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December 20, 2014

December 20, 2014

Posted by **dino** on Monday, October 22, 2012 at 1:59pm.

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 2:59pmcurrent 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( (n-50)^2 - 2500) + 15000

= -2(n-50)^2 + 5000 + 15000

= -2(n-50)^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 300-2(49) = 202

revenue is 99(202) = 19998 which is less than 20000

at 101 members cost per member is 300-2(51) = 198

revenue is 101(198) = 19998 , which is again less than 20000

My answer is correct.

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