Maximize `Z = 4x + y` subject to : `x + y le 50, x, y ge 0`.

Answer

To solve the problem of maximizing \( Z = 4x + y \) subject to the constraints \( x + y \leq 50 \) and \( x, y \geq 0 \), we can follow these steps:

### Step 1: Identify the Constraints
The constraints given are:
1. \( x + y \leq 50 \)
2. \( x \geq 0 \)
3. \( y \geq 0 \)

### Step 2: Graph the Constraints
To graph the constraint \( x + y = 50 \):
- Find the x-intercept by setting \( y = 0 \):
\[ x + 0 = 50 \implies x = 50 \quad \text{(Point A: (50, 0))} \]
- Find the y-intercept by setting \( x = 0 \):
\[ 0 + y = 50 \implies y = 50 \quad \text{(Point B: (0, 50))} \]

Now, we can plot these points on a graph.

### Step 3: Determine the Feasible Region
The feasible region is defined by the area where all constraints are satisfied. Since \( x \) and \( y \) must be non-negative, we only consider the first quadrant. The line \( x + y = 50 \) divides the first quadrant, and we are interested in the area below this line.

### Step 4: Identify the Corner Points
The corner points of the feasible region are:
1. Point A: \( (50, 0) \)
2. Point B: \( (0, 50) \)
3. Point C: \( (0, 0) \)

### Step 5: Evaluate the Objective Function at Each Corner Point
Now we will evaluate \( Z = 4x + y \) at each of the corner points:

- At Point A \( (50, 0) \):
\[ Z = 4(50) + 0 = 200 \]

- At Point B \( (0, 50) \):
\[ Z = 4(0) + 50 = 50 \]

- At Point C \( (0, 0) \):
\[ Z = 4(0) + 0 = 0 \]

### Step 6: Determine the Maximum Value
From the evaluations:
- \( Z \) at Point A is 200.
- \( Z \) at Point B is 50.
- \( Z \) at Point C is 0.

The maximum value of \( Z \) occurs at Point A \( (50, 0) \) and is equal to 200.

### Conclusion
Thus, the maximum value of \( Z = 4x + y \) subject to the given constraints is:
\[ \boxed{200} \]