I've been given a homework question I'm finding really tough, any help would be much appreciated!
Sydtown Airport Corporation (SAC) is concerned about increased flight traffic congestion at the airport. An external consultantís report suggests that the airport is more likely to experience significant congestion between October and June than at any other time of the year. Based on estimates, demand is Qd1= 450 - 0.2P , where Qd1 is quantity demanded for runway timeslots between October and June. Demand during the remaining months of the year is Qd2 = 218.75 - 0.125P , where Qd2 is quantity demanded for runway timeslots between July and September. SAC incurs an additional cost of $950 each time one of the many different airlines utilises the runway, provided 70 or fewer airplanes use the runway on a given day. When more than 70 airplanes use the airport runway, the additional cost incurred by SAC is $5 billion (the cost of building an additional runway and terminal). SAC currently charges airlines a fixed fee of $1,412.50 each time the runway is utilised.
Devise a pricing plan that would enhance SACís profitability.
Again, I would so, so appreciated a hand with this question.
What price will generate a daily demand of 70 slots for each of the periods?
Originally Posted by Kellychu977
Just out interest, how would you do that? And why 70?
My question is the same as james's. How would you Find that? And Do you set it at seventy because that's when it starts going up?
Thanks so much,
Between Oct and June the revenre per day is:
R=Qd1 . p - 950 . Qd1=(450-0.2p)p-950(450-0.2p)
Now find p that maximises this renvenue. If the corresponding Qd1 for this price is greater than 70 check the revenue for a price that generates 70 flights as this will the maximum revenue and the corresponding price.
Then repeat for July to September.
Sorry to be a nonsense, but I have worked this out and I keep getting different prices. Would you be able to post up your solution so i can compare my answers.
R is maximised at p=1600. This corresponds to Qd1=130, which is greater than 70. So profit is maximised at the price that gives Qd1=70 which is p=1900
Originally Posted by CaptainBlack
Show us what you are trying to do so that we can spot what difficulties you are having.
Originally Posted by knightly
How did you solve R=Qd1 . p - 950 . Qd1=(450-0.2p)p-950(450-0.2p)?
Here is how capitanblack did P=1600
so MR=-.4p+640 ,maximize pt MR=0
-.4p+640 = 0
1) Is R represent revenue or profit ? because 950 is cost of production , why do we put that in revenue function ,so it looks more like a profit function .
2)below is what i think ....
TR = P.Q =2250Q-5Q^2
MC=950 (as question mentioned each time an airline use of the runway cost SAC $950 )
set MR = MC
3)When do we use P=MC , MR=MC or MR=0 to maximize profit ?
It is profit, revenue was a mistake in the original (it is a bit more complicated than that but we may as well call it profit (we are ignoring a lot of fixed costs as far as I can see))
Originally Posted by s92225
Where should we incorporate the fixed fee of $1,412.50?
It seems we are ignoring fixed costs.
That is the existing price charged to the airlines for each utilisation. So has no relevance to the question other than to act as noise.
Originally Posted by knightly
I worked out qd2 as 50 with p = 1350.
If this is the point where marginal revenue is 0, then is it the most profitable price for SAC for the period?
Originally Posted by jameslee