A manufacturer has 3 machines I, II and III installed in his factory. Machines I and II are capable of being operated for at most 12 hrs whereas machine III is capable to operate 5 hrs a day. He produced only two items M and N each requiring the use of all 3 machines. Number of hours for producing 1 unit of M and N each on the 3 machines are given below:
| Items | No. of hours in each machine | ||
| I | II | III | |
| M | 1 | 2 | 1 |
| N | 2 | 1 | 1.25 |
He makes a profit of Rs. 600 on M and Rs. 400 on N. If x is the number of M items and y is the number of N items, then
A. \(Z=600x+400y\)
B. \(x+2y \ge 12\)
C. \(x+\displaystyle\frac{5}{4}y \le 5\)
D. \(2x+y \le 12\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A, C and D only
The profit is the objective function, and each machine gives one constraint: hours used on that machine ≤ hours available. Each machine is read down its own column of the table.
Profit: \(Z = 600x+400y\)
Machine I (1 h per M, 2 h per N, at most 12 h): \(x+2y \le 12\)
Machine II (2 h per M, 1 h per N, at most 12 h): \(2x+y \le 12\)
Machine III (1 h per M, 1.25 h per N, 5 h): \(x+1.25y \le 5\)
\(\displaystyle 1.25 = \frac{5}{4}\), so \(x+\frac{5}{4}y \le 5\)
So \(Z = 600x+400y\), and the constraints are \(x+2y \le 12\), \(2x+y \le 12\) and \(x+\frac{5}{4}y \le 5\) (with \(x, y \ge 0\)).
Checking the statements
(A) \(Z=600x+400y\) — Correct. Rs. 600 per M and Rs. 400 per N.
(B) \(x+2y \ge 12\) — Incorrect. Machine I can run at most 12 hours, so the sign is ≤, not ≥.
(C) \(x+\frac{5}{4}y \le 5\) — Correct. This is the machine III limit, with 1.25 = 5/4.
(D) \(2x+y \le 12\) — Correct. This is the machine II limit: 2 h per M, 1 h per N.