Maximum of `Z= 4x + y` subject to:
`x + y le 50, 3x + y le 90, x ge 0, y ge 0 ` is ........at .........

Answer

To solve the problem of maximizing \( Z = 4x + y \) subject to the given constraints, we will follow these steps:

### Step 1: Identify the Constraints
The constraints given are:
1. \( x + y \leq 50 \)
2. \( 3x + y \leq 90 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert Constraints to Equations
To find the boundary lines of the constraints, we convert the inequalities into equations:
1. \( x + y = 50 \)
2. \( 3x + y = 90 \)

### Step 3: Find Intercepts of Each Line
- For \( x + y = 50 \):
- When \( x = 0 \): \( y = 50 \) (Point: \( (0, 50) \))
- When \( y = 0 \): \( x = 50 \) (Point: \( (50, 0) \))

- For \( 3x + y = 90 \):
- When \( x = 0 \): \( y = 90 \) (Point: \( (0, 90) \))
- When \( y = 0 \): \( x = 30 \) (Point: \( (30, 0) \))

### Step 4: Find the Intersection of the Lines
To find the intersection of the lines \( x + y = 50 \) and \( 3x + y = 90 \), we can solve the equations simultaneously:
1. From \( x + y = 50 \), we can express \( y \) as \( y = 50 - x \).
2. Substitute \( y \) in the second equation:
\[ 3x + (50 - x) = 90 \implies 2x + 50 = 90 \implies 2x = 40 \implies x = 20 \]
3. Substitute \( x = 20 \) back into \( y = 50 - x \):
\[ y = 50 - 20 = 30 \]
Thus, the intersection point is \( (20, 30) \).

### Step 5: Determine the Feasible Region
The feasible region is bounded by the lines and the axes. The vertices of the feasible region are:
1. \( (0, 50) \)
2. \( (50, 0) \)
3. \( (20, 30) \)
4. \( (30, 0) \)

### Step 6: Evaluate the Objective Function at Each Vertex
Now, we will evaluate \( Z = 4x + y \) at each vertex:
1. At \( (0, 50) \):
\[ Z = 4(0) + 50 = 50 \]
2. At \( (50, 0) \):
\[ Z = 4(50) + 0 = 200 \]
3. At \( (20, 30) \):
\[ Z = 4(20) + 30 = 80 + 30 = 110 \]
4. At \( (30, 0) \):
\[ Z = 4(30) + 0 = 120 \]

### Step 7: Identify the Maximum Value
From the evaluations:
- \( Z(0, 50) = 50 \)
- \( Z(50, 0) = 200 \)
- \( Z(20, 30) = 110 \)
- \( Z(30, 0) = 120 \)

The maximum value of \( Z \) is \( 200 \) at the point \( (50, 0) \).

### Conclusion
The maximum of \( Z = 4x + y \) subject to the given constraints is \( 200 \) at the point \( (50, 0) \).

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