Posted by dino on Monday, October 22, 2012 at 1:59pm.
can anyone please help me set this problem up, I can do the math but I have tried for hours...
Air fare club offers membership at $300.00, at least 50 people must join.For every member over 50 their fare will be reduced by $2 for every member. Due to space limits only 125 passengers will be allowed. how many memberships will maximize the revenue?

algebra  Reiny, Monday, October 22, 2012 at 2:59pm
current rate = $300
number of members = 50
let the number of additonal member be n
so actual number of members = n+50
cost per member = 300  2n
Revenue = number of members x membership fee
= (n+50)(300  2n)
= 300n  2n^2 + 15000 100n
= 2n^2 + 200n + 15000
= 2(n^2  100n + 2500  2500 ) + 15000
= 2( (n50)^2  2500) + 15000
= 2(n50)^2 + 5000 + 15000
= 2(n50)^2 + 20000
This is a parabola opening downward with a vertex of (50,20000)
So the maximum of Revenue is $20,000 when n = 50
So 50 addition to the original 50 should join.
The maximum revenue is obtained when 100 members are in.
check:
at 100 members cost is 300  2(50) = 200
revenue is 100(200) = 20000
at 99 members cost per member is 3002(49) = 202
revenue is 99(202) = 19998 which is less than 20000
at 101 members cost per member is 3002(51) = 198
revenue is 101(198) = 19998 , which is again less than 20000
My answer is correct.
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