem Statement
Minimize z= 20x1 + 40x2 Sub to the constraints, 36x1 + 6x2 ≥ 108 3x1 + 12x2 ≥ 36 20x1 + 10x2 ≥ 100 x1,x2 ≥ 0 Solve the lpp method for simplex method
Solution
To minimize the objective function subject to the given constraints using the simplex method, follow these steps:
Step 1: Convert Inequalities to Equalities
First, convert the inequalities to equalities by introducing slack variables and .
36x_1 + 6x_2 - s_1 &= 108, \quad (s_1 \geq 0) \\ 3x_1 + 12x_2 - s_2 &= 36, \quad (s_2 \geq 0) \\ 20x_1 + 10x_2 - s_3 &= 100, \quad (s_3 \geq 0) \\ \end{aligned}$$ The objective function remains: $$z = 20x_1 + 40x_2$$ ### Step 2: Set Up the Initial Simplex Tableau Write the initial simplex tableau. Here, $$z$$ is converted to $$-z$$ for the simplex method (since we are minimizing). $$\begin{array}{c|cccccc|c} & x_1 & x_2 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline \text{P1} & 36 & 6 & -1 & 0 & 0 & 108 \\ \text{P2} & 3 & 12 & 0 & -1 & 0 & 36 \\ \text{P3} & 20 & 10 & 0 & 0 & -1 & 100 \\ \hline z & -20 & -40 & 0 & 0 & 0 & 0 \\ \end{array}$$ ### Step 3: Perform the Simplex Algorithm Apply the simplex method iteratively to find the optimal solution. #### Iteration 1: - **Entering Variable:** $$x_2$$ (most negative coefficient in $$z$$-row). - **Leaving Variable:** Determine using the minimum ratio test. $$\frac{108}{6} = 18, \quad \frac{36}{12} = 3, \quad \frac{100}{10} = 10$$ The smallest ratio is 3, so the leaving variable is $$s_2$$. Perform row operations to make $$x_2$$ the basic variable for $$s_2$$: $$\begin{array}{c|cccccc|c} & x_1 & x_2 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline \text{P1} & 34.5 & 0 & -1 & 0.5 & 0 & 90 \\ \text{P2} & 0.25 & 1 & 0 & -0.0833 & 0 & 3 \\ \text{P3} & 17.5 & 0 & 0 & 0.8333 & -1 & 70 \\ \hline z & -10 & 0 & 0 & -3.3333 & 0 & 120 \\ \end{array}$$ #### Iteration 2: - **Entering Variable:** $$x_1$$ (most negative coefficient in $$z$$-row). - **Leaving Variable:** Determine using the minimum ratio test. $$\frac{90}{34.5} \approx 2.61, \quad \frac{3}{0.25} = 12, \quad \frac{70}{17.5} = 4$$ The smallest ratio is approximately 2.61, so the leaving variable is $$s_1$$. Perform row operations to make $$x_1$$ the basic variable for $$s_1$$: $$\begin{array}{c|cccccc|c} & x_1 & x_2 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline \text{P1} & 1 & 0 & -0.02899 & 0.01449 & 0 & 2.61 \\ \text{P2} & 0 & 1 & -0.00725 & -0.08985 & 0 & 2.82 \\ \text{P3} & 0 & 0 & 0.50725 & 0.50725 & -1 & 24.25 \\ \hline z & 0 & 0 & -0.2899 & -3.1449 & 0 & 121.57 \\ \end{array}$$ #### Iteration 3: The $$z$$-row has no negative coefficients, indicating the optimal solution has been reached. ### Step 4: Read the Optimal Solution From the final tableau, the values of the variables are: $$x_1 = 2.61, \quad x_2 = 2.82, \quad z = 121.57$$ ### Summary The optimal solution to minimize $$z = 20x_1 + 40x_2$$ under the given constraints is: $$\boxed{z = 121.57 \text{ at } (x_1, x_2) = (2.61, 2.82)}$$ Do you need more details or have any questions? Here are some questions you might find useful: 1. How do you solve linear programming problems graphically? 2. What is the dual of a linear programming problem? 3. How does the Big M method work in linear programming? 4. Can you explain the differences between the primal and dual simplex methods? 5. What are the applications of linear programming in real-world scenarios? **Tip:** Always double-check the calculations and the pivot operations in each iteration of the simplex method to ensure accuracy.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Programming
Simplex Method
Optimization
Formulas
Simplex method tableau
Theorems
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Suitable Grade Level
College Level
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