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# Another Optimization Problem (I'm really trying. Honestly!)?

This is almost embarrassing. I never have this much trouble with math.

A cruise line offers a trip for $2000 per passenger. If at least 100 passengers sign up, the price is reduced for all the passengers by $10 for every additional passenger (beyond 100) who goes on the trip. The boat can accommodate 250 passengers. What number of passengers maximizes the cruise line's total revenue? What price does each passenger pay then?

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- mohanrao dLv 79 years agoFavorite Answer
let the total number of passengers = 100 + x

The price of trip per passenger = 2000 - 10x

Total Revenue, R(x) = (100 + x)(2000 - 10x)

differentiate and equate it to zero

(100 + x)(-10) + (2000 - 10x) = 0

=> -1000 - 10x + 2000 - 10x = 0

=> 20x = 1000

x = 50

Thus 150 passengers maximize the revenue.

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