A hospital ward needs at least 10 nurses on duty in a day, with at least 3 of them on the night shift, and at most 8 nurses can be put on the day shift. A day-shift nurse costs Rs. 1,500 and a night-shift nurse costs Rs. 1,800 a day. If x nurses work the day shift and y work the night shift, the least daily cost is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Rs. 15,900
The problem is
Minimise \(Z = 1500x + 1800y\)
subject to
\(x + y \ge 10\)
\(y \ge 3\)
\(x \le 8\)
\(x \ge 0\)
The feasible region is unbounded (y has no upper limit). Its corner points are
\(x = 0\) on \(x + y = 10\): (0, 10)
\(y = 3\) on \(x + y = 10\): (7, 3)
\(x = 8,\ y = 3\): (8, 3)
Values of Z:
\(Z(0, 10) = 18000\)
\(Z(7, 3) = 10500 + 5400 = 15900\)
\(Z(8, 3) = 12000 + 5400 = 17400\)
The half-plane \(1500x + 1800y \lt 15900\) has no point in common with the region, because
\(Z = 1500(x + y) + 300y \ge 15000 + 900 = 15900\)
So the least cost is Rs. 15,900, with 7 day-shift and 3 night-shift nurses.
Why the other options are incorrect
Option (1) Rs. 15,000 — Incorrect. This needs 10 day-shift nurses and none at night, which breaks both \(x \le 8\) and \(y \ge 3\).
Option (3) Rs. 17,400 — Incorrect. This is the cost at (8, 3), with one nurse more than needed.
Option (4) Rs. 18,000 — Incorrect. This is the cost at (0, 10), all on the dearer night shift.