What number of passengers will maximize revenue?

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SUMMARY

The discussion centers on maximizing revenue for New Horizons Travel's flight package, where the fare is set at $400 per passenger and an additional $8 per unsold seat. The flight accommodates a maximum of 120 passengers and will be canceled if fewer than 50 passengers book. The optimal number of passengers to maximize revenue is determined to be 85, derived from the revenue equation: MONEY = kn + (120 - n)k'n, where k = 400 and k' = 8. The first derivative of this equation is set to zero to find the maximum revenue point.

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New Horizons Travel adversities a package plan for a vacation. The fare for the flight is $400/person plus $8.00/person for each unsold seat on the plane. THe plane holds 120 passengers and the flight will be canceled if there are fewer than 50 passengers. What number of passengers will maximize revenue?

i understand this is a revenue quesitosn but how do u start it off?

thanks.
 
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MONEY=kn+(120-n)k'n,
where k=400, k'=8.

Derive with respect to n and find out for what value of n will the first derivation be zero. This must be a maximum for obviouse reasons. The answer you should get is 85, which is (thankfully) inside your constraints.
 
thanks for your help. I got the right answer.
 
how was the equation gotten
pls expalin
 
how was the equation gotten

Every person needs to pay 400 dollars for their seat (kn) plus every person need s to pay 8 dolars for each empty seat ((120-n)k'n)
 

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