Asmall cruising ship that can hold up to 68 people provides three-day excursions to groups of 46 or more. if the group contains 46 people, each person pays $70. the cost per person for all members of the party is reduced by $1 for each person in excess of 46. find the size of the group that maximizes income for the owners of the ship.

58 passengers

Step-by-step explanation:

Capacity of the cruise ship =  68 people

Minimum number for an excursion =  46 people

Maximum cost per person = $ 70

Let the additional passenger = y

y minimum =1

y maximum =  68- 46

                   = 22

y  can be represented in the inequality below:

1 ≤ y ≤ 22

New ticket cost of excursion for every added passenger = (70-y)

Total passengers = (46+y)

Income  (I) =  (70-y) (46+y)

                = 3220 + 70y -46y - y²

                = 3220 +24y  -y² (1)

To maximize the income function I (y) in equation (1), dI/dy =0 , and (1) becomes :

24-2y = 0

2y = 24

y =12

So the total number of passengers that maximizes income is :

 = (46+y)

 =  (46+12

= 58 passengers

so then what are you asking?

step-by-step explanation: