Solve the given linear programming problem using the table method. The table of basic solutions is provided.
Maximize
Pequals=1515x 1x1plus+1111x 2x2
subject to
x 1x1plus+x 2x2less than or equals≤6
x 1x1plus+4x 2x2less than or equals≤12
x 1x1​, x 2x2greater than or equals≥0
Question content area bottom left
Part 1
x 1x1
x 2x2
s 1s1
s 2s2
​Feasible?
0
0
6
12
Yes
0
6
0
minus−12
No
0
3
3
0
Yes
6
0
0
6
Yes
12
0
minus−6
0
No
4
2
0
0
Yes
 
.
.
.
Question content area right
Part 1
Select the correct choice below​ and, if​ necessary, fill in the answer boxes within your choice.
A.
The function is maximized when x 1x1equals=enter your response here​, x 2x2equals=enter your response here s 1s1equals=enter your response here​, and s 2s2equals=enter your response here​, where Pequals=enter your response here.
​(Simplify your​ answers.)
B.
The system has no solution.

To solve this linear programming problem using the table method, let's first analyze the constraints and objective function.

### Problem:
**Maximize**  
\[ P = 15x_1 + 11x_2 \]

**Subject to the constraints**:  
\[
\begin{align*}
x_1 + x_2 &\leq 6 \\
x_1 + 4x_2 &\leq 12 \\
x_1, x_2 &\geq 0
\end{align*}
\]
Also, there are slack variables \(s_1\) and \(s_2\) associated with these inequalities, defined as follows:
\[
x_1 + x_2 + s_1 = 6 \quad \text{and} \quad x_1 + 4x_2 + s_2 = 12
\]

The table of basic solutions provided gives several potential feasible and non-feasible solutions, and we will evaluate them based on the objective function.

### Table of solutions:

| \(x_1\) | \(x_2\) | \(s_1\) | \(s_2\) | Feasible? |
|--------|--------|--------|--------|-----------|
| 0      | 0      | 6      | 12     | Yes       |
| 0      | 6      | 0      | -12    | No        |
| 0      | 3      | 3      | 0      | Yes       |
| 6      | 0      | 0      | 6      | Yes       |
| 12     | 0      | -6     | 0      | No        |
| 4      | 2      | 0      | 0      | Yes       |

### Evaluating feasible solutions:

1. **\( x_1 = 0, x_2 = 0 \)**:
   \[
   P = 15(0) + 11(0) = 0
   \]
   Feasible.

2. **\( x_1 = 0, x_2 = 3 \)**:
   \[
   P = 15(0) + 11(3) = 33
   \]
   Feasible.

3. **\( x_1 = 6, x_2 = 0 \)**:
   \[
   P = 15(6) + 11(0) = 90
   \]
   Feasible.

4. **\( x_1 = 4, x_2 = 2 \)**:
   \[
   P = 15(4) + 11(2) = 60 + 22 = 82
   \]
   Feasible.

### Conclusion:
The maximum value of \( P \) is achieved when \( x_1 = 6 \) and \( x_2 = 0 \), which gives \( P = 90 \).

Thus, the correct solution is:

\[
\text{The function is maximized when } x_1 = 6, x_2 = 0, s_1 = 0, s_2 = 6, \text{ where } P = 90.
\]

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:
1. How do slack variables help in linear programming problems?
2. What are the graphical methods of solving linear programming problems?
3. Can the simplex method be applied to this problem, and how would it work?
4. What is the dual of this linear programming problem, and how can it be solved?
5. How does degeneracy affect the solution in linear programming?

**Tip:** Always check the feasibility of solutions in linear programming problems by verifying the constraints.

Solve the given linear programming problem using the table method. The table of basic solutions is provided. Maximize P = 15x1 + 11x2 subject to x1 + x2 ≤ 6, x1 + 4x2 ≤ 12, x1, x2 ≥ 0.

Learn how to solve a linear programming problem using the table method. This example maximizes P = 15x1 + 11x2, subject to constraints x1 + x2 ≤ 6 and x1 + 4x2 ≤ 12, with slack variables s1 and s2. Explore feasible solutions and find the optimal value of the objective function. Suitable for Grades 10-12.

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