Use the simplex method to solve the linear programming problem.
Maximize
z equals 10 x 1 plus 24 x 2
subject​ to:
12 x 1 plus4 x 2
less than or equals 52
6 x 1 plusx 2
less than or equals 20
4 x 1 plus4 x 2
less than or equals 56
with
x 1greater than or equals​0, x 2greater than or equals0

We are given a linear programming problem to solve using the **Simplex Method**. The goal is to **maximize** the objective function subject to a set of constraints.

### Objective Function:
\[
\text{Maximize } z = 10x_1 + 24x_2
\]

### Constraints:
1. \(12x_1 + 4x_2 \leq 52\)
2. \(6x_1 + x_2 \leq 20\)
3. \(4x_1 + 4x_2 \leq 56\)

Additionally, the non-negativity constraints are:
\[
x_1 \geq 0, \quad x_2 \geq 0
\]

### Steps to solve using the Simplex Method:

1. **Convert inequalities to equalities by introducing slack variables** for each constraint:
   - For \(12x_1 + 4x_2 \leq 52\), introduce slack variable \(s_1\), giving:  
     \[
     12x_1 + 4x_2 + s_1 = 52
     \]
   - For \(6x_1 + x_2 \leq 20\), introduce slack variable \(s_2\), giving:  
     \[
     6x_1 + x_2 + s_2 = 20
     \]
   - For \(4x_1 + 4x_2 \leq 56\), introduce slack variable \(s_3\), giving:  
     \[
     4x_1 + 4x_2 + s_3 = 56
     \]

2. **Rewrite the objective function** in terms of the decision variables and slack variables:
   \[
   z = 10x_1 + 24x_2 + 0s_1 + 0s_2 + 0s_3
   \]

3. **Set up the initial Simplex tableau**:

\[
\begin{array}{c|cccccc|c}
\text{Basis} & x_1 & x_2 & s_1 & s_2 & s_3 & z & \text{RHS} \\
\hline
s_1 & 12 & 4 & 1 & 0 & 0 & 0 & 52 \\
s_2 & 6  & 1 & 0 & 1 & 0 & 0 & 20 \\
s_3 & 4  & 4 & 0 & 0 & 1 & 0 & 56 \\
\hline
z   & -10 & -24 & 0 & 0 & 0 & 1 & 0 \\
\end{array}
\]

4. **Determine the entering variable** (most negative value in the objective row):
   - The most negative coefficient is \(-24\) (associated with \(x_2\)). So, \(x_2\) enters the basis.

5. **Determine the leaving variable** using the minimum ratio test**:
   \[
   \frac{\text{RHS}}{\text{Coefficient of } x_2} = \min\left(\frac{52}{4}, \frac{20}{1}, \frac{56}{4}\right) = \min(13, 20, 14) = 13
   \]
   Therefore, the row associated with \(s_1\) (the first row) leaves the basis.

6. **Pivot on the element in the first row and second column** to perform Gaussian elimination and update the tableau.After performing the pivot operation where \( x_2 \) enters and \( s_1 \) leaves, the updated Simplex tableau is:

\[
\begin{array}{c|cccccc|c}
\text{Basis} & x_1 & x_2 & s_1 & s_2 & s_3 & z & \text{RHS} \\
\hline
x_2 & 3 & 1 & 0.25 & 0 & 0 & 0 & 13 \\
s_2 & 3 & 0 & -0.25 & 1 & 0 & 0 & 7 \\
s_3 & -8 & 0 & -1 & 0 & 1 & 0 & 4 \\
z   & 62 & 0 & 6 & 0 & 0 & 1 & 312 \\
\end{array}
\]

### Interpretation:
- The new objective function value is \( z = 312 \).
- The current solution is \( x_1 = 0 \), \( x_2 = 13 \), and slack variables \( s_1 = 0 \), \( s_2 = 7 \), and \( s_3 = 4 \).

### Next Step:
Since all the coefficients in the objective row (excluding \( z \)) are non-negative, we have reached the optimal solution.

### Optimal Solution:
\[
x_1 = 0, \quad x_2 = 13, \quad z = 312
\]

Would you like more details on any step, or do you have any further questions?

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Here are 5 related questions:
1. How would the solution change if one of the constraints is modified?
2. How can we handle a minimization problem using the Simplex Method?
3. Can you explain how to identify if a problem has multiple optimal solutions?
4. What are the degeneracy and unboundedness conditions in the Simplex Method?
5. How would the introduction of new variables affect the Simplex tableau?

**Tip:** In a maximization problem, the Simplex Method terminates when there are no more negative coefficients in the objective row.

Use the simplex method to solve the linear programming problem. Maximize z = 10x1 + 24x2 subject to: 12x1 + 4x2 ≤ 52, 6x1 + x2 ≤ 20, 4x1 + 4x2 ≤ 56 with x1 ≥ 0, x2 ≥ 0.

Solve the linear programming problem using the Simplex Method. Maximize the objective function z = 10x1 + 24x2 subject to a set of constraints, including 12x1 + 4x2 ≤ 52, 6x1 + x2 ≤ 20, and 4x1 + 4x2 ≤ 56. This step-by-step guide walks through each phase of the Simplex algorithm, from converting inequalities to slack variables to interpreting the optimal solution.

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