Mathematics / Economic

Arrrrroooooooooooogah! Homework alarm! Arrrooooooooooooooogah!

What about the Atheists who want to visit Jerusalem?
Del

This sounds like a linear programming exercise from Junior School (that's 7 - 11 yo in the UK).

Anyway, this question can not be answered, because we don't know whose revenue is to be maximised and what their profit margin is.

Are you telling us this nice Christian lady is going to pocket the 25 difference if more than 100 people go?

Do you leave Abe behind to look after the business and maximize the revenue?

The way I see it, if 200 people go the last person gets paid 1,700 for going ( 800 - (100x25) ).
Note to self, be sure to book late.
Del
(Not sure if Atheist cats are allowed)

If maximum revenue is what is wanted, then a full plane is the answer.

If maximum revenue means greatest profit (no pun intended) per person, then 100 or less flyers is the answer.

or fewer.

You're too young to travel alone. Go ask your mother.

Passenger 131 will pay 25

passenger 132 will go free

The revenue will be equal for 131 and 132

So make it 132 and give someone a free trip - the Jerusalem crowd will see to it that he will be relieved of the 800 xxx he saved by being last.

If the rest of the seats are filled by normal paying people the maximum revenue will be at 100 club members .

The person who made the stupid offer should be fired.

Will this be an international flight?

Del, take note:

Traveling Internationally?

If traveling with your pet internationally, then you should attach an extra copy of your pet's health certificate to the container.

From:

Pet Cargo Travel - Pet airline Cargo Crates - PetTravel.com

Yes.

Would it be greener if they went by bus?

^{[This one might out-do the Bath Breaking Thread!]}

One of these?... Looks pretty green to me.

Del

Just my era....drool

Thank you for your contributions. Since this problem is mathematics/Economics I would think we can approach it as follows: The total amount the minimum passengers number of 100 pay @ 800 each = 80000 Each additional passenger pays 25 less than the previous passenger that is: The 101th passenger pays 775 i.e 800-25. The 102th passenger pays 750 i.e 775-25 and so on. However, this progression will reach a limiting value of zero. That is: 80000+ sum[800-25n] for n taken values from 1, 2, .... . Where n final value is what make the parentheses zero. That analysis make our n to have a terminal value of 32. Thus the total passenger is = 132 and total passenger payment is 92,400 as the maximum revenue . Have a great day!

You too!

However, this progression will reach a limiting value of zero.

That is not stated in the question, andzero is not a mathematical limit.
Del

The club got greedy and overbooked as many as they could to maximise profits. Unfortunately this led to 300 people trying to get on the chartered flight designed only for 220 (the charter's fine print did not say anywhere that one or more people could'nt travel for free). A quick bribe to the charter company (again to maximise profits) and all 300 people were on their way.

Tragedy struck when the overloaded flight failed to clear a mountain resulting in the deaths of everyone on board.

Welcome to the real world where situations similar to this happen and big business is put before the safety of consumers (sometimes with tragic results).