A cylinder manufacturer makes small and large cylinders from a large piece of cardboard. The large cylinder requires 4 sq. m and small cylinder requires 3 sq. m of cardboard. The manufacturer is required to make at least 3 large cylinder and at least twice as many small cylinder as large cylinders. If 60 sq. m of cardboard is in the stock and profit on the small and large cylinders are Rs. 25 and Rs 35 respectively. How many of each type of cylinders to made to maximize the profit. Also, find the maximum profit.

Answer

To solve the problem of maximizing the profit from the production of large and small cylinders, we can follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = number of large cylinders
- \( y \) = number of small cylinders

### Step 2: Formulate the Objective Function
The profit from large cylinders is Rs. 35 each, and from small cylinders is Rs. 25 each. Therefore, the objective function to maximize profit \( Z \) is:
\[ Z = 35x + 25y \]

### Step 3: Set Up Constraints
1. **Cardboard Constraint**: The large cylinder requires 4 sq. m and the small cylinder requires 3 sq. m of cardboard. The total available cardboard is 60 sq. m:
\[ 4x + 3y \leq 60 \]

2. **Minimum Large Cylinders**: The manufacturer must produce at least 3 large cylinders:
\[ x \geq 3 \]

3. **Small Cylinders Requirement**: The manufacturer must produce at least twice as many small cylinders as large cylinders:
\[ y \geq 2x \]

4. **Non-negativity Constraints**: The number of cylinders cannot be negative:
\[ x \geq 0, \quad y \geq 0 \]

### Step 4: Graph the Constraints
To find the feasible region, we need to graph the constraints:

1. **From the cardboard constraint** \( 4x + 3y = 60 \):
- When \( x = 0 \): \( 3y = 60 \) → \( y = 20 \) (point (0, 20))
- When \( y = 0 \): \( 4x = 60 \) → \( x = 15 \) (point (15, 0))

2. **From the minimum large cylinders constraint** \( x = 3 \):
- This is a vertical line at \( x = 3 \).

3. **From the small cylinders requirement** \( y = 2x \):
- This is a straight line through the origin with slope 2.

### Step 5: Identify Corner Points
Now we find the intersection points of these lines to identify the corner points of the feasible region:

1. Intersection of \( 4x + 3y = 60 \) and \( x = 3 \):
\[ 4(3) + 3y = 60 \implies 12 + 3y = 60 \implies 3y = 48 \implies y = 16 \quad \text{(Point (3, 16))} \]

2. Intersection of \( 4x + 3y = 60 \) and \( y = 2x \):
\[ 4x + 3(2x) = 60 \implies 4x + 6x = 60 \implies 10x = 60 \implies x = 6 \quad \text{and} \quad y = 2(6) = 12 \quad \text{(Point (6, 12))} \]

3. Intersection of \( x = 3 \) and \( y = 2x \):
\[ y = 2(3) = 6 \quad \text{(Point (3, 6))} \]

### Step 6: Evaluate the Objective Function at Each Corner Point
Now we evaluate the profit function \( Z = 35x + 25y \) at each corner point:

1. At \( (3, 16) \):
\[ Z = 35(3) + 25(16) = 105 + 400 = 505 \]

2. At \( (6, 12) \):
\[ Z = 35(6) + 25(12) = 210 + 300 = 510 \]

3. At \( (3, 6) \):
\[ Z = 35(3) + 25(6) = 105 + 150 = 255 \]

### Step 7: Determine Maximum Profit
The maximum profit occurs at the point \( (6, 12) \) with a profit of Rs. 510.

### Final Answer
To maximize profit, the manufacturer should produce:
- **6 large cylinders**
- **12 small cylinders**

The maximum profit will be **Rs. 510**.