Maximum value of `Z=3x+4y`
subject to `x-y le -1, -x+y le 0, x ge 0 , y ge 0` is

Options

A. 1
B. 4
C. 6
D. 7

Answer

To solve the linear programming problem of maximizing \( Z = 3x + 4y \) subject to the constraints \( x - y \leq -1 \), \( -x + y \leq 0 \), \( x \geq 0 \), and \( y \geq 0 \), we will follow these steps:

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

### Step 2: Rewrite the Constraints
We can rewrite the inequalities in a more usable form:
1. \( y \geq x + 1 \) (from \( x - y \leq -1 \))
2. \( y \leq x \) (from \( -x + y \leq 0 \))
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 3: Graph the Constraints
To find the feasible region, we will graph the constraints on the coordinate plane.

1. **For \( y \geq x + 1 \)**:
- The line \( y = x + 1 \) has a y-intercept at (0, 1) and a slope of 1. The region above this line is included.

2. **For \( y \leq x \)**:
- The line \( y = x \) passes through the origin (0, 0) and has a slope of 1. The region below this line is included.

3. **For \( x \geq 0 \)**:
- This indicates that we are only considering the right half of the graph (where x is non-negative).

4. **For \( y \geq 0 \)**:
- This indicates that we are only considering the upper half of the graph (where y is non-negative).

### Step 4: Determine the Feasible Region
Now, we will find the intersection points of the lines to determine the vertices of the feasible region:

- The line \( y = x + 1 \) intersects the line \( y = x \) at:
\[ x + 1 = x \implies \text{No intersection} \]

- The line \( y = x + 1 \) intersects the x-axis (where \( y = 0 \)):
\[ 0 = x + 1 \implies x = -1 \quad (\text{not feasible since } x \geq 0) \]

- The line \( y = x \) intersects the y-axis (where \( x = 0 \)):
\[ y = 0 \quad \text{(point (0, 0))} \]

### Step 5: Analyze the Feasible Region
From the above analysis, we see that:
- The line \( y = x + 1 \) is above the line \( y = x \) and does not intersect within the feasible region defined by \( x \geq 0 \) and \( y \geq 0 \).
- The only point we have is (0, 0), which does not satisfy \( y \geq x + 1 \).

### Conclusion
Since there are no points that satisfy all constraints simultaneously, we conclude that there is **no feasible region**. Therefore, there is **no solution** for the given linear programming problem, and thus no maximum value of \( Z \).