Hotel Room Occupancy Revenue Maximization

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 = $2000(0) + 3000(20) + 3500(5) = 60000 + 17500 = 77,500$.