A recycling plant runs Process I for x hours and Process II for y hours a day. One hour of Process I gives 4 tonnes of paper and 1 tonne of plastic, and one hour of Process II gives 1 tonne of paper and 2 tonnes of plastic. The plant must produce at least 12 tonnes of paper and at least 10 tonnes of plastic a day. Process I costs Rs. 500 and Process II costs Rs. 300 per hour, and the daily cost is to be minimised. Which of the following statements are correct?
(A) The feasible region is unbounded.
(B) The corner points of the feasible region are (0, 12), (2, 4) and (10, 0).
(C) The least daily cost is Rs. 3,600, when only Process II is run.
(D) Running each process for 4 hours a day is a feasible plan.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (D) only
The problem is
Minimise \(Z = 500x + 300y\)
subject to
\(4x + y \ge 12\) (paper)
\(x + 2y \ge 10\) (plastic)
\(x \ge 0,\ y \ge 0\)
Both constraints are “at least”, so the feasible region lies away from the origin and is unbounded. The two lines meet where, putting \(y = 12 - 4x\),
\(x + 24 - 8x = 10\)
\(x = 2,\ y = 4\)
The corner points are (0, 12), (2, 4) and (10, 0). Values of Z:
\(Z(0, 12) = 3600\)
\(Z(2, 4) = 1000 + 1200 = 2200\)
\(Z(10, 0) = 5000\)
The half-plane \(500x + 300y \lt 2200\), that is \(5x + 3y \lt 22\), has no point in common with the region, so the least cost is Rs. 2,200 at (2, 4).
So the region is unbounded with corners (0, 12), (2, 4), (10, 0), and the least cost is Rs. 2,200 at (2, 4).
Checking the statements
(A) The feasible region is unbounded. — Correct. Both requirements are “at least” conditions, so x and y can be as large as we like.
(B) The corner points of the feasible region are (0, 12), (2, 4) and (10, 0). — Correct. (2, 4) is where \(4x + y = 12\) meets \(x + 2y = 10\).
(C) The least daily cost is Rs. 3,600, when only Process II is run. — Incorrect. \(Z(2, 4) = 2200 \lt 3600\).
(D) Running each process for 4 hours a day is a feasible plan. — Correct. \(16 + 4 = 20 \ge 12\) and \(4 + 8 = 12 \ge 10\).