The hourly operating cost of an airplane, which seats up to 405 passengers, is estimated to be $5,500. If an airline charges each passenger a fare of $100 per hour of flight, find the hourly profit P it earns operating the airplane as a function of the number of passengers x. (Be sure to specify the domain.)
What is the least number of passengers it must carry in order to make a profit? HINT [The cost function is constant (Variable cost = 0).]
Profit: P(x)=100x-5500
I solved it:
P(x)=0
100x-5500=0
5500/100=55 Passengers.
It is wrong, I don't understand.
Thank You
x>55
To find the minimum number of passengers required to make a profit, you need to find the value of x when P(x) is greater than zero. In other words, you need to find the value of x when the revenue (100x) exceeds the cost ($5,500).
Profit, P(x), can be calculated using the formula P(x) = 100x - 5500, as you mentioned. To find the minimum number of passengers required to make a profit, you need to find the value of x where P(x) > 0.
Setting P(x) greater than zero, we have:
100x - 5500 > 0
Now, let's solve for x:
100x > 5500
x > 5500/100
x > 55
So, the minimum number of passengers required to make a profit is 56 (since you can't have fractional passengers).
Therefore, for x ≥ 56 passengers, the airline will make a profit.
I hope this clears up any confusion. If you have any further questions, feel free to ask!
I'm sorry, but it seems like there may have been some confusion. To find the least number of passengers the airline must carry in order to make a profit, we need to set the profit equal to or greater than zero (P(x) ≥ 0), rather than setting it equal to zero (P(x) = 0).
The profit function is given as:
Profit: P(x) = 100x - 5500
To find the least number of passengers, we need to solve the inequality P(x) ≥ 0:
100x - 5500 ≥ 0
Now, let's solve for x:
100x ≥ 5500
x ≥ 5500/100
x ≥ 55
Therefore, the least number of passengers the airline must carry in order to make a profit is 55 (or more) passengers.