Hotel Room Revenue Maximization

Let $x, y, z$ be the number of single, double, and triple occupancy rooms used, respectively. We have $x + y + z = 25$ and $x + 2y + 3z = 55$. Subtracting the first equation from the second gives $y + 2z = 30$, so $y = 30 - 2z$. Substituting into $x + y + z = 25$ gives $x + (30 - 2z) + z = 25 \implies x = z - 5$. Since $x \ge 0$, 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 to minimize $z$. The minimum non-negative value for $z$ such that $x \ge 0$ is $z = 5$. Then $x = 0$ and $y = 30 - 2(5) = 20$. Maximum Revenue = $20 \times 3000 + 5 \times 3500 = 60000 + 17500 = \text{Rs. } 77,500$.