A company manufactures two types of toys A and B. A toy of type A requires 5 minute for cutting and 10 minute for assembling. A toy of type B requires 8 minute for cutting and 8 minutes for assembling. There are 3 hours available for cutting and 4 hours available for assembling th etoys in a day. The profit is Rs 50 each on a toy of Type A and Rs 60 each on a toy of type B. How many toys of each type should the company manufacture in a day to maximize the profit ? Use linear programming to find the solution.

Answer

To solve the problem using linear programming, we will follow these steps:

### Step 1: Define the Variables
Let:
- \( x \) = number of toys of type A manufactured
- \( y \) = number of toys of type B manufactured

### Step 2: Formulate the Objective Function
The profit from each toy is given as follows:
- Profit from toy A = Rs 50
- Profit from toy B = Rs 60

Thus, the objective function to maximize profit \( Z \) can be expressed as:
\[ Z = 50x + 60y \]

### Step 3: Set Up the Constraints
We need to consider the time constraints for cutting and assembling:

1. **Cutting Time Constraint**:
- Time required for cutting toy A = 5 minutes
- Time required for cutting toy B = 8 minutes
- Total available cutting time = 3 hours = 180 minutes

Therefore, the constraint for cutting is:
\[ 5x + 8y \leq 180 \]

2. **Assembling Time Constraint**:
- Time required for assembling toy A = 10 minutes
- Time required for assembling toy B = 8 minutes
- Total available assembling time = 4 hours = 240 minutes

Therefore, the constraint for assembling is:
\[ 10x + 8y \leq 240 \]

3. **Non-negativity Constraints**:
\[ x \geq 0 \]
\[ y \geq 0 \]

### Step 4: Solve the Inequalities
We will convert the inequalities into equations to find the corner points.

1. From the cutting time constraint:
\[ 5x + 8y = 180 \]
- When \( x = 0 \): \( 8y = 180 \) → \( y = 22.5 \) (Point A: \( (0, 22.5) \))
- When \( y = 0 \): \( 5x = 180 \) → \( x = 36 \) (Point B: \( (36, 0) \))

2. From the assembling time constraint:
\[ 10x + 8y = 240 \]
- When \( x = 0 \): \( 8y = 240 \) → \( y = 30 \) (Point C: \( (0, 30) \))
- When \( y = 0 \): \( 10x = 240 \) → \( x = 24 \) (Point D: \( (24, 0) \))

### Step 5: Find Intersection Points
To find the intersection of the two constraints:
1. Solve the equations:
\[ 5x + 8y = 180 \quad (1) \]
\[ 10x + 8y = 240 \quad (2) \]

Subtract (1) from (2):
\[ (10x + 8y) - (5x + 8y) = 240 - 180 \]
\[ 5x = 60 \implies x = 12 \]

Substitute \( x = 12 \) back into (1):
\[ 5(12) + 8y = 180 \implies 60 + 8y = 180 \implies 8y = 120 \implies y = 15 \]
Thus, the intersection point is \( (12, 15) \).

### Step 6: Evaluate the Objective Function at Each Corner Point
1. At \( (0, 22.5) \):
\[ Z = 50(0) + 60(22.5) = 1350 \]
2. At \( (36, 0) \):
\[ Z = 50(36) + 60(0) = 1800 \]
3. At \( (0, 30) \):
\[ Z = 50(0) + 60(30) = 1800 \]
4. At \( (24, 0) \):
\[ Z = 50(24) + 60(0) = 1200 \]
5. At \( (12, 15) \):
\[ Z = 50(12) + 60(15) = 600 + 900 = 1500 \]

### Step 7: Conclusion
The maximum profit occurs at the point \( (12, 15) \) with a profit of Rs 1500. Therefore, the company should manufacture:
- 12 toys of type A
- 15 toys of type B