Is there a detailed solved problem of a simplex maximization problem?

Yes, here is a detailed solved example of a simplex maximization problem:

Maximize: Z = 3x1 + 2x2

Subject to:
2x1 + x2 ≤ 10
x1 + 2x2 ≤ 8
x1, x2 ≥ 0

Step 1: Convert the inequalities into equations by adding slack variables:
2x1 + x2 + s1 = 10
x1 + 2x2 + s2 = 8

Step 2: Write the initial tableau:
| x1 | x2 | s1 | s2 | RHS |
| 2 | 1 | 1 | 0 | 10 |
| 1 | 2 | 0 | 1 | 8 |
| -3 | -2 | 0 | 0 | 0 |

Step 3: Choose the entering variable (most negative coefficient in the objective function):
Entering variable: x1

Step 4: Choose the leaving variable (minimum ratio test):
Minimum ratio for row 1: 10/2 = 5
Minimum ratio for row 2: 8/1 = 8

Leaving variable: s1

Step 5: Perform row operations to pivot:
| x1 | x2 | s1 | s2 | RHS |
| 1 | 0 | 1 | -2 | 2 |
| 0 | 1 | -2 | 3 | 6 |
| 0 | 0 | 3 | 5 | 30 |

Step 6: Repeat steps 3-5 until the solution is optimal:
Entering variable: x2
Leaving variable: s2

Pivot:
| x1 | x2 | s1 | s2 | RHS |
| 1 | 0 | 1 | -2 | 2 |
| 0 | 1 | -2 | 3 | 6 |
| 0 | 0 | 3 | 5 | 30 |

The optimal solution is x1 = 2, x2 = 6, with a maximum value of Z = 22.

Problem:

Maximize the objective function:

$$Z = 5x_1 + 3x_2$$

Subject to the following constraints:

$$x_1 + 2x_2 \leq 10$$

$$2x_1 + x_2 \leq 12$$

$$x_1, x_2 \geq 0$$

Step 1: Convert to Standard Form

Add slack variables to convert the inequalities to equations:

$$x_1 + 2x_2 + s_1 = 10$$

$$2x_1 + x_2 + s_2 = 12$$

Step 2: Construct the Initial Simplex Tableau

| Variable | Initial Basis | RHS |
|---|---|---|
| Z | 0 | 0 |
| x_1 | 1 | 10 |
| x_2 | 0 | 12 |
| s_1 | 0 | 10 |
| s_2 | 1 | 12 |

Step 3: Identify the Entering Variable

The most negative coefficient in the objective function row is -5 for x_1. Therefore, x_1 enters the basis.

Step 4: Identify the Leaving Variable

To determine the leaving variable, calculate the ratios of the RHS to the coefficients of the entering variable in the constraint rows. The smallest non-negative ratio determines the leaving variable.

| Variable | Ratio |
|---|---|
| x_1 | 10/1 = 10 |
| x_2 | 12/2 = 6 |
| s_1 | 10/1 = 10 |
| s_2 | 12/2 = 6 |

Since both s_1 and s_2 have a ratio of 10, we can choose either as the leaving variable. Let's choose s_1.

Step 5: Perform the Pivot Operation

Replace s_1 in the basis with x_1 by dividing Row 1 by the coefficient of x_1 in Row 1 (1).

| Variable | New Basis | RHS |
|---|---|---|
| Z | 0 | 0 |
| x_1 | 1 | 10 |
| x_2 | 2 | 8 |
| s_1 | 0 | 10 |
| s_2 | 0 | 108 |

Step 6: Repeat Steps 3-5 until Optimality

- Entering Variable: x_2
- Leaving Variable: s_2

| Variable | New Basis | RHS |
|---|---|---|
| Z | 0 | 0 |
| x_1 | 2 | 4 |
| x_2 | 1 | 20 |
| s_1 | 0 | 20 |
| s_2 | 0 | 0 |

All coefficients in the objective function row are non-negative, indicating optimality has been reached.

Optimal Solution:

$$x_1 = 4$$

$$x_2 = 20$$

$$Z = 140$$