# math

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The owner of a luxury motor yacht that sails among the 4000 Greek islands charges \$960/person/day if exactly 20 people sign up for the cruise. However, if more than 20 people sign up (up to the maximum capacity of 100) for the cruise, then each fare is reduced by \$8 for each additional passenger.

Assuming at least 20 people sign up for the cruise, determine how many passengers will result in the maximum revenue for the owner of the yacht.

? passengers

What is the maximum revenue?

\$ ?

What would be the fare/passenger in this case? (Round your answer to the nearest dollar.)

? dollars per passenger

• math -

current: 20 passengers at \$960 each

let the number of addional passengers be x
cost per passenger = 960 - 8x

revenue (R) = (20+x)(960-8x)
= 19200 - 160x + 960x -8 x^2
dR/dx = -160 + 960 - 16x = 0 for a max of R
16x = 800
x = 50

There should be an additional 50 or a total of 70 passengers
the cost per passenger would be 960-8(50) or \$560 per day

check:

try a number around 70 passengers
for 69 passengers
cost = 960-8(49) = 568
R = 39192

for 70 passengers
cost = 960-8(50) = 560
R = 39200 ---- the highest

for 71 passengers
cost = 960-51(8) = 552
R = 39192

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