The iinear programming problem Maximise `Z=x_1 +x_2 `
Subject to constraints
` x_1+2x_2 le le 2000`
` x_1+ x_2 le 15000`
` x_2 le 600`
` x_1 ge has `

Options

A. no feasible solution
B. unique optimal solution
C. a finite number of optimal solutions
D. infinite number of optimal solutions

Explanation

To solve the linear programming problem, we need to maximize the objective function \( Z = x_1 + x_2 \) subject to the given constraints. Let's go through the steps systematically.

### Step 1: Identify the Constraints
The constraints given are:
1. \( x_1 + 2x_2 \leq 2000 \)
2. \( x_1 + x_2 \leq 1500 \)
3. \( x_2 \leq 600 \)
4. \( x_1 \geq 0 \)
5. \( x_2 \geq 0 \)

### Step 2: Convert Constraints to Equations
To find the boundary lines for the constraints, we convert the inequalities into equations:
1. \( x_1 + 2x_2 = 2000 \)
2. \( x_1 + x_2 = 1500 \)
3. \( x_2 = 600 \)

### Step 3: Find Intercepts of Each Equation
For each equation, we find the intercepts:

1. **For \( x_1 + 2x_2 = 2000 \)**:
- When \( x_1 = 0 \): \( 2x_2 = 2000 \) → \( x_2 = 1000 \) → Point (0, 1000)
- When \( x_2 = 0 \): \( x_1 = 2000 \) → Point (2000, 0)

2. **For \( x_1 + x_2 = 1500 \)**:
- When \( x_1 = 0 \): \( x_2 = 1500 \) → Point (0, 1500)
- When \( x_2 = 0 \): \( x_1 = 1500 \) → Point (1500, 0)

3. **For \( x_2 = 600 \)**:
- This is a horizontal line at \( x_2 = 600 \).

### Step 4: Determine Points of Intersection
Next, we find the points of intersection of the lines:
1. **Intersection of \( x_1 + 2x_2 = 2000 \) and \( x_1 + x_2 = 1500 \)**:
- Subtract the second equation from the first:
\[ (x_1 + 2x_2) - (x_1 + x_2) = 2000 - 1500 \implies x_2 = 500 \]
- Substitute \( x_2 = 500 \) into \( x_1 + x_2 = 1500 \):
\[ x_1 + 500 = 1500 \implies x_1 = 1000 \]
- Intersection point: (1000, 500)

2. **Intersection of \( x_1 + x_2 = 1500 \) and \( x_2 = 600 \)**:
- Substitute \( x_2 = 600 \) into \( x_1 + x_2 = 1500 \):
\[ x_1 + 600 = 1500 \implies x_1 = 900 \]
- Intersection point: (900, 600)

3. **Intersection of \( x_1 + 2x_2 = 2000 \) and \( x_2 = 600 \)**:
- Substitute \( x_2 = 600 \) into \( x_1 + 2x_2 = 2000 \):
\[ x_1 + 1200 = 2000 \implies x_1 = 800 \]
- Intersection point: (800, 600)

### Step 5: Identify Feasible Region
The feasible region is bounded by the points:
- A: (0, 1000)
- B: (2000, 0)
- C: (0, 1500)
- D: (1500, 0)
- E: (1000, 500)
- F: (900, 600)
- G: (800, 600)

### Step 6: Evaluate the Objective Function at Each Vertex
Now we evaluate \( Z = x_1 + x_2 \) at each vertex:
1. At (0, 1000): \( Z = 0 + 1000 = 1000 \)
2. At (2000, 0): \( Z = 2000 + 0 = 2000 \)
3. At (0, 1500): \( Z = 0 + 1500 = 1500 \)
4. At (1500, 0): \( Z = 1500 + 0 = 1500 \)
5. At (1000, 500): \( Z = 1000 + 500 = 1500 \)
6. At (900, 600): \( Z = 900 + 600 = 1500 \)
7. At (800, 600): \( Z = 800 + 600 = 1400 \)

### Step 7: Determine Maximum Value
The maximum value of \( Z \) occurs at points (2000, 0) and (0, 1500), both yielding \( Z = 2000 \).

### Conclusion
The maximum value of \( Z \) is **2000** at the point **(2000, 0)**.

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