A firm must produce at least 600 units of a part. Machine M1 makes 20 units an hour at a cost of Rs. 300 an hour and can be used for at most 24 hours. Machine M2 makes 30 units an hour at a cost of Rs. 400 an hour and can be used for at most 15 hours. Let M1 be used for x hours and M2 for y hours.
(A) The cost to be minimised is \(Z = 300x + 400y\).
(B) The production constraint is \(20x + 30y \le 600\).
(C) The least cost of the order is Rs. 8250.
(D) In the least-cost plan, M1 is used for 24 hours.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (C) only
At least 600 units are needed, so the problem is
Minimise \(Z = 300x + 400y\)
subject to
\(20x + 30y \ge 600\), that is \(2x + 3y \ge 60\)
\(x \le 24,\ y \le 15\)
\(x \ge 0,\ y \ge 0\)
The line \(2x + 3y = 60\) meets \(y = 15\) at \(x = 7.5\), and meets \(x = 24\) at \(y = 4\).
The corner points are (7.5, 15), (24, 4) and (24, 15). Values of Z:
\(Z(7.5, 15) = 2250 + 6000 = 8250\)
\(Z(24, 4) = 7200 + 1600 = 8800\)
\(Z(24, 15) = 7200 + 6000 = 13200\)
So the least cost is Rs. 8250, with M1 used for 7.5 hours and M2 for 15 hours.
Checking the statements
(A) The cost to be minimised is \(Z = 300x + 400y\). — Correct. x hours at Rs. 300 and y hours at Rs. 400.
(B) The production constraint is \(20x + 30y \le 600\). — Incorrect. At least 600 units are needed, so it is \(20x + 30y \ge 600\).
(C) The least cost of the order is Rs. 8250. — Correct. \(Z(7.5, 15) = 8250\) is the smallest corner value.
(D) In the least-cost plan, M1 is used for 24 hours. — Incorrect. M1 is used for only 7.5 hours.