A guest house lets deluxe rooms and standard rooms. It can let at most 30 rooms a night. A deluxe room needs 2 hours of housekeeping and a standard room needs 1 hour, and 40 hours of housekeeping are available each day. The profit is Rs. 1500 on a deluxe room and Rs. 1000 on a standard room for a night. Let x deluxe rooms and y standard rooms be let in a night.
(A) The housekeeping constraint is \(2x + y \le 40\).
(B) (15, 15) is a corner point of the feasible region.
(C) The maximum profit for a night is Rs. 35000.
(D) In the most profitable plan all 30 rooms are let.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (C) and (D) only
The problem is
Maximise \(Z = 1500x + 1000y\)
subject to
\(x + y \le 30\) (rooms)
\(2x + y \le 40\) (housekeeping)
\(x \ge 0,\ y \ge 0\)
The two lines meet where, subtracting,
\(x = 10,\ y = 20\)
The corner points are (0, 0), (20, 0), (10, 20) and (0, 30). Values of Z:
\(Z(0, 0) = 0\)
\(Z(20, 0) = 30000\)
\(Z(10, 20) = 15000 + 20000 = 35000\)
\(Z(0, 30) = 30000\)
So the corners are (0, 0), (20, 0), (10, 20), (0, 30), and the maximum profit is Rs. 35000 at (10, 20).
Checking the statements
(A) The housekeeping constraint is \(2x + y \le 40\). — Correct. x deluxe rooms take 2x hours and y standard rooms take y hours, at most 40 in all.
(B) (15, 15) is a corner point of the feasible region. — Incorrect. \(2(15) + 15 = 45 \gt 40\), so (15, 15) is not even feasible.
(C) The maximum profit for a night is Rs. 35000. — Correct. \(Z(10, 20) = 35000\) is the largest corner value.
(D) In the most profitable plan all 30 rooms are let. — Correct. \(10 + 20 = 30\) rooms.