SATHEE: Shortcut Methods

JEE Exam:

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.

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:

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

Constraints:

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

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:

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:

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

Constraints:

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.

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.

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.