Solve the linear programming problem below using the simplex algorithm. Include all the solution steps.
min 𝑍 = 5𝑥 − 4𝑦
subject to 𝑥 + 𝑦 ≤ 5
2𝑥 − 𝑦 = 16
𝑥 ≥ 0
𝑦 ∈ ℝ

To solve the given linear programming problem using the simplex algorithm, we'll follow the steps systematically. Here is the problem statement:

**Objective function:**  
\[
\text{Minimize } Z = 5x - 4y
\]

**Subject to the constraints:**  
1. \(x + y \leq 5\)
2. \(2x - y = 16\)
3. \(x \geq 0\)
4. \(y \in \mathbb{R}\)

Since \(y\) is not required to be non-negative, we'll have to use a modified simplex approach and carefully handle the equality and inequality constraints.

---

### Step 1: Convert constraints into standard form

**1. Convert the inequality into an equation:**  
For the constraint \(x + y \leq 5\), introduce a slack variable \(s_1 \geq 0\):
\[
x + y + s_1 = 5
\]

**2. Write the equality constraint directly:**  
\[
2x - y = 16
\]

So, the problem now looks like:

**Objective function:**  
\[
Z = 5x - 4y
\]

**Subject to:**  
\[
x + y + s_1 = 5 \tag{1}
\]
\[
2x - y = 16 \tag{2}
\]
\[
x \geq 0, \; s_1 \geq 0, \; y \in \mathbb{R}
\]

---

### Step 2: Set up the initial simplex tableau

We introduce the following variables:
- \(x\): Decision variable
- \(y\): Decision variable
- \(s_1\): Slack variable
- \(Z\): Objective function value

To convert the minimization problem into a maximization problem, we rewrite \(Z\):
\[
\text{Maximize } Z' = -Z = -5x + 4y
\]

The initial tableau is:

| Basic | \(x\) | \(y\) | \(s_1\) | RHS |
|-------|------|------|--------|-----|
| \(s_1\) | 1  | 1  | 1      | 5   |
|   \(y\) | 2  | -1 | 0      | 16  |
| \(Z'\) | -5 | 4  | 0      | 0   |

---

### Step 3: Perform simplex iterations

1. **Determine the entering variable**:  
   The most negative value in the \(Z'\) row is \(-5\), corresponding to \(x\). So, \(x\) will enter the basis.

2. **Determine the leaving variable**:  
   Perform the ratio test to find the leaving variable:
   - For row 1: \(\frac{5}{1} = 5\)
   - For row 2: \(\frac{16}{2} = 8\)

   Since 5 is the smallest ratio, \(s_1\) leaves the basis.

3. **Pivot to form a new tableau**:  
   Pivot on the element in the \(x\) column of row 1, which is 1.

| Basic | \(x\) | \(y\) | \(s_1\) | RHS |
|-------|------|------|--------|-----|
|   \(x\) | 1  | 1  | 1      | 5   |
|   \(y\) | 0  | -3 | -2     | 6   |
| \(Z'\) | 0  | 9  | 5      | 25  |

---

### Step 4: Check for optimality

Since all the coefficients in the \(Z'\) row are now non-negative, the current solution is optimal.

### Step 5: Extract the solution

From the final tableau:

- Basic variables:
  - \(x = 5\)
  - \(y = 6\)
- Slack variable:
  - \(s_1 = 0\)

Substitute into the original objective function:

\[
Z = 5x - 4y = 5(5) - 4(6) = 25 - 24 = 1
\]

---

### Final Answer:

The solution to the linear programming problem is:
- \(x = 5\)
- \(y = 6\)
- Minimum value of \(Z\) is 1.

Would you like a detailed explanation of any part? Here are some related questions you might find interesting:

1. How do we determine the entering and leaving variables in the simplex method?
2. What are slack variables and why are they used in linear programming?
3. How do we convert a minimization problem into a maximization problem for the simplex method?
4. What is the significance of the pivot element in the simplex tableau?
5. Can we solve linear programming problems graphically and how does it compare with the simplex method?

**Tip:** Always double-check your final tableau for non-negative coefficients in the objective function row to confirm optimality!

Learn how to solve a linear programming problem using the simplex algorithm with detailed solution steps. This problem involves minimizing Z = 5x - 4y subject to constraints x + y ≤ 5 and 2x - y = 16. Ideal for students studying optimization techniques at the undergraduate level.

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