If Z=2x + y subject to `x le 4, y le 6, x+y ge 6, x ge 0, y ge 0`, then the maximum value of Z is

Options

A. 18
B. 10
C. 6
D. 14

Answer

To solve the problem, we need to maximize the function \( Z = 2x + y \) subject to the given constraints. Let's go through the steps systematically.

### Step 1: Identify the constraints
The constraints given are:
1. \( x \leq 4 \)
2. \( y \leq 6 \)
3. \( x + y \geq 6 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)

### Step 2: Graph the constraints
We will graph the constraints on the coordinate plane to find the feasible region.

1. **Graph \( x = 4 \)**: This is a vertical line at \( x = 4 \).
2. **Graph \( y = 6 \)**: This is a horizontal line at \( y = 6 \).
3. **Graph \( x + y = 6 \)**: This line can be rewritten as \( y = 6 - x \). It intersects the axes at \( (6, 0) \) and \( (0, 6) \).
4. **Graph \( x \geq 0 \)**: This indicates the right half of the plane (to the right of the y-axis).
5. **Graph \( y \geq 0 \)**: This indicates the upper half of the plane (above the x-axis).

### Step 3: Determine the feasible region
The feasible region is the area where all the constraints overlap. We need to find the intersection points of the lines to identify the vertices of the feasible region.

### Step 4: Find the vertices of the feasible region
1. **Intersection of \( x = 4 \) and \( y = 6 \)**: This gives the point \( (4, 6) \).
2. **Intersection of \( x + y = 6 \) and \( x = 4 \)**: Substitute \( x = 4 \) into \( x + y = 6 \):
\[ 4 + y = 6 \implies y = 2 \quad \text{(Point: (4, 2))} \]
3. **Intersection of \( x + y = 6 \) and \( y = 6 \)**: Substitute \( y = 6 \) into \( x + y = 6 \):
\[ x + 6 = 6 \implies x = 0 \quad \text{(Point: (0, 6))} \]

### Step 5: Evaluate \( Z = 2x + y \) at each vertex
Now we will evaluate \( Z \) at each of the vertices found:
1. At \( (4, 6) \):
\[ Z = 2(4) + 6 = 8 + 6 = 14 \]
2. At \( (4, 2) \):
\[ Z = 2(4) + 2 = 8 + 2 = 10 \]
3. At \( (0, 6) \):
\[ Z = 2(0) + 6 = 0 + 6 = 6 \]

### Step 6: Determine the maximum value of \( Z \)
The maximum value of \( Z \) occurs at the point \( (4, 6) \):
\[ \text{Maximum value of } Z = 14 \]

### Final Answer
The maximum value of \( Z \) is \( \boxed{14} \).