Minimise `Z = 3x + 5y`
such that `x + 3y ge 3, x + y ge 12, x , y ge 0`

Answer

To solve the linear programming problem of minimizing \( Z = 3x + 5y \) subject to the constraints \( x + 3y \geq 3 \), \( x + y \geq 12 \), and \( x, y \geq 0 \), we can follow these steps:

### Step 1: Identify the Constraints
We have the following constraints:
1. \( x + 3y \geq 3 \)
2. \( x + y \geq 12 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To find the boundary lines of the constraints, we convert the inequalities to equations:
1. \( x + 3y = 3 \)
2. \( x + y = 12 \)

### Step 3: Find Intercepts of the Lines
For the first equation \( x + 3y = 3 \):
- Set \( x = 0 \): \( 3y = 3 \) → \( y = 1 \) (Point A: \( (0, 1) \))
- Set \( y = 0 \): \( x = 3 \) (Point B: \( (3, 0) \))

For the second equation \( x + y = 12 \):
- Set \( x = 0 \): \( y = 12 \) (Point C: \( (0, 12) \))
- Set \( y = 0 \): \( x = 12 \) (Point D: \( (12, 0) \))

### Step 4: Plot the Lines and Identify the Feasible Region
Plot the points A, B, C, and D on a graph. The feasible region is determined by the inequalities:
- For \( x + 3y \geq 3 \), the region is above the line.
- For \( x + y \geq 12 \), the region is above the line.

### Step 5: Check Feasibility of the Origin
Check if the origin (0,0) satisfies the constraints:
- \( 0 + 3(0) \geq 3 \) → False
- \( 0 + 0 \geq 12 \) → False

Since the origin does not satisfy the constraints, the feasible region is above the lines.

### Step 6: Identify the Corner Points of the Feasible Region
The corner points of the feasible region are:
- Point C: \( (0, 12) \)
- Point D: \( (12, 0) \)

### Step 7: Evaluate the Objective Function at the Corner Points
Now we evaluate \( Z = 3x + 5y \) at the corner points:
1. At Point C \( (0, 12) \):
\[ Z = 3(0) + 5(12) = 60 \]
2. At Point D \( (12, 0) \):
\[ Z = 3(12) + 5(0) = 36 \]

### Step 8: Determine the Minimum Value
The minimum value of \( Z \) occurs at Point D \( (12, 0) \), where \( Z = 36 \).

### Final Answer
The minimum value of \( Z \) is **36** at the point **(12, 0)**.

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