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smanzy · Apr 28, 2009, 04:57 PM

The captain of a river boat cruise charges $36 per person, including lunch. The cruise averages 300 customers per day. The captain is considering increasing the price. A survey of customers indicates that for every $2 increase, there would be 10 fewer customers. What increase in price would maximize revenue?

Perito · Apr 29, 2009, 06:46 AM

You need to try to set it up yourself.

$$36/person \, \times \, 300\, people = 10800$

$$38/person \, \times \, 290\, people = 11020$

in general

$(36 + 2n) \, \times \, (300-10n) = T$ (T = total)

where n is the number of $2 steps.

$-20n^2+240n+10800=T$

differentiate and set to zero to get maximum (or possibly a minimum) $\frac {dT}{dn}=0$

Hopefully you know calculus. If not, post back and we'll try to solve it some other way.

$-40n+240=0$

$n=6$

The price to charge is, therefore, $48/person, and you'll have 240 people on the boat. You will also save money on lunch :D

You can check using 4 and 7 to assure yourself that n=6 is a maximum (it is).