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

To minimize the objective function \( z = 20x_1 + 40x_2 \) 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 \( s_1, s_2, \) and \( s_3 \).

\[
\begin{aligned}
36x_1 + 6x_2 - s_1 &= 108, \quad (s_1 \geq 0) \\
3x_1 + 12x_2 - s_2 &= 36, \quad (s_2 \geq 0) \\
20x_1 + 10x_2 - s_3 &= 100, \quad (s_3 \geq 0) \\
\end{aligned}
\]

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.

Learn how to minimize z= 20x1 + 40x2 subject to linear constraints using the simplex method. Follow a step-by-step solution with detailed simplex tableau iterations. Suitable for college-level students and professionals.

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