A manufactures makes two products, A and B. Product A sell at Rs. 200 each and takes `1/2` hours to make. Product B shell at Rs. 300 each and takes 1 hour to make. There is a permanent order for 14 of product A and 16 of product B.A working week consists of 40 hours of production and the weekly turnover must not be less than Rs. 10000. If the profit on each of the product A is Rs.20 and on product B, it is Rs.30 then how many of each should be produced so that the profit is maximum? Asso, find the maximum profit.

Explanation

To solve the problem of maximizing profit for the manufacturer producing products A and B, we will follow these steps:

### Step 1: Define the Variables
Let:
- \( x \) = number of units produced of product A
- \( y \) = number of units produced of product B

### Step 2: Formulate the Objective Function
The profit from product A is Rs. 20 per unit, and from product B is Rs. 30 per unit. Therefore, the objective function to maximize profit \( Z \) is:
\[ Z = 20x + 30y \]

### Step 3: Set Up the Constraints
1. **Time Constraint**: Product A takes \( \frac{1}{2} \) hours and product B takes 1 hour. The total production time cannot exceed 40 hours:
\[ \frac{1}{2}x + y \leq 40 \]
This can be rewritten as:
\[ x + 2y \leq 80 \]

2. **Turnover Constraint**: The total turnover must not be less than Rs. 10,000:
\[ 200x + 300y \geq 10000 \]
This can be simplified to:
\[ 2x + 3y \geq 100 \]

3. **Demand Constraints**: There is a permanent order for 14 units of product A and 16 units of product B:
\[ x \geq 14 \]
\[ y \geq 16 \]

4. **Non-negativity Constraints**:
\[ x \geq 0, \quad y \geq 0 \]

### Step 4: Graph the Constraints
To find the feasible region, we will graph the constraints on the coordinate plane.

1. **Constraint 1**: \( x + 2y \leq 80 \)
- When \( x = 0 \), \( y = 40 \) (point (0, 40))
- When \( y = 0 \), \( x = 80 \) (point (80, 0))

2. **Constraint 2**: \( 2x + 3y \geq 100 \)
- When \( x = 0 \), \( y = \frac{100}{3} \approx 33.33 \) (point (0, 33.33))
- When \( y = 0 \), \( x = 50 \) (point (50, 0))

3. **Demand Constraints**:
- \( x = 14 \) (vertical line)
- \( y = 16 \) (horizontal line)

### Step 5: Identify the Feasible Region
The feasible region is determined by the intersection of all the constraints. The vertices of this region will be calculated next.

### Step 6: Find the Intersection Points
1. **Intersection of \( x + 2y = 80 \) and \( 2x + 3y = 100 \)**:
- From \( x + 2y = 80 \), express \( x = 80 - 2y \).
- Substitute into \( 2(80 - 2y) + 3y = 100 \):
\[ 160 - 4y + 3y = 100 \implies -y = -60 \implies y = 60 \]
- Substitute \( y = 60 \) back to find \( x \):
\[ x + 2(60) = 80 \implies x + 120 = 80 \implies x = -40 \quad \text{(not feasible)} \]

2. **Intersection of \( x + 2y = 80 \) and \( y = 16 \)**:
\[ x + 2(16) = 80 \implies x + 32 = 80 \implies x = 48 \quad \text{(point (48, 16))} \]

3. **Intersection of \( 2x + 3y = 100 \) and \( y = 16 \)**:
\[ 2x + 3(16) = 100 \implies 2x + 48 = 100 \implies 2x = 52 \implies x = 26 \quad \text{(point (26, 16))} \]

4. **Intersection of \( x = 14 \) and \( 2x + 3y = 100 \)**:
\[ 2(14) + 3y = 100 \implies 28 + 3y = 100 \implies 3y = 72 \implies y = 24 \quad \text{(point (14, 24))} \]

5. **Intersection of \( x = 14 \) and \( x + 2y = 80 \)**:
\[ 14 + 2y = 80 \implies 2y = 66 \implies y = 33 \quad \text{(point (14, 33))} \]

### Step 7: Evaluate the Objective Function at Each Vertex
Now we evaluate \( Z = 20x + 30y \) at each of the feasible points:
1. At (48, 16):
\[ Z = 20(48) + 30(16) = 960 + 480 = 1440 \]
2. At (26, 16):
\[ Z = 20(26) + 30(16) = 520 + 480 = 1000 \]
3. At (14, 24):
\[ Z = 20(14) + 30(24) = 280 + 720 = 1000 \]
4. At (14, 33):
\[ Z = 20(14) + 30(33) = 280 + 990 = 1270 \]

### Step 8: Determine the Maximum Profit
The maximum profit occurs at (48, 16) with a profit of Rs. 1440.

### Conclusion
To maximize profit, the manufacturer should produce:
- **48 units of product A**
- **16 units of product B**

### Maximum Profit
The maximum profit is **Rs. 1440**.

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