### Shortcut Methods

**JEE Exam:**

**Shortcut Method:**

- Calculate the ratio of profit to labor hours required for each product.
- Choose the product with a higher ratio.
- Allocate all the available labor hours to that product.
- If any labor hours are left, allocate them to the other product.
- Calculate the maximum profit based on the allocated labor hours.

**Solution:**

- Profit/Labor Hour Ratio for Product A = 3/3 = 1
- Profit/Labor Hour Ratio for Product B = 5/4 = 1.25
- Since Product B has a higher ratio, allocate all 120 labor hours to Product B.
- No labor hours left, so all 180 units of raw material are allocated to product B.
- Maximum Profit = 5 * 180 = 900 units.
- Conclusion: Produce only Product B to maximize profit.

**Shortcut Method:**

- Convert the given information into a linear programming problem.
- Use a graphical method (e.g., the corner point method) to find the optimal solution.

**Detailed Solution:**

**Step 1:**Convert the problem into a linear programming problem.

Objective Function (Maximize Profit): Z = 5x + 6y

Constraints:

- Fertilizer: 2x + 3y ≤ 240
- Water: 4x + 6y ≤ 2880
- Land Availability: x + y ≤ 10

x ≥ 0, y ≥ 0 (Non-negative constraints)

**Step 2:**Use the corner point method.- Find the corner points: (0, 0), (0, 10), (8, 0), (6, 4).
- Evaluate the objective function at each corner point.
- The maximum profit occurs at the corner point (6, 4).
- Conclusion: Plant 6 acres of wheat and 4 acres of corn to maximize profit.

**Shortcut Method:**

- Calculate the ratio of profit to units of raw material required for each product.
- Choose the product with a higher ratio.
- Allocate all the available raw material to that product.
- If any raw material is left, allocate it to the other product.
- Calculate the maximum profit based on the allocated raw materials.

**Solution:**

- Profit/Raw Material Ratio for Product C1 = 2/4 = 0.5
- Profit/Raw Material Ratio for Product C2 = 3/6 = 0.5
- Since both products have the same ratio, choose either one. Assume we choose C1.
- Allocate all 240 units of raw material to Product C1.
- No raw material left, so all 360 labor hours are allocated to product C1.
- Maximum Profit = 2 * 240 = 480 units.
- Conclusion: Produce only Product C1 to maximize profit.

**CBSE Board Exam:**

**Shortcut Method:**

- Convert the given information into a linear programming problem.
- Use a graphical method (e.g., the corner point method) to find the optimal solution.

**Detailed Solution:**

**Step 1:**Convert the problem into a linear programming problem.

Objective Function (Maximize Profit): Z = 4x + 6y

Constraints:

- Fertilizer: 3x + 4y ≤ 180
- Water: 2x + 3y ≤ 360
- Land Availability: x + y ≤ 6

x ≥ 0, y ≥ 0 (Non-negative constraints)

**Step 2:**Use the corner point method.- Find the corner points: (0, 0), (0, 6), (4, 2), (2, 4).
- Evaluate the objective function at each corner point.
- The maximum profit occurs at the corner point (4, 2).
- Conclusion: Plant 4 hectares of wheat and 2 hectares of barley to maximize profit.

**Shortcut Method:**

- Calculate the ratio of profit to machine hours required for each product.
- Choose the product with a higher ratio.
- Allocate all the available machine hours to that product.
- If any machine hours are left, allocate them to the other product.
- Calculate the maximum profit based on the allocated machine hours.

**Solution:**

- Profit/Machine Hour Ratio for Product A = 4/2 = 2
- Profit/Machine Hour Ratio for Product B = 6/3 = 2
- Since both products have the same ratio, choose either one. Assume we choose A.
- Allocate all 120 machine hours to product A.
- No machine hours left, so all 180 labor hours can only be allocated to Product A.
- Maximum Profit = 4 * 120 = 480 units.
- Conclusion: Produce only Product A to maximize profit.

**Shortcut Method:**

- Let the amount invested in Scheme A be
`x`

units and the amount invested in Scheme B be`10,000 - x`

units. - The total interest earned from Scheme A is
`0.1x`

, and that from Scheme B is`0.12(10,000 - x)`

. - The total interest earned is
`0.1x + 0.12(10,000 - x) = 0.1x + 1200 - 0.12x = 1200 - 0.02x`

. - To maximize the total interest, we need to minimize
`0.02x`

. - Since
`x`

can take any value between 0 and 10,000, the minimum value of`0.02x`

is 0, which occurs at`x = 10,000`

. - Therefore, invest all the money in Scheme B to maximize the total interest.