To solve the problem of maximizing \( Z = 11x + 7y \) subject to the constraints \( 2x + y \leq 6 \), \( x \leq 2 \), \( x \geq 0 \), and \( y \geq 0 \), we will follow these steps:
### Step 1: Identify the constraints
The constraints given are:
1. \( 2x + y \leq 6 \)
2. \( x \leq 2 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)
### Step 2: Convert inequalities to equations
To find the boundary lines, we convert the inequalities into equations:
1. \( 2x + y = 6 \)
2. \( x = 2 \)
3. \( x = 0 \) (y-axis)
4. \( y = 0 \) (x-axis)
### Step 3: Find the intercepts of the lines
For the line \( 2x + y = 6 \):
- Set \( x = 0 \): \( y = 6 \) (Intercept at (0, 6))
- Set \( y = 0 \): \( 2x = 6 \) → \( x = 3 \) (Intercept at (3, 0))
### Step 4: Plot the lines and identify the feasible region
- The line \( x = 2 \) is a vertical line that intersects the x-axis at (2, 0).
- The lines \( x = 0 \) and \( y = 0 \) are the axes.
- The feasible region is bounded by the lines and lies in the first quadrant (where \( x \geq 0 \) and \( y \geq 0 \)).
### Step 5: Identify the corner points of the feasible region
The corner points of the feasible region are:
1. \( (0, 0) \)
2. \( (2, 0) \)
3. \( (2, 2) \) (where \( x = 2 \) intersects \( 2x + y = 6 \))
4. \( (0, 6) \) (the y-intercept of the line \( 2x + y = 6 \))
### Step 6: Evaluate the objective function at each corner point
Now we will calculate \( Z = 11x + 7y \) at each corner point:
1. At \( (0, 0) \): \( Z = 11(0) + 7(0) = 0 \)
2. At \( (2, 0) \): \( Z = 11(2) + 7(0) = 22 \)
3. At \( (2, 2) \): \( Z = 11(2) + 7(2) = 22 + 14 = 36 \)
4. At \( (0, 6) \): \( Z = 11(0) + 7(6) = 42 \)
### Step 7: Determine the maximum value of \( Z \)
From the calculations:
- \( Z(0, 0) = 0 \)
- \( Z(2, 0) = 22 \)
- \( Z(2, 2) = 36 \)
- \( Z(0, 6) = 42 \)
The maximum value of \( Z \) is \( 42 \) at the point \( (0, 6) \).
### Final Answer
The maximum value of \( Z = 11x + 7y \) subject to the given constraints is **42**.
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