Question

There are 25 rooms in a hotel. Each room can accommodate at 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?

• A. Rs. 72500
• B. Rs. 77500
• C. Rs. 87500
• D. Rs. 82500
• E. Rs. 92500

Hint:

Formulate constraints as linear equations and maximize revenue using the given conditions.

Answer 1

The Correct Option is B

Step 1: To maximize revenue, we want as many triple occupancy rooms as possible (highest per room revenue? Actually per person: Single = 2000/person, Double = 1500/person, Triple ≈ 1166.67/person. But per room, Triple gives 3500, Double 3000, Single 2000. So triple rooms give highest revenue per room.
Step 2: 55 people, each triple room holds 3. Maximum triple rooms = $\lfloor 55/3 \rfloor = 18$ triple rooms (54 peopl(e), with 1 person left.
Step 3: Revenue = $18 \times 3500 = 63000$ + 1 single room = 2000 = 65000. But we have 25 rooms available, so we can use more rooms.
Step 4: Alternatively, use 17 triple rooms (51 peopl(e), remaining 4 people. These 4 can be in 2 double rooms (3000 each = 6000) or 1 triple + 1 single? But triple already used. 4 people = 2 double rooms (4 peopl(e). Revenue = $17 × 3500 + 2 × 3000 = 59500 + 6000 = 65500.
Step 5: Try to maximize: Let t = triple, d = double, s = single. t+d+s = 25, 3t+2d+s = 55. Subtract: (3t+2d+s) - (t+d+s) = 55-25 → 2t+d = 30.
Step 6: Revenue R = 3500t + 3000d + 2000s. Since s = 25 - t - d, R = 3500t + 3000d + 2000(25-t-(d) = 3500t + 3000d + 50000 - 2000t - 2000d = 1500t + 1000d + 50000.
Step 7: Constraint: 2t + d = 30 → d = 30 - 2t. Also t ≥ 0, d ≥ 0 → t ≤ 15.
Step 8: R = 1500t + 1000(30-2t) + 50000 = 1500t + 30000 - 2000t + 50000 = 80000 - 500t.
Step 9: R is maximum when t is minimum. Minimum t = 0? Then d=30, but d cannot exceed 25 (rooms). So constraints: d ≤ 25 → 30-2t ≤ 25 → 5 ≤ 2t → t ≥ 2.5 → t ≥ 3.
Step 10: Also s ≥ 0 → 25 - t - (30-2t) ≥ 0 → 25 - t - 30 + 2t ≥ 0 → t - 5 ≥ 0 → t ≥ 5.
Step 11: Minimum t = 5. Then d = 30-10=20, s = 25-5-20=0. Check: 3×5+2×20=15+40=55 people.
Step 12: Revenue = 3500×5 + 3000×20 = 17500 + 60000 = 77500.
Step 13: Final Answer: Rs. 77500.