A manufacturer produces two products, A and B. Both products are processed on two different machines. The available capacity of the first machine is 12 hours and of second machine is 9 hours per day. Each unit of product A requires 3 hours on both machines and each unit of product B requires 2 hours on first machine and 1 hour on second machine. Each unit of product A is sold at Rs. 7 profit and that of product B at a profit of Rs. 4. A LPP to maximise the profit is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( Maximise\ Z=7x+4y\ \ Subject\ to\ x\ge 0,y\ge 0\ 3x+2y\le 12 \)
\( 3x+y\le 9 \)
Let x units of A and y units of B be produced per day.
Objective: a profit of Rs. 7 on each A and Rs. 4 on each B:
\( \displaystyle Z=7x+4y \text{ (to be maximised)} \)
First machine (3 hours for A, 2 hours for B, at most 12 hours):
\( \displaystyle 3x+2y\le 12 \)
Second machine (3 hours for A, 1 hour for B, at most 9 hours):
\( \displaystyle 3x+y\le 9 \)
Production cannot be negative: \( x\ge 0,\ y\ge 0 \). This is Option (2).
Why the other options are incorrect
Option (1) \( Maximise\ Z=3x+2y\ \ Subject\ to\ \ x\ge 0,y\ge 0\ \ 7x+4y\le 12 \)
\( 3x+y\le 9 \) — Incorrect. The profits and the machine hours are interchanged: 3x + 2y is the time on the first machine, not the profit, and 7x + 4y is the profit, not a time.Option (3) \( Maximise\ Z=3x+y\ \ Subject\ to\ \ x\ge 0,y\ge 0\ \ 7x+4y\le 9 \)
\( 3x+2y\le 12 \) — Incorrect. The objective 3x + y is the time on the second machine, and 7x + 4y ≤ 9 mixes the profit with the capacity of the second machine.Option (4) \( Maximise\ Z=7x+4y\ \ Subject\ to\ x\ge 0,y\ge 0\ \ 3x+2y\ge 12\ \ 3x+y\ge 9 \) — Incorrect. The capacities are upper limits (“available capacity”), so the constraints must be ≤, not ≥.