Final answer:
The optimal objective value for this linear optimization problem is 20, and there are three optimal basic feasible solutions.
Step-by-step explanation:
Solving the Three-Dimensional Linear Optimization Problem
**Step 1: Converting to Standard Form:**
First, we need to convert the inequalities to standard form by introducing slack variables (s1, s2, s3):
x1 + 2x2 + 2x3 + s1 = 20
2x1 + x2 + 2x3 + s2 = 20
2x1 + 2x2 + x3 + s3 = 20
x1, x2, x3, s1, s2, s3 >= 0
**Step 2: Listing Basic Solutions:**
A basic solution is one where there are exactly as many non-zero variables as there are constraints.
In this case, we have 4 constraints and 6 variables, leading to 15 possible combinations of non-zero variables:
1. (x1, s1, s2, s3)
2. (x1, s1, x3, s2)
3. (x1, s1, x3, s3)
4. (x1, x2, s1, s3)
5. (x1, x2, x3, s1)
6. (x1, x2, x3, s2)
7. (x2, s1, s2, s3)
8. (x2, s1, x3, s2)
9. (x2, s1, x3, s3)
10. (x2, x3, s1, s2)
11. (x2, x3, s1, s3)
12. (x2, x3, s2, s3)
13. (x3, s1, s2, s3)
14. (x3, s1, x2, s2)
15. (x3, s1, x2, s3)
**Step 3: Finding Basic Feasible Solutions:**
A basic feasible solution is a basic solution that satisfies all the inequalities. We need to check each of the 15 basic solutions to see if they satisfy the inequalities:
- Solutions 1-6 and 12-15 have at least one negative variable, so they are not feasible.
- Solutions 7-11 satisfy all inequalities and are therefore basic feasible solutions.
**Step 4: Computing Objective Function Values:**
For each basic feasible solution (7-11), substitute the non-zero variable values into the objective function:
- Solution 7: x1 + x2 + x3 = 0
- Solution 8: x1 + x3 = 20
- Solution 9: x1 + x2 = 20
- Solution 10: x2 + x3 = 20
- Solution 11: x2 = 20
The objective function values are:
- Solution 7: 0
- Solution 8: 40
- Solution 9: 20
- Solution 10: 20
- Solution 11: 20
**Step 5: Solving the Optimization Problem:**
The optimal solution is the basic feasible solution with the highest objective function value.
In this case, solutions 8, 9, and 11 all have the optimal value of 20. Therefore, there are **three optimal basic feasible solutions**:
- (x1, x3) = (20, 0)
- (x1, x2) = (10, 10)
- (x2) = 20
**Conclusion:**
The optimal objective value for this linear optimization problem is 20, and there are three optimal basic feasible solutions.