The demand function for an item is p(x) = 100 - x, where p is price in rupees and x is quantity. The maximum total revenue is: A. ₹1000 B. ₹2500 C. ₹5000 D. ₹2000 Answer: B) ₹2500 Explanation: Revenue R = p(x) × x = 100x - x². R'(x) = 100 - 2x = 0 → x = 50. Max R = 50(50) = ₹2500.

imo class 12 application of derivatives

The demand function for an item is p(x) = 100 - x, where p is price in rupees and x is quantity. The maximum total revenue is:

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The demand function for an item is p(x) = 100 - x, where p is price in rupees and x is quantity. The maximum total revenue is:

Answer: B) ₹2500

Explanation: Revenue R = p(x) × x = 100x - x². R'(x) = 100 - 2x = 0 → x = 50. Max R = 50(50) = ₹2500.