The point at which the maximum value of `Z = 4x + 6y` subject to the constraints `3x + 2y le 12, x + y ge 4, x ge 0, y ge 0` is obtained at the point

Options

A. (0,6)
B. (6,0)
C. (0,4)
D. (4,0)

Answer

To solve the problem of maximizing \( Z = 4x + 6y \) subject to the constraints \( 3x + 2y \leq 12 \), \( x + y \geq 4 \), \( x \geq 0 \), and \( y \geq 0 \), we will follow these steps:

### Step 1: Graph the Constraints
Start by graphing the constraints to find the feasible region.

1. **For \( 3x + 2y = 12 \)**:
- Set \( y = 0 \): \( 3x = 12 \) → \( x = 4 \) → Point \( (4, 0) \)
- Set \( x = 0 \): \( 2y = 12 \) → \( y = 6 \) → Point \( (0, 6) \)
- The line intersects the axes at points \( (4, 0) \) and \( (0, 6) \).

2. **For \( x + y = 4 \)**:
- Set \( y = 0 \): \( x = 4 \) → Point \( (4, 0) \)
- Set \( x = 0 \): \( y = 4 \) → Point \( (0, 4) \)
- The line intersects the axes at points \( (4, 0) \) and \( (0, 4) \).

### Step 2: Identify the Feasible Region
The feasible region is bounded by the lines \( 3x + 2y = 12 \) and \( x + y = 4 \), along with the axes.

- The region satisfying \( 3x + 2y \leq 12 \) is below the line \( 3x + 2y = 12 \).
- The region satisfying \( x + y \geq 4 \) is above the line \( x + y = 4 \).
- The non-negativity constraints \( x \geq 0 \) and \( y \geq 0 \) restrict the region to the first quadrant.

### Step 3: Find the Corner Points
The corner points of the feasible region can be found by solving the equations of the lines:

1. **Intersection of \( 3x + 2y = 12 \) and \( x + y = 4 \)**:
- From \( x + y = 4 \), we can express \( y = 4 - x \).
- Substitute into \( 3x + 2(4 - x) = 12 \):
\[ 3x + 8 - 2x = 12 \implies x + 8 = 12 \implies x = 4 \]
- Substitute \( x = 4 \) back into \( y = 4 - x \):
\[ y = 4 - 4 = 0 \implies (4, 0) \]

2. **Check the other corner points**:
- \( (0, 6) \) from the line \( 3x + 2y = 12 \).
- \( (0, 4) \) from the line \( x + y = 4 \).

Thus, the corner points are \( (0, 6) \), \( (0, 4) \), and \( (4, 0) \).

### Step 4: Evaluate \( Z \) at Each Corner Point
Now, we will evaluate \( Z = 4x + 6y \) at each corner point:

1. At \( (0, 6) \):
\[ Z = 4(0) + 6(6) = 36 \]

2. At \( (0, 4) \):
\[ Z = 4(0) + 6(4) = 24 \]

3. At \( (4, 0) \):
\[ Z = 4(4) + 6(0) = 16 \]

### Step 5: Determine the Maximum Value
The maximum value of \( Z \) occurs at the point \( (0, 6) \) with \( Z = 36 \).

### Conclusion
The point at which the maximum value of \( Z = 4x + 6y \) is obtained is \( (0, 6) \).

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