3. Solve the following linear programming model graphically: Maximize $$ \mathrm{Z}=3 \mathrm{X}_{1}+6 \mathrm{X}_{2} $$ Subject to: $$ 3 \mathrm{X}_{1}+2 \mathrm{X}_{2} \leq 18 $$ $$ \mathrm{X}_{1}+\mathrm{X}_{2} \geq 5 $$ $$ \mathrm{X}_{1} \leq 4 $$ $$ \mathrm{X}_{1}, \mathrm{X}^{2} \geq 0 $$

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To solve the given linear programming (LP) problem graphically, we'll follow these steps:

1. **Graph the Constraints**
2. **Identify the Feasible Region**
3. **Determine the Corner Points**
4. **Evaluate the Objective Function at Each Corner Point**
5. **Find the Optimal Solution**

Let's go through each step in detail.

---

## 1. Graph the Constraints

First, let's rewrite each inequality to make plotting easier.

### Constraint 1:
$$ 3X_1 + 2X_2 \leq 18 $$

**To find the boundary line:**  
Set $$ 3X_1 + 2X_2 = 18 $$

- **X₁-intercept** ($$ X_2 = 0 $$):
  $$ 3X_1 = 18 \Rightarrow X_1 = 6 $$
  
- **X₂-intercept** ($$ X_1 = 0 $$):
  $$ 2X_2 = 18 \Rightarrow X_2 = 9 $$

**Boundary Line:** Connect (6, 0) and (0, 9)

### Constraint 2:
$$ X_1 + X_2 \geq 5 $$

**To find the boundary line:**  
Set $$ X_1 + X_2 = 5 $$

- **X₁-intercept** ($$ X₂ = 0 $$):
  $$ X_1 = 5 $$
  
- **X₂-intercept** ($$ X_1 = 0 $$):
  $$ X_2 = 5 $$

**Boundary Line:** Connect (5, 0) and (0, 5)

### Constraint 3:
$$ X_1 \leq 4 $$

**Boundary Line:** A vertical line at $$ X_1 = 4 $$

### Non-negativity Constraints:
$$ X_1 \geq 0 $$
$$ X_2 \geq 0 $$

---

## 2. Identify the Feasible Region

The feasible region is the set of all points that satisfy all constraints simultaneously. Here's how each constraint affects the region:

- **Constraint 1 ($$ 3X_1 + 2X_2 \leq 18 $$):** Shade below the line $$ 3X_1 + 2X_2 = 18 $$.
  
- **Constraint 2 ($$ X_1 + X_2 \geq 5 $$):** Shade above the line $$ X_1 + X_2 = 5 $$.
  
- **Constraint 3 ($$ X_1 \leq 4 $$):** Shade to the left of the line $$ X_1 = 4 $$.
  
- **Non-negativity Constraints:** Consider only the first quadrant where $$ X_1 \geq 0 $$ and $$ X_2 \geq 0 $$.

The intersection of these shaded regions forms the **feasible region**.

---

## 3. Determine the Corner Points

The optimal solutions of an LP problem lie at the **corner points** (vertices) of the feasible region. Let's find these points by solving the equations where the boundary lines intersect.

1. **Intersection of $$ 3X_1 + 2X_2 = 18 $$ and $$ X_1 + X_2 = 5 $$:**

$$
   \begin{cases}
   3X_1 + 2X_2 = 18 \
   X_1 + X_2 = 5
   \end{cases}
   $$

**Solve the second equation for $$ X_1 $$:**
   $$ X_1 = 5 - X_2 $$

**Substitute into the first equation:**
   $$ 3(5 - X_2) + 2X_2 = 18 $$
   $$ 15 - 3X_2 + 2X_2 = 18 $$
   $$ -X_2 = 3 $$
   $$ X_2 = -3 $$

Since $$ X_2 $$ cannot be negative, there's **no feasible intersection** between these two lines within the first quadrant.

2. **Intersection of $$ 3X_1 + 2X_2 = 18 $$ and $$ X_1 = 4 $$:**

Substitute $$ X_1 = 4 $$ into $$ 3X_1 + 2X_2 = 18 $$:
   $$ 3(4) + 2X_2 = 18 $$
   $$ 12 + 2X_2 = 18 $$
   $$ 2X_2 = 6 $$
   $$ X_2 = 3 $$

**Point:** (4, 3)

3. **Intersection of $$ X_1 + X_2 = 5 $$ and $$ X_1 = 4 $$:**

Substitute $$ X_1 = 4 $$ into $$ X_1 + X_2 = 5 $$:
   $$ 4 + X_2 = 5 $$
   $$ X_2 = 1 $$

**Point:** (4, 1)

4. **Intersection with Axes:**

- **For $$ X_1 = 0 $$:**
     - From $$ 3X_1 + 2X_2 = 18 $$: (0, 9)
     - From $$ X_1 + X_2 = 5 $$: (0, 5)
   
   - **For $$ X_2 = 0 $$:**
     - From $$ 3X_1 + 2X_2 = 18 $$: (6, 0)
     - From $$ X_1 + X_2 = 5 $$: (5, 0)

Among these, only the points that satisfy all constraints are part of the feasible region.

**Feasible Corner Points:**
   - (4, 3)
   - (4, 1)
   - (5, 0)
   - (0, 5)

**Note:** Points like (6, 0) and (0, 9) are **not** feasible because they don't satisfy all constraints simultaneously.

---

## 4. Evaluate the Objective Function at Each Corner Point

The objective is to **maximize** $$ Z = 3X_1 + 6X_2 $$.

Let's calculate $$ Z $$ for each feasible corner point:

1. **At (0, 5):**
   $$
   Z = 3(0) + 6(5) = 0 + 30 = 30
   $$

2. **At (4, 1):**
   $$
   Z = 3(4) + 6(1) = 12 + 6 = 18
   $$

3. **At (4, 3):**
   $$
   Z = 3(4) + 6(3) = 12 + 18 = 30
   $$

4. **At (5, 0):**
   $$
   Z = 3(5) + 6(0) = 15 + 0 = 15
   $$

---

## 5. Find the Optimal Solution

Comparing the values of $$ Z $$ at each corner point:

- **(0, 5):** $$ Z = 30 $$
- **(4, 1):** $$ Z = 18 $$
- **(4, 3):** $$ Z = 30 $$
- **(5, 0):** $$ Z = 15 $$

The maximum value of $$ Z $$ is **30**, achieved at two points:

- **Point A:** (0, 5)
- **Point B:** (4, 3)

Both points are optimal solutions for this LP problem, yielding the same maximum value of the objective function.

---

## Conclusion

The graphical solution shows that the maximum value $$ Z = 30 $$ is attained at two optimal points within the feasible region:

1. **(0, 5)**
2. **(4, 3)**

Thus, the optimal solutions are:
- $$ X_1 = 0 $$, $$ X_2 = 5 $$
- $$ X_1 = 4 $$, $$ X_2 = 3 $$

Both solutions maximize the objective function $$ Z = 3X_1 + 6X_2 $$ under the given constraints.
The maximum value of $$ Z = 3X_1 + 6X_2 $$ is 30, achieved at two points: (0, 5) and (4, 3).

3. Solve the following linear programming model graphically: Maximize \( \mathrm{Z}=3 \mathrm{X}_{1}+6 \mathrm{X}_{2} \) Subject to: \( 3 \mathrm{X}_{1}+2 \mathrm{X}_{2} \leq 18 \) \( \mathrm{X}_{1}+\mathrm{X}_{2} \geq 5 \) \( \mathrm{X}_{1} \leq 4 \) \( \mathrm{X}_{1}, \mathrm{X}^{2} \geq 0 \)

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