Solved by the simplex method: Linear programming questions involving the minimization of three variables.

To solve a linear programming problem involving the minimization of three variables using the simplex method, we need to set up the problem in standard form and then apply the simplex method to find the optimal solution.

Let's consider the following linear programming problem:

Minimize:
Z = 3x + 2y + 4z

Subject to:
2x + y + z ≥ 10
x + 2y + 3z ≥ 15
x, y, z ≥ 0

To solve this problem using the simplex method, we first need to convert the problem into standard form by introducing slack variables:

Minimize:
Z = 3x + 2y + 4z

Subject to:
2x + y + z + s1 = 10
x + 2y + 3z + s2 = 15
x, y, z, s1, s2 ≥ 0

Now, we can set up the initial simplex tableau:

| x | y | z | s1 | s2 | RHS |
|---|---|---|----|----|-----|
| 2 | 1 | 1 | 1 | 0 | 10 |
| 1 | 2 | 3 | 0 | 1 | 15 |
| 3 | 2 | 4 | 0 | 0 | 0 |

Next, we apply the simplex method to find the optimal solution. We perform row operations to pivot and update the tableau until we reach the optimal solution.

After applying the simplex method, we find the optimal solution:
x = 0, y = 5, z = 0, Z = 10

Therefore, the minimum value of Z is 10, and the optimal values of x, y, and z are 0, 5, and 0, respectively.

Problem:

Minimize the objective function:

```
Z = 2x + 3y + 4z
```

Subject to the constraints:

```
x + 2y + 3z >= 6
2x + y + 2z >= 5
x, y, z >= 0
```

Solution:

Step 1: Write the problem in standard form

Add slack variables s1 and s2 to convert the inequality constraints to equalities:

```
x + 2y + 3z + s1 = 6
2x + y + 2z + s2 = 5
```

Step 2: Construct the initial simplex tableau

The initial tableau is:

```
| x | y | z | s1 | s2 | RHS |
|---|---|---|---|---|---|
| 1 | 2 | 3 | 1 | 0 | 6 |
| 2 | 1 | 2 | 0 | 1 | 5 |
|---------------------------|
| 2 | 3 | 4 | 0 | 0 | Z |
```

Step 3: Find the pivot column

The most negative coefficient in the bottom row is -4, which is in column z. So, z is the entering variable.

Step 4: Find the pivot row

To find the pivot row, we divide the RHS values by the corresponding coefficients of the entering variable in each row:

| Row | RHS / -4 |
|---|---|
| 1 | 6 / -4 = -1.5 |
| 2 | 5 / -4 = -1.25 |

The row with the smallest non-negative quotient is row 1. So, row 1 is the pivot row.

Step 5: Pivot

Pivot on the element -4 in row 1, column z:

```
| x | y | z | s1 | s2 | RHS |
|---|---|---|---|---|---|
| 1 | 1/4 | 0 | 1/4 | -3/4 | 1 |
| 2 | 1/2 | 1 | -1/2 | 1/2 | 2 |
|---------------------------|
| 2 | 2 | 0 | 1 | -2 | Z |
```

Step 6: Check for optimality

All coefficients in the bottom row are non-negative. Therefore, the optimal solution has been reached.

Step 7: Read the optimal solution

The optimal values of the decision variables are:

```
x = 1, y = 2, z = 0
```

Step 8: Find the optimal value of the objective function

Substituting the optimal solutions into the objective function, we get:

```
Z = 2(1) + 3(2) + 4(0) = 10
```

Therefore, the minimum value of the objective function is Z = 10.