To solve the linear programming problem and find the maximum value of \( Z = 3x + 5y \) under the given constraints, we will follow these steps:
### Step 1: Identify the Constraints
The constraints given are:
1. \( x + 2y \leq 2000 \)
2. \( x + y \leq 1500 \)
3. \( y \leq 600 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)
### Step 2: Convert Inequalities into Equations
To find the feasible region, we convert the inequalities into equations:
1. \( x + 2y = 2000 \)
2. \( x + y = 1500 \)
3. \( y = 600 \)
### Step 3: Find the Intersection Points
Next, we will find the intersection points of these lines:
1. **Intersection of \( x + 2y = 2000 \) and \( x + y = 1500 \)**:
- From \( x + y = 1500 \), we can express \( x = 1500 - y \).
- Substitute into the first equation:
\[
1500 - y + 2y = 2000 \implies 1500 + y = 2000 \implies y = 500
\]
- Substitute \( y = 500 \) back to find \( x \):
\[
x = 1500 - 500 = 1000
\]
- So, the intersection point is \( (1000, 500) \).
2. **Intersection of \( x + 2y = 2000 \) and \( y = 600 \)**:
- Substitute \( y = 600 \) into the first equation:
\[
x + 2(600) = 2000 \implies x + 1200 = 2000 \implies x = 800
\]
- So, the intersection point is \( (800, 600) \).
3. **Intersection of \( x + y = 1500 \) and \( y = 600 \)**:
- Substitute \( y = 600 \) into the second equation:
\[
x + 600 = 1500 \implies x = 900
\]
- So, the intersection point is \( (900, 600) \).
### Step 4: Identify the Feasible Region
The feasible region is determined by the constraints and the intersection points found. The vertices of the feasible region are:
1. \( (0, 0) \)
2. \( (1000, 500) \)
3. \( (800, 600) \)
4. \( (900, 600) \)
### Step 5: Evaluate the Objective Function at Each Vertex
Now we will evaluate \( Z = 3x + 5y \) at each vertex:
1. At \( (0, 0) \):
\[
Z = 3(0) + 5(0) = 0
\]
2. At \( (1000, 500) \):
\[
Z = 3(1000) + 5(500) = 3000 + 2500 = 5500
\]
3. At \( (800, 600) \):
\[
Z = 3(800) + 5(600) = 2400 + 3000 = 5400
\]
4. At \( (900, 600) \):
\[
Z = 3(900) + 5(600) = 2700 + 3000 = 5700
\]
### Step 6: Determine the Maximum Value
The maximum value of \( Z \) occurs at the point \( (1000, 500) \) and is:
\[
\text{Maximum } Z = 5500
\]
### Final Answer
The maximum value of \( Z = 3x + 5y \) under the given constraints is **5500**.
