To find the minimum value of \( Z = 100x + 50y \) subject to the constraints \( x + y \leq 300 \), \( 3x + y \leq 600 \), \( y \leq x + 200 \), \( x \geq 0 \), and \( y \geq 0 \), we will solve this graphically.
### Step 1: Convert inequalities to equations
We first convert the inequalities into equations to find the boundary lines:
1. \( x + y = 300 \) (Equation 1)
2. \( 3x + y = 600 \) (Equation 2)
3. \( y = x + 200 \) (Equation 3)
### Step 2: Find intercepts of each equation
- For **Equation 1** \( x + y = 300 \):
- When \( x = 0 \), \( y = 300 \) (Point: \( (0, 300) \))
- When \( y = 0 \), \( x = 300 \) (Point: \( (300, 0) \))
- For **Equation 2** \( 3x + y = 600 \):
- When \( x = 0 \), \( y = 600 \) (Point: \( (0, 600) \))
- When \( y = 0 \), \( x = 200 \) (Point: \( (200, 0) \))
- For **Equation 3** \( y = x + 200 \):
- When \( x = 0 \), \( y = 200 \) (Point: \( (0, 200) \))
- When \( y = 0 \), \( x = -200 \) (not valid since \( x \geq 0 \))
### Step 3: Plot the lines on a graph
We plot the lines based on the intercepts found:
- Line from \( (0, 300) \) to \( (300, 0) \)
- Line from \( (0, 600) \) to \( (200, 0) \)
- Line from \( (0, 200) \) with a slope of 1 (intersects the y-axis at 200)
### Step 4: Identify feasible region
The feasible region is where all these inequalities overlap, considering the constraints \( x \geq 0 \) and \( y \geq 0 \).
### Step 5: Find corner points of the feasible region
We find the points of intersection of the lines:
1. **Intersection of Equation 1 and Equation 2**:
\[
x + y = 300 \\
3x + y = 600
\]
Subtracting the first from the second:
\[
2x = 300 \implies x = 150 \\
y = 300 - 150 = 150 \quad \text{(Point: } (150, 150)\text{)}
\]
2. **Intersection of Equation 1 and Equation 3**:
\[
x + y = 300 \\
y = x + 200
\]
Substituting \( y \):
\[
x + (x + 200) = 300 \implies 2x + 200 = 300 \implies 2x = 100 \implies x = 50 \\
y = 50 + 200 = 250 \quad \text{(Point: } (50, 250)\text{)}
\]
3. **Intersection of Equation 2 and Equation 3**:
\[
3x + y = 600 \\
y = x + 200
\]
Substituting \( y \):
\[
3x + (x + 200) = 600 \implies 4x + 200 = 600 \implies 4x = 400 \implies x = 100 \\
y = 100 + 200 = 300 \quad \text{(Point: } (100, 300)\text{)}
\]
### Step 6: Evaluate \( Z \) at each corner point
Now we evaluate \( Z \) at the corner points:
1. At \( (0, 600) \): \( Z = 100(0) + 50(600) = 30000 \)
2. At \( (100, 300) \): \( Z = 100(100) + 50(300) = 25000 \)
3. At \( (150, 150) \): \( Z = 100(150) + 50(150) = 22500 \)
4. At \( (300, 0) \): \( Z = 100(300) + 50(0) = 30000 \)
### Step 7: Determine the minimum value
The minimum value of \( Z \) occurs at the point \( (150, 150) \) with \( Z = 22500 \).
### Final Answer
The minimum value of \( Z \) is \( 22500 \) at the point \( (150, 150) \).
