An aviation fuel manufacturer sells two types of fuel A and B. Type A fuel is 25 % grade 1 gasoline, 25 % of grade 2 gasoline and 50 % of grade 3 gasoline. Type B fuel is 50 % of grade 2 gasoline and 50 % of grade 3 gasoline. Available for production are 500 liters per hour grade 1 and 200 liters per hour of grade 2 and grade 3 each. Costs are 60 paise per liter for grade 1, 120 paise for grade 2 and100 paise for grade 3. Type A can be sold at Rs. 7.50 per liter and B can be sold at Rs. 9.00 per liter. How much of each fuel should be made and sold to maximise the profit.

To solve this problem, we need to maximize the profit from selling two types of fuel, A and B, under the constraints of available gasoline grades and their costs.

### Step 1: Define the Variables

Let:
- \( x \) = liters of Type A fuel produced per hour.
- \( y \) = liters of Type B fuel produced per hour.

### Step 2: Determine the Constraints

Type A fuel consists of:
- 25% Grade 1 gasoline: \( 0.25x \) liters
- 25% Grade 2 gasoline: \( 0.25x \) liters
- 50% Grade 3 gasoline: \( 0.5x \) liters

Type B fuel consists of:
- 50% Grade 2 gasoline: \( 0.5y \) liters
- 50% Grade 3 gasoline: \( 0.5y \) liters

The total available gasoline per hour is:
- Grade 1: 500 liters
- Grade 2: 200 liters
- Grade 3: 200 liters

From the above, the constraints are:
1. \( 0.25x \leq 500 \)
2. \( 0.25x + 0.5y \leq 200 \) (for Grade 2 gasoline)
3. \( 0.5x + 0.5y \leq 200 \) (for Grade 3 gasoline)

We also have non-negativity constraints:
4. \( x \geq 0 \)
5. \( y \geq 0 \)

### Step 3: Formulate the Objective Function

The profit function is calculated by considering the selling price and the cost of production:

- Selling price for Type A: Rs. 7.50 per liter
- Selling price for Type B: Rs. 9.00 per liter

Cost per liter for each grade of gasoline:
- Grade 1: 60 paise (0.60 Rs)
- Grade 2: 120 paise (1.20 Rs)
- Grade 3: 100 paise (1.00 Rs)

The cost of producing Type A and B fuel can be calculated as follows:

- Cost to produce 1 liter of Type A:
  \[
  0.25 \times 0.60 + 0.25 \times 1.20 + 0.5 \times 1.00 = 0.15 + 0.30 + 0.50 = 0.95 \text{ Rs per liter}
  \]

- Cost to produce 1 liter of Type B:
  \[
  0.5 \times 1.20 + 0.5 \times 1.00 = 0.60 + 0.50 = 1.10 \text{ Rs per liter}
  \]

Profit per liter:
- Profit for Type A: \( 7.50 - 0.95 = 6.55 \) Rs per liter
- Profit for Type B: \( 9.00 - 1.10 = 7.90 \) Rs per liter

Total profit function \( P \) to maximize:
\[
P = 6.55x + 7.90y
\]

### Step 4: Set Up the Linear Programming Problem

The problem now is to maximize:
\[
P = 6.55x + 7.90y
\]

Subject to:
\[
\begin{cases}
0.25x \leq 500 \\
0.25x + 0.5y \leq 200 \\
0.5x + 0.5y \leq 200 \\
x \geq 0, \; y \geq 0
\]
  
### Step 5: Solve the Linear Programming Problem

To solve, we first simplify the constraints:

1. \( x \leq 2000 \)
2. \( x + 2y \leq 800 \)
3. \( x + y \leq 400 \)

We can graph these constraints to find the feasible region or use the Simplex Method to find the optimal solution.

Let's compute this using the constraints.The optimal solution to maximize the profit is:

- **Type A fuel**: \( x = 0 \) liters per hour
- **Type B fuel**: \( y = 400 \) liters per hour

### Maximum Profit
The maximum profit is Rs. **3160** per hour.

### Explanation
- All available resources should be allocated to producing Type B fuel, which yields a higher profit margin of Rs. 7.90 per liter compared to Type A fuel at Rs. 6.55 per liter.
- This solution uses up all 200 liters of Grade 2 and Grade 3 gasoline (which are required for Type B fuel).

Would you like more details, or do you have any further questions?

### Related Questions
1. What would the profit be if we changed the selling price of Type A fuel?
2. How does the profit change if the costs of the gasoline grades change?
3. What happens if the availability of one type of gasoline changes?
4. How would the constraints change if we introduce a new fuel type?
5. What other methods can we use to solve this optimization problem?

**Tip:** In problems like this, always check the cost-to-profit ratio to determine which product provides the best return on investment!

Learn how to maximize profit from aviation fuel production using linear programming. This problem involves optimizing the production of two types of fuels, A and B, based on gasoline grades and costs. Understand the constraints and the objective function to achieve maximum profitability.

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