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August 30, 2014

August 30, 2014

Posted by **Britney** on Wednesday, November 14, 2012 at 11:17pm.

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 -
**Reiny**, Thursday, November 15, 2012 at 7:30amcurrent: 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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