Question

A manufacturing company produces two items, A and B. Machine I (max 640 mins): A takes 20, B takes 15. Machine II (max 500 mins): A takes 5, B takes 8. Profit: A is Rs 25, B is Rs 18. Formulation is:

• A. Max $z=25x+18y$ s.t. $20x+15y \le 640, 5x+8y \le 500, x,y \ge 0$
• B. Max $z=25x+18y$ s.t. $20x+15y \le 640, 5x+8y \le 500, x,y \ge 0$
• C. Max $z=25x+18y$ s.t. $20x+5y \le 8, 5x+8y \le 10, x,y \ge 0$
• D. Max $z=25x+18y$ s.t. $4x+3y \le 128, 5x+8y \le 500, x,y \ge 0$

Hint:

Always convert all time units to the same unit (minutes) before formulating.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to construct a Linear Programming Problem (LPP) to maximize total profit based on given constraints of machine time.

Step 2: Key Formula or Approach:
Convert all time units to the same unit (minutes).
$z = \text{Total Profit}$. Constraints are $\sum \text{time}_i \times \text{units} \le \text{Max time}$.

Step 3: Detailed Explanation:
Objective function: $z = 25x + 18y$ (Maximize).
Constraint for Machine I:
Maximum time $= 10$ hours $40$ minutes $= 10 \times 60 + 40 = 640$ minutes.
Time taken per item A $= 20$ min, B $= 15$ min.
Constraint: $20x + 15y \le 640$.
Constraint for Machine II:
Maximum time $= 8$ hours $20$ minutes $= 8 \times 60 + 20 = 500$ minutes.
Time taken per item A $= 5$ min, B $= 8$ min.
Constraint: $5x + 8y \le 500$.
Non-negativity: $x \ge 0, y \ge 0$.

Step 4: Final Answer:
The formulation is given in option (A).