Minimise Z=13x-15y subject to the constraints ` x+y le 7, 2x-3y+6 ge 0, x ge 0, x ge 0, and y ge 0`

Explanation

To solve the linear programming problem of minimizing \( Z = 13x - 15y \) subject to the constraints:

1. \( x + y \leq 7 \)
2. \( 2x - 3y + 6 \geq 0 \) (which can be rewritten as \( 2x - 3y \geq -6 \))
3. \( x \geq 0 \)
4. \( y \geq 0 \)

we will follow these steps:

### Step 1: Identify the constraints
We have the following constraints:
- \( x + y \leq 7 \)
- \( 2x - 3y \geq -6 \)
- \( x \geq 0 \)
- \( y \geq 0 \)

### Step 2: Rewrite the inequalities
The second constraint can be rewritten as:
- \( 2x - 3y \geq -6 \) can be rearranged to \( 2x - 3y = -6 \).

### Step 3: Find the intersection points of the constraints
To find the feasible region, we need to find the intersection points of the lines defined by the equations of the constraints.

1. **For \( x + y = 7 \)**:
- If \( x = 0 \), then \( y = 7 \) (point \( (0, 7) \)).
- If \( y = 0 \), then \( x = 7 \) (point \( (7, 0) \)).

2. **For \( 2x - 3y = -6 \)**:
- If \( x = 0 \), then \( -3y = -6 \) → \( y = 2 \) (point \( (0, 2) \)).
- If \( y = 0 \), then \( 2x = -6 \) → \( x = -3 \) (not feasible since \( x \geq 0 \)).

### Step 4: Find the intersection of the two lines
To find the intersection point of \( x + y = 7 \) and \( 2x - 3y = -6 \):
1. From \( x + y = 7 \), we can express \( y \) as \( y = 7 - x \).
2. Substitute \( y \) in \( 2x - 3(7 - x) = -6 \):
\[ 2x - 21 + 3x = -6 \implies 5x - 21 = -6 \implies 5x = 15 \implies x = 3. \]
3. Substitute \( x = 3 \) back into \( y = 7 - x \):
\[ y = 7 - 3 = 4. \]
So, the intersection point is \( (3, 4) \).

### Step 5: Identify the feasible region
The feasible region is bounded by the points:
- \( (0, 0) \)
- \( (7, 0) \)
- \( (0, 2) \)
- \( (3, 4) \)

### Step 6: Evaluate the objective function at the vertices
Now we evaluate \( Z = 13x - 15y \) at each of the vertices:
1. At \( (0, 0) \):
\[ Z = 13(0) - 15(0) = 0. \]
2. At \( (7, 0) \):
\[ Z = 13(7) - 15(0) = 91. \]
3. At \( (0, 2) \):
\[ Z = 13(0) - 15(2) = -30. \]
4. At \( (3, 4) \):
\[ Z = 13(3) - 15(4) = 39 - 60 = -21. \]

### Step 7: Determine the minimum value
The minimum value of \( Z \) occurs at \( (0, 2) \) where \( Z = -30 \).

### Final Answer
The minimum value of \( Z \) is \( -30 \) at the point \( (0, 2) \).

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