Consider the following linear program:
Maximize x1 + x2
Subject to: x1 − 2x2 ≤ 4
x1 ≤ 5
2x1 − x2 ≤ 6
x1, x2 ≥ 0.
Use the simplex method to find the optimal solution to this LP or show that it does not exist. When constructing the canonical plateau, please include -z (rather than z) in the first column and the first row. So, the values in the first row should be 1 1 1 0 0 0 0

We are tasked with solving the following linear program using the simplex method:

\[
\text{Maximize } z = x_1 + x_2
\]
subject to the constraints:
\[
x_1 - 2x_2 \leq 4
\]
\[
x_1 \leq 5
\]
\[
2x_1 - x_2 \leq 6
\]
\[
x_1, x_2 \geq 0
\]

### Step 1: Convert to standard form
We first convert the inequalities into equalities by introducing slack variables \(s_1\), \(s_2\), and \(s_3\). This gives:

1. \(x_1 - 2x_2 + s_1 = 4\) (slack variable \(s_1\) for the first constraint)
2. \(x_1 + s_2 = 5\) (slack variable \(s_2\) for the second constraint)
3. \(2x_1 - x_2 + s_3 = 6\) (slack variable \(s_3\) for the third constraint)

The objective function is to maximize \(z = x_1 + x_2\), which we rewrite in terms of minimizing \(-z\). Thus, the objective function becomes:
\[
\text{Minimize } -z = -(x_1 + x_2)
\]

### Step 2: Construct the initial simplex tableau
The decision variables are \(x_1\), \(x_2\), and the slack variables are \(s_1\), \(s_2\), and \(s_3\). The tableau will include the coefficients of these variables in each constraint and the objective function.

We will include the objective function row in the form \(-z + x_1 + x_2 = 0\), as requested. The first row should contain the coefficients of \(x_1\), \(x_2\), and the slack variables along with the right-hand side constants.

The initial simplex tableau is:

\[
\begin{array}{c|cccccc|c}
-z & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
\hline
s_1 & 1 & -2 & 1 & 0 & 0 & 0 & 4 \\
s_2 & 1 & 0 & 0 & 1 & 0 & 0 & 5 \\
s_3 & 2 & -1 & 0 & 0 & 1 & 0 & 6 \\
\end{array}
\]

### Step 3: Apply the simplex method

#### Pivoting
To start optimizing, we look for the entering variable by selecting the most negative coefficient in the objective row (top row). Here, both coefficients of \(x_1\) and \(x_2\) are positive (1 and 1), meaning we already have a tableau that satisfies the optimality condition. This suggests that the current basic solution is optimal.

#### Basic solution
The basic variables are \(s_1\), \(s_2\), and \(s_3\). Since the decision variables \(x_1\) and \(x_2\) are not part of the basic variables in this tableau, their values are zero. The values of the slack variables are given by the right-hand side column:

- \(s_1 = 4\)
- \(s_2 = 5\)
- \(s_3 = 6\)

Thus, the optimal solution is:
\[
x_1 = 0, \quad x_2 = 0, \quad \text{and} \quad z = 0.
\]

### Step 4: Interpret the result
At the optimal solution, both decision variables \(x_1\) and \(x_2\) are zero, meaning that the maximum value of the objective function \(z = x_1 + x_2\) is also zero.

This solution is optimal because there are no negative entries in the top row of the tableau, which indicates that further improvements are not possible under the simplex method.

---

Would you like further clarification on any steps or more details on the simplex method? Here are some follow-up questions that might interest you:

1. What is the significance of slack variables in linear programming?
2. How does the simplex method ensure that the solution improves with each iteration?
3. What happens if there are ties for the entering or leaving variable in the simplex method?
4. How would the solution change if one of the constraints were modified?
5. What is the difference between a feasible solution and an optimal solution in linear programming?

**Tip:** In linear programming, the number of iterations required by the simplex method depends on the problem's constraints and the shape of the feasible region.

Consider the following linear program: Maximize x1 + x2 Subject to: x1 − 2x2 ≤ 4 x1 ≤ 5 2x1 − x2 ≤ 6 x1, x2 ≥ 0. Use the simplex method to find the optimal solution to this LP or show that it does not exist. When constructing the canonical plateau, please include -z (rather than z) in the first column and the first row. So, the values in the first row should be 1 1 1 0 0 0 0

Solve the linear programming problem: Maximize x1 + x2 subject to constraints x1 − 2x2 ≤ 4, x1 ≤ 5, 2x1 − x2 ≤ 6 using the simplex method. Learn how to apply slack variables and find the optimal solution. Suitable for undergraduate students studying optimization techniques.

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