Solve the following integer linear programming problem optimally using
Branch-and-bound technique
Maximize Z = 5x1 + 10xz
Subject to
3x1 + 4x2 ≤ 24
3x1 + 6x2≤6
X1, X2 ≥ 0 and integers.

To solve the given integer linear programming problem optimally using the Branch-and-Bound technique, let's proceed with the following steps:

### Problem Statement:
Maximize \( Z = 5x_1 + 10x_2 \)

Subject to:
\[
3x_1 + 4x_2 \leq 24
\]
\[
3x_1 + 6x_2 \leq 6
\]
\[
x_1, x_2 \geq 0 \quad \text{and integers}
\]

### Step 1: Solve the Linear Programming Relaxation
First, we solve the problem without the integer constraints (i.e., \( x_1 \) and \( x_2 \) are allowed to take fractional values).

#### Objective Function:
Maximize \( Z = 5x_1 + 10x_2 \)

#### Constraints:
\[
3x_1 + 4x_2 \leq 24
\]
\[
3x_1 + 6x_2 \leq 6
\]
\[
x_1, x_2 \geq 0
\]

### Step 2: Graphical Method or Simplex Method
Given the simplicity of the problem, we can solve it graphically or by using the simplex method. However, considering this as an integer linear programming problem, let’s do the graphical approach:

#### Constraint 1: 
\[
3x_1 + 4x_2 \leq 24
\]
This line passes through the points \((x_1, x_2)\) where:
- \( x_1 = 0 \) gives \( x_2 = 6 \)
- \( x_2 = 0 \) gives \( x_1 = 8 \)

#### Constraint 2:
\[
3x_1 + 6x_2 \leq 6
\]
This line passes through the points:
- \( x_1 = 0 \) gives \( x_2 = 1 \)
- \( x_2 = 0 \) gives \( x_1 = 2 \)

### Feasible Region:
The feasible region is bounded by the intersection of the two constraints and the axes \(x_1\) and \(x_2\) being non-negative.

### Step 3: Identify the Vertices of the Feasible Region
The intersection points (vertices) of the feasible region can be calculated as follows:

1. Intersection of \( 3x_1 + 4x_2 = 24 \) and \( 3x_1 + 6x_2 = 6 \):

\[
3x_1 + 4x_2 = 24
\]
\[
3x_1 + 6x_2 = 6
\]
Subtracting the second equation from the first:
\[
-2x_2 = 18 \quad \text{which gives} \quad x_2 = -9 \quad \text{(not feasible since } x_2 \geq 0)
\]

Thus, no intersection between these two lines in the feasible region.

2. Other vertices are:
   - \( (0, 0) \)
   - \( (2, 0) \)
   - \( (0, 1) \)

### Step 4: Evaluate the Objective Function at Each Vertex
We evaluate the objective function \( Z = 5x_1 + 10x_2 \) at each feasible vertex:

1. At \( (0, 0) \):
   \[
   Z = 5(0) + 10(0) = 0
   \]

2. At \( (2, 0) \):
   \[
   Z = 5(2) + 10(0) = 10
   \]

3. At \( (0, 1) \):
   \[
   Z = 5(0) + 10(1) = 10
   \]

### Step 5: Conclusion
The maximum value of \( Z \) occurs at both \( (2, 0) \) and \( (0, 1) \), with \( Z = 10 \).

Since both \( (2, 0) \) and \( (0, 1) \) are feasible and integer solutions, the optimal solution is:
\[
(x_1, x_2) = (2, 0) \quad \text{or} \quad (0, 1)
\]
with a maximum objective function value of \( Z = 10 \).

### Summary:
The optimal solution is either \( (x_1, x_2) = (2, 0) \) or \( (0, 1) \), with a maximum value \( Z = 10 \).

---

Would you like to see further details, or have any questions? Here are some related questions you might find useful:

1. What is the Branch-and-Bound technique?
2. How do you graphically represent the feasible region?
3. How is the Simplex method used for linear programming?
4. What are integer programming problems used for?
5. How can linear programming be applied in real-world situations?

**Tip:** In integer programming, the Branch-and-Bound method is highly effective when the feasible region is large and complex.

Explore how to solve the integer linear programming problem optimally using the Branch-and-Bound technique. Learn about the objective function Z = 5x1 + 10x2, constraints, and the feasible region. Find the maximum value of Z with integer solutions.

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