Question

Two tailors A and B earn ₹150 and ₹200 per day respectively. A can stitch 6 shirts and 4 pants per day, while B can stitch 10 shirts and 4 pants per day. If the tailors A and B work for x and y days respectively. To maximize the earning for producing at least 60 shirts and 32 pants, the LPP is:

• A. Maximize Z = 150x + 200y, subject to 6x + 10y \(\geq\) 60, 4x + 4y \(\geq\) 32, x, y \(\geq\) 0
• B. Maximize Z = 150x + 200y, subject to 6x + 10y \(\leq\) 60, 4x + 4y \(\leq\) 32, x, y \(\geq\) 0
• C. Maximize Z = 150x + 200y, subject to 6x + 4y \(\geq\) 60, 10x + 4y \(\geq\) 32, x, y \(\geq\) 0
• D. Maximize Z = 150x + 200y, subject to 6x + 10y \(\geq\) 60, 4x + 4y \(\leq\) 32, x, y \(\geq\) 0

Answer 1

The Correct Option is A

To solve this problem, we need to formulate a Linear Programming Problem (LPP) to maximize the earnings of two tailors, A and B. Given:

• Tailor A earns ₹150/day and can stitch 6 shirts and 4 pants per day.
• Tailor B earns ₹200/day and can stitch 10 shirts and 4 pants per day.

Objective: Maximize the total earnings, Z, from working x days for tailor A and y days for tailor B.

Objective Function: Maximize Z = 150x + 200y

Constraints:

• To produce at least 60 shirts: 6x + 10y ≥ 60
• To produce at least 32 pants: 4x + 4y ≥ 32
• Non-negativity constraints: x, y ≥ 0

Therefore, the correct formulation of the LPP is:

Maximize Z = 150x + 200y, subject to:

• 6x + 10y ≥ 60
• 4x + 4y ≥ 32
• x, y ≥ 0