The linear programming problem Maximise `Z= x_(1)+x_(2)`
Subject to constraints
`{:(x_(1)+2x_(2) le 2000),(x_(1)+x_(2) le 1500),(x_(2) le 600),(x_(1) ge 0):}` has

Options

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

Answer

To solve the linear programming problem of maximizing \( Z = x_1 + x_2 \) subject to the given constraints, we will follow these steps:

### 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. \( x_1 \geq 0 \)

### Step 2: Convert Inequalities to Equations
To graph the constraints, we convert the inequalities into equations:
1. \( x_1 + 2x_2 = 2000 \)
2. \( x_1 + x_2 = 1500 \)
3. \( x_2 = 600 \)
4. \( x_1 = 0 \)

### Step 3: Find Intercepts for Each Constraint
**For the first constraint \( x_1 + 2x_2 = 2000 \):**
- If \( x_1 = 0 \), then \( 2x_2 = 2000 \) → \( x_2 = 1000 \) (Point: \( (0, 1000) \))
- If \( x_2 = 0 \), then \( x_1 = 2000 \) (Point: \( (2000, 0) \))

**For the second constraint \( x_1 + x_2 = 1500 \):**
- If \( x_1 = 0 \), then \( x_2 = 1500 \) (Point: \( (0, 1500) \))
- If \( x_2 = 0 \), then \( x_1 = 1500 \) (Point: \( (1500, 0) \))

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

### Step 4: Graph the Constraints
Plot the points on a graph:
- Plot \( (0, 1000) \) and \( (2000, 0) \) for the first constraint.
- Plot \( (0, 1500) \) and \( (1500, 0) \) for the second constraint.
- Draw a horizontal line at \( x_2 = 600 \).

### Step 5: Identify the Feasible Region
The feasible region is the area that satisfies all constraints simultaneously. It will be bounded by the lines we have drawn. The feasible region will be the area where all the constraints overlap.

### Step 6: Determine the Corner Points
Identify the corner points of the feasible region:
1. Intersection of \( x_1 + 2x_2 = 2000 \) and \( x_2 = 600 \):
- Substitute \( x_2 = 600 \) into \( x_1 + 2(600) = 2000 \) → \( x_1 + 1200 = 2000 \) → \( x_1 = 800 \) (Point: \( (800, 600) \))

2. Intersection of \( x_1 + x_2 = 1500 \) and \( x_2 = 600 \):
- Substitute \( x_2 = 600 \) into \( x_1 + 600 = 1500 \) → \( x_1 = 900 \) (Point: \( (900, 600) \))

3. Intersection of \( x_1 + x_2 = 1500 \) and \( x_1 + 2x_2 = 2000 \):
- Solve the equations:
- From \( x_1 + x_2 = 1500 \), we get \( x_1 = 1500 - x_2 \).
- Substitute into \( x_1 + 2x_2 = 2000 \):
- \( (1500 - x_2) + 2x_2 = 2000 \) → \( 1500 + x_2 = 2000 \) → \( x_2 = 500 \) → \( x_1 = 1000 \) (Point: \( (1000, 500) \))

### Step 7: Evaluate the Objective Function at Each Corner Point
Evaluate \( Z = x_1 + x_2 \) at each corner point:
1. At \( (800, 600) \): \( Z = 800 + 600 = 1400 \)
2. At \( (900, 600) \): \( Z = 900 + 600 = 1500 \)
3. At \( (1000, 500) \): \( Z = 1000 + 500 = 1500 \)

### Step 8: Determine the Maximum Value
The maximum value of \( Z \) is \( 1500 \) which occurs at the points \( (900, 600) \) and \( (1000, 500) \).

### Conclusion
The linear programming problem has an infinite number of optimal solutions along the line segment connecting the points \( (900, 600) \) and \( (1000, 500) \).

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