Maximize `Z = 2x + 3y` subject to `x + 2y le 6, x ge 4, y ge 0` is :

Options

A. 6 at (6,0)
B. 6 at (0,6)
C. 12 at (6,0)
D. 12 at (0,6)

Answer

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

### Step 1: Identify the Constraints
We have the following constraints:
1. \( x + 2y \leq 6 \)
2. \( x \geq 4 \)
3. \( y \geq 0 \)

### Step 2: Convert the Inequalities to Equations
To graph the constraints, we convert the inequalities to equations:
1. \( x + 2y = 6 \)
2. \( x = 4 \)
3. \( y = 0 \)

### Step 3: Find Intercepts for the First Constraint
For the equation \( x + 2y = 6 \):
- To find the x-intercept, set \( y = 0 \):
\[ x + 2(0) = 6 \implies x = 6 \quad \text{(Point: (6, 0))} \]
- To find the y-intercept, set \( x = 0 \):
\[ 0 + 2y = 6 \implies y = 3 \quad \text{(Point: (0, 3))} \]

### Step 4: Graph the Constraints
- The line \( x + 2y = 6 \) passes through points (6, 0) and (0, 3).
- The line \( x = 4 \) is a vertical line at \( x = 4 \).
- The line \( y = 0 \) is the x-axis.

### Step 5: Determine the Feasible Region
The feasible region is where all constraints overlap:
- The area below the line \( x + 2y = 6 \).
- To the right of the line \( x = 4 \).
- Above the x-axis (where \( y \geq 0 \)).

### Step 6: Identify Corner Points of the Feasible Region
The corner points of the feasible region can be found by determining the intersections of the lines:
1. Intersection of \( x + 2y = 6 \) and \( x = 4 \):
\[ 4 + 2y = 6 \implies 2y = 2 \implies y = 1 \quad \text{(Point: (4, 1))} \]
2. The points (6, 0) and (4, 0) are also corner points.

### Step 7: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( Z = 2x + 3y \) at each corner point:
1. At (4, 1):
\[ Z = 2(4) + 3(1) = 8 + 3 = 11 \]
2. At (4, 0):
\[ Z = 2(4) + 3(0) = 8 + 0 = 8 \]
3. At (6, 0):
\[ Z = 2(6) + 3(0) = 12 + 0 = 12 \]

### Step 8: Determine the Maximum Value
The maximum value of \( Z \) occurs at the point (6, 0):
\[ \text{Maximum } Z = 12 \]

### Conclusion
The maximum value of \( Z = 2x + 3y \) subject to the given constraints is **12** at the point **(6, 0)**.

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