Question

A cruise line offers a trip for dollar-sign 1800 per passenger. If at least 100 passengers sign up, the price is reduced for all the passengers by dollar-sign Baseline 10 for every additional passenger (beyond 100) who goes on the trip. The boat can accommodate 250 passengers. What number of passengers maximizes the cruise line’s total revenue? What price does each passenger pay then?

Answer 1

Answer:

(a) 140

(b) $1,400

Explanation:

Price of the trip = $1,800 per person

The boat can accommodate 250 passengers.

Since the price of the ticket is reduced by $10 per person if at least 100 passengers sign up.

P(n) = n{1,800 - (n - 100)10}

      = n{1,800 - 10n + 1000

P(n) = 2,800n - $10n^{2}$

We have to maximize P(n) = 800n - $10n^{2}$

subject to 0≤ n ≤250

P'(n) = 2,800 - 20n

P"(n) = -20

For critical points, solve the equation P'(n) = 0

2,800 - 20n = 0

n = 140

P"(140) = -20 < 0

n = 140 is a point of maxima.

Thus, the number of passenger is 140.

Revenue = P(140)

               = 2,800(140) - $10(140)^{2}$

               = 392,000 - 196,000

               = $196,000

Therefore,

Price paid by each passenger:

$=\frac{Revenue}{number\ of\ passenger}$

$=\frac{196,000}{140}$

= $1,400