Maximize `Z = x + 2y` subject to `x + y ge 5, x ge 0, y ge 0` is :

Options

A. 5 at (0,5)
B. 10 at (0,5)
C. 5 at (5,0)
D. 10 at (5,10)

Answer

To solve the problem of maximizing \( Z = x + 2y \) subject to the constraints \( x + y \geq 5 \), \( x \geq 0 \), and \( y \geq 0 \), we can follow these steps:

### Step 1: Identify the Objective Function and Constraints
The objective function is given as:
\[ Z = x + 2y \]
The constraints are:
1. \( x + y \geq 5 \)
2. \( x \geq 0 \)
3. \( y \geq 0 \)

### Step 2: Graph the Constraints
To graph the constraints, we first rewrite the inequality \( x + y \geq 5 \) as an equation:
\[ x + y = 5 \]
To find the intercepts:
- When \( x = 0 \), \( y = 5 \) (point \( (0, 5) \))
- When \( y = 0 \), \( x = 5 \) (point \( (5, 0) \))

Now, plot these points on the graph and draw the line connecting them. Since the inequality is \( \geq \), we shade the area above this line.

### Step 3: Identify the Feasible Region
The feasible region is where all the constraints overlap. Since \( x \geq 0 \) and \( y \geq 0 \), we are only interested in the first quadrant. The feasible region is the area above the line \( x + y = 5 \) in the first quadrant.

### Step 4: Analyze the Feasible Region
The feasible region is unbounded because as \( y \) increases, \( x \) can take on larger values without any upper limit. Therefore, we can conclude that the solution is unbounded.

### Step 5: Conclusion
Since the feasible region is unbounded, the maximum value of \( Z = x + 2y \) can increase indefinitely as we move along the line \( x + y = 5 \) towards infinity. Thus, there is no maximum value for \( Z \) in this case.

### Summary
The solution to the problem is that the maximum value of \( Z = x + 2y \) is unbounded.

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