Use the simplex method to minimize Z = X1 + 10X2 + 7X3 subject to X1 + 3X2 + 2X3 ≤ 10 -X1 - 5X2 - X3 ≥ -8 X1, X2, X3 ≥ 0

To use the simplex method, we first need to convert the inequalities into equations by introducing slack variables:

1. X1 + 3X2 + 2X3 + X4 = 10
2. -X1 - 5X2 - X3 - X5 = -8

Now we can write the initial simplex tableau:

| X1 | X2 | X3 | X4 | X5 | RHS |
|----|----|----|----|----|-----|
| 1 | 3 | 2 | 1 | 0 | 10 |
| -1 | -5 | -1 | 0 | 1 | -8 |
| -1 | -10| -7 | 0 | 0 | 0 |

Next, we need to choose the entering variable. Since X1 has the most negative coefficient in the objective function, we choose X1 as the entering variable. To determine the leaving variable, we calculate the minimum ratio test for each constraint:

10/1 = 10
-8/-1 = 8

The minimum ratio is 8, so X5 leaves the basis. Perform row operations to pivot X1 into the basis:

| X1 | X2 | X3 | X4 | X5 | RHS |
|----|----|----|----|----|-----|
| 0 | -2 | 1 | 1 | 1 | 2 |
| 0 | -3 | 0 | 1 | 1 | 0 |
| 1 | 10 | 7 | 0 | 0 | 0 |

Now, X2 enters the basis. Calculate the minimum ratio test:

2/-2 = -1
0/-3 = N/A

Since the minimum ratio is negative, the solution is unbounded. Therefore, the optimal solution is not bounded and the problem is infeasible.

Step 1: Write the problem in standard form

Maximize Z' = -Z = -X1 - 10X2 - 7X3
Subject to
X1 + 3X2 + 2X3 + S1 = 10
-X1 - 5X2 - X3 + S2 = -8
X1, X2, X3, S1, S2 ≥ 0

Step 2: Create the initial simplex tableau

| | X1 | X2 | X3 | S1 | S2 | RHS |
|---|---|---|---|---|---|---|
| Z' | -1 | -10 | -7 | 0 | 0 | 0 |
| X1 | 1 | 3 | 2 | 1 | 0 | 10 |
| X2 | -1 | -5 | -1 | 0 | 1 | -8 |

Step 3: Select the entering variable

The most negative coefficient in the Z' row is -10, which corresponds to X2. So, X2 is the entering variable.

Step 4: Select the leaving variable

To determine the leaving variable, divide each positive number in the RHS column by the corresponding coefficient of the entering variable in the pivot row.

| Variable | RHS | Coefficient | Ratio |
|---|---|---|---|
| X1 | 10 | 1 | 10 |
| S2 | -8 | -1 | 8 |

The smallest ratio is 8, so S2 is the leaving variable.

Step 5: Perform the pivot operation

Divide the pivot row (row 2) by the pivot element (-1).
Subtract 3 times row 2 from row 1.
Subtract 5 times row 2 from row 3.

| | X1 | X2 | X3 | S1 | S2 | RHS |
|---|---|---|---|---|---|---|
| Z' | -1 | 0 | -23 | 3 | 0 | -10 |
| X1 | 4 | 1 | 4 | 1 | 0 | 2 |
| X2 | 0 | 1 | 4 | 0 | 1 | -2 |

Step 6: Check for optimality

There are no negative coefficients in the Z' row, so the current solution is optimal.

Step 7: Extract the solution

The optimal solution is X1 = 2, X2 = -2 (but it is non-negative, so it's set to 0), X3 = 0.

Step 8: Calculate the optimal objective value

Z' = -10 = -Z = Z
Therefore, the minimum value of Z is 10.