Question

There are 25 rooms in a hotel. Each room can accommodate at the most three people. For each room, the single occupancy charge is Rs. 2000 per day, the double occupancy charge is Rs. 3000 per day, and the triple occupancy charge is Rs. 3500 per day.

If there are 55 people staying in the hotel today, what is the maximum possible revenue from room occupancy charges today?

Accepted Answer

C. Let $x, y, z$ be the number of single, double, and triple occupancy rooms used, respectively. We are given: 1) $x + y + z = 25$ 2) $x + 2y + 3z = 55$ Subtracting (1) from (2): $y + 2z = 30 \implies y = 30 - 2z$ Since $x + y + z = 25 \implies x + (30 - 2z) + z = 25 \implies x = z - 5$. For $x \ge 0$, we must have $z \ge 5$. To maximize revenue $R = 2000x + 3000y + 3500z$: $R = 2000(z - 5) + 3000(30 - 2z) + 3500z = 2000z - 10000 + 90000 - 6000z + 3500z = 80000 - 500z$ To maximize $R$, we need the minimum possible integer $z \ge 5$, which is $z = 5$. Then $x = 0$ and $y = 20$. Maximum Revenue = $2000(0) + 3000(20) + 3500(5) = 60000 + 17500 = 77,500$.

Hotel Room Occupancy Revenue Maximization

Explanation

Practise this under exam conditions