Total seats per day \(=200\times 3=600\).
Overall occupancy \(=80\%\), so total tickets sold
\[0.8\times 600=480.\]
Ticket prices:
First two shows: Rs. 250 each (maximum \((2\times200=400) seats\)
Late-night show: Rs. 200 (maximum (200) seats)
To maximize revenue, sell as many tickets as possible at the higher price.
Thus, fill both Rs. 250 shows completely:
\[400 \text{ tickets at Rs. }250,\]
and the remaining
\[480-400=80 \text{ tickets at Rs. }200.\]
\[400\times250 + 80\times200 = 100000 + 16000 = 116000.\]
\[\text{Daily fixed cost} = 10000, \quad \text{Show cost} = 3\times5000=15000.\]
\[\text{Total cost} = 25000.\]
\[116000 - 25000 = 91000.\]
\[\boxed{\text{Maximum profit} = \text{Rs. }91{,}000}\]
Hence, the correct answer is option (4).
Given information:
Calculate total costs: $$\text{Total cost} = 10,000 + 3 \times 5,000 = 10,000 + 15,000 = \text{Rs. } 25,000$$
Calculate total seats occupied: $$\text{Total seats available} = 200 \times 3 = 600$$ $$\text{Total seats occupied} = 0.80 \times 600 = 480$$
Maximize revenue: To maximize profit, fill the higher-priced shows first.
Calculate revenue: $$\text{Revenue} = 200 \times 250 + 200 \times 250 + 80 \times 200$$ $$= 50,000 + 50,000 + 16,000 = \text{Rs. } 1,16,000$$
Calculate profit: $$\text{Profit} = \text{Revenue} - \text{Cost} = 1,16,000 - 25,000 = \text{Rs. } 91,000$$
Answer: (4) Rs. 91,000