Minimize using the simplex method: Z = 8x1 + 10x2 + 7x3 Subject to: x1 + 3x2 + 2x3 ≤ 10 -x1 - 5x2 - x3 ≥ -8 x1, x2, x3 ≥ 0

To minimize Z = 8x1 + 10x2 + 7x3, subject to the constraints:
1. x1 + 3x2 + 2x3 ≤ 10
2. -x1 - 5x2 - x3 ≥ -8
3. x1, x2, x3 ≥ 0

We first convert the inequalities to equations by introducing slack variables:
1. x1 + 3x2 + 2x3 + x4 = 10
2. -x1 - 5x2 - x3 - x5 = -8

The initial simplex tableau is:
```
| x1 | x2 | x3 | x4 | x5 | RHS |
|------|------|------|------|------|-------|
| 1 | 3 | 2 | 1 | 0 | 10 |
| -1 | -5 | -1 | 0 | 1 | -8 |
| -8 | -10 | -7 | 0 | 0 | 0 |
```

We choose x1 as the entering variable and x5 as the leaving variable. Perform row operations to pivot:
```
| x1 | x2 | x3 | x4 | x5 | RHS |
|------|------|------|------|------|-------|
| 0 | -2 | 1 | 1 | 1 | 2 |
| 1 | 3 | 2 | 1 | 0 | 10 |
| 0 | 30 | 23 | 8 | 8 | 80 |
```

The next pivot is x2 entering and x4 leaving:
```
| x1 | x2 | x3 | x4 | x5 | RHS |
|------|------|------|------|------|-------|
| 0 | 0 | 5 | -2 | 3 | 8 |
| 1 | 0 | -4 | 5 | -3 | 2 |
| 0 | 1 | 23/30 | 8/30 | 8/30 | 8/3 |
```

The optimal solution is x1 = 2, x2 = 8/3, x3 = 8/3 with Z = 8(2) + 10(8/3) + 7(8/3) = 64/3.

Step 1: Convert to Standard Form

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

x1 3x2 2x3 + s1 = 10
-x1 - 5x2 - x3 + s2 = -8
x1, x2, x3, s1, s2 ≥ 0

Step 2: Construct the Initial Simplex Tableau

| Variable | x1 | x2 | x3 | s1 | s2 | RHS |
|---|---|---|---|---|---|---|
| Z | -8 | -10 | -7 | 0 | 0 | 0 |
| s1 | 1 | 3 | 2 | 1 | 0 | 10 |
| s2 | -1 | -5 | -1 | 0 | 1 | -8 |

Step 3: Choose the Entering Variable

The variable with the most negative coefficient in the Z-row is x1, so x1 enters the basis.

Step 4: Choose the Leaving Variable

To determine the leaving variable, calculate the ratios of the RHS to the coefficients of x1 in the constraint rows:

Ratio for s1: 10 / 1 = 10
Ratio for s2: -8 / -1 = 8

The smallest ratio is 8, so s2 leaves the basis.

Step 5: Perform the Pivot Operation

Divide the row of s2 by its pivot element (-1) to get:

| Variable | x1 | x2 | x3 | s1 | s2 | RHS |
|---|---|---|---|---|---|---|
| Z | -8 | -10 | -7 | 0 | 0 | 0 |
| s1 | 1 | 3 | 2 | 1 | 0 | 10 |
| x1 | 1 | 5 | 1 | 0 | -1 | 8 |

Step 6: Repeat Steps 3-5 until an Optimal Solution is Reached

x2 enters the basis, s1 leaves the basis.
x3 enters the basis, x2 leaves the basis.

The final simplex tableau is:

| Variable | x1 | x2 | x3 | s1 | s2 | RHS |
|---|---|---|---|---|---|---|
| Z | 0 | 0 | 0 | 0 | 0 | 68 |
| s1 | 0 | 0 | -1 | 1 | 0 | 2 |
| x2 | 0 | 0 | 2 | -1 | 0 | 4 |
| x3 | 0 | 0 | 2 | 0 | -1 | 10 |

Optimal Solution:

Z = 68
x1 = 0
x2 = 4
x3 = 10