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 solution step by step.
### Step 1: Identify the Objective Function and Constraints
The objective function is:
\[
z = x_1 + x_2
\]
The constraints 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: Graph the Constraints
To graph the constraints, we need to convert each inequality into an equation and 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 \( x_2 = 600 \).
4. **For \( x_1 \geq 0 \)**:
- This indicates that we are only considering the right side of the y-axis.
### Step 3: Determine the Feasible Region
Plot the lines on a graph and shade the feasible region that satisfies all constraints. The feasible region is where all the shaded areas overlap.
### Step 4: Find the Corner Points of the Feasible Region
The corner points of the feasible region can be found by solving the equations of the lines where they intersect.
1. **Intersection of \( x_1 + 2x_2 = 2000 \) and \( x_2 = 600 \)**:
\[
x_1 + 2(600) = 2000 \implies x_1 + 1200 = 2000 \implies x_1 = 800 \quad \text{(Point: (800, 600))}
\]
2. **Intersection of \( x_1 + x_2 = 1500 \) and \( x_2 = 600 \)**:
\[
x_1 + 600 = 1500 \implies x_1 = 900 \quad \text{(Point: (900, 600))}
\]
3. **Intersection of \( x_1 + 2x_2 = 2000 \) and \( x_1 + x_2 = 1500 \)**:
\[
x_1 + 2x_2 = 2000 \quad \text{and} \quad x_1 + x_2 = 1500
\]
Subtract the second equation from the first:
\[
x_2 = 500 \implies x_1 + 500 = 1500 \implies x_1 = 1000 \quad \text{(Point: (1000, 500))}
\]
4. **Intersection of \( x_1 + x_2 = 1500 \) and \( x_1 = 0 \)**:
\[
0 + x_2 = 1500 \implies x_2 = 1500 \quad \text{(Point: (0, 1500))}
\]
5. **Intersection of \( x_1 = 0 \) and \( x_2 = 600 \)**:
\[
(0, 600)
\]
### Step 5: Evaluate the Objective Function at the Corner Points
Now we evaluate \( z = x_1 + x_2 \) at the corner points:
1. \( (800, 600) \): \( z = 800 + 600 = 1400 \)
2. \( (900, 600) \): \( z = 900 + 600 = 1500 \)
3. \( (1000, 500) \): \( z = 1000 + 500 = 1500 \)
4. \( (0, 600) \): \( z = 0 + 600 = 600 \)
### Step 6: Determine the Maximum Value
The maximum value of \( z \) occurs at the points \( (900, 600) \) and \( (1000, 500) \), both yielding \( z = 1500 \).
### Conclusion
The linear programming problem has infinite solutions along the line segment connecting the points \( (900, 600) \) and \( (1000, 500) \) where \( z = 1500 \).
