(8.13.10) An airline sells all the tickets for a certain route at the same price
ID: 2831795 • Letter: #
Question
(8.13.10) An airline sells all the tickets for a certain route at the same price. If it charges $250 per ticket it sells 5,000 tickets. For every $10 the ticket price is reduced, an extra thousand tickets are sold. It costs the airline $100 to fly a person. What price will generate the greatest profit for the airline?
(8.13.11) A commuter railway has 800 passengers per day and charges each one $2. For each 2 cents that the fare is increased, 5 fewer people will go by train. What is the greatest profit for the airline?
(8.13.12) A poster is to have a total area of 500cm^2. There is a margin round the edges of 6cm at the top and 4cm at the sides and bottom where nothing is printed. What width should the poster be in order to have the largest printed area?
(8.13.13) A sports field is to have the shape of a retangle with semi-circles put on the 2 ends. It must have a perimeter of 1000m. What is the maximum area possible for the retangular part.
Explanation / Answer
(8.3.10)
Locate the cost and revenue functions. When solving the maximize profit in calculus, the problem will generally provide you with the cost and revenue function to start off, but will ask you to solve for "x." In a maximize profit problem, the "x" represents the number of units you must produce to generate the most profit.
Plug your cost and revenue functions into the maximize profit equation: P(x) = R(x) - C(x) where "R(x)" is the revenue function and "C(x)" is the cost function.
Initially the revenue would be (5000 denoted as 5k) 5k*250 =1250k and cost to airlines would be 5k*100=500k
similarly, if ticket price is reduced by $10 then revenue would be (5k+1000)*(250-10) = 6k*240=1440k and cost would be 6k*100=600k
If price is further reduced by $10 then revenue would be (5k+1000+1000)*(250-10-10)=7k*230=1610k and cost would be 7k*100=700k
And the series would go on like this.
Converting Revenue and Cost into equation forms..
Revenue function R(x) = (5k+1000t)*(250-10t) where 'k' represents as above and 't' ranges from 0 to infinite
Cost function C(x) = (5k+1000t)*100 where 'k' and 't' represents as above.
P(x) = R(x) - C(x)
P(x) = {(5k+1000t)*(250-10t)} - {(5k+1000t)*100}
P(x) = 750k + 100kt -10kt2
This P(x) is the equation for profit.
our equation was P(x) = 750k + 100kt - 10kt2,
The derivative P' (x) = 100k - 20kt
the derivative set to zero would be:
0 = 100k - 20kt
20kt = 100k
t = 5.
Putting 't' in (250-10t) gives (250-50) = 200
So if the price of the ticket is 200, airline will get maximum profits.
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