An airport is located next to a housing development. Where X is the number of pl

Question

An airport is located next to a housing development. Where X is the number of planes that land per day and Y is the number of houses in the housing development, profits of the airport are 20X - X 2 and profits of the developer are 28Y - Y2 - XY. Let H1 be the number of houses built if a single profit-maximizing company owns the airport and the housing development. Let H2 be the number of houses built if the airport and the housing development are operated independently and the airport has to pay the developer the total “damages” XY done by the planes to the developer’s profits.
a. H1 =12 and H2 =14
b. H1 = H2 = 12
c. H1 =14 and H2 = 13
d. H1 = 14 and H2 = 12
e. H1 = 13 and H2 = 17

Correct answer A.

Explanation / Answer

X is the number of planes and Y is the number of houses.

Profits of the airport = 20X -X2

And profits of the developer = 28Y - Y2 - XY

Now, the first case is , H1 be the number of houses built if a single profit maximising company owns the airport and the housing development :

This implies that total profit of airport and developer has to maximise:

So,Total profit = (20X - X2 )+ (28Y - Y2- XY)

To maximise total profit partial dfferentiate it with respect to X and with respect to Y and equate them to zero. Then we get these equations;

By partial differentiate Total profit function w.r.t X = 20 - 2X -Y = 0 ------(1)

By partial differentiate w.r.t Y = 28 - 2Y - X = 0 -----(2)

Now, multiply equation (1) by 2 , we get

40- 4X - 2Y = 0 ----(3)

Now, solve equation (2) and (3) , we get

X = 4

Put X=4 in equation(1) ,we get

20 -2(4) -Y= 0

Y = 12 (number of houses )

So, this is the case when single profit maximising firm owns the airport and developer , therefore Y=12 =H1 = 12 houses.

And now, the second case, H2 be the number of houses built if the airport and the housing developement are operated independently and the airport has to pay the developer the total damages XY done by the planes to the developer's profits:

Earlier developer profits = 28Y -Y2 - XY

After paying XY amount by the planes to the developer, then developer profits =

28Y- Y2 -XY +XY

28Y -Y2

Now, to maximise profit differentiate it w.r.t Y and eqiuate it equal to zero.

28 - 2Y = 0

2Y = 28

Y = 14 (Number of houses)

Therefore, H2 = Y = 14 houses in the second case.

Hence the final answer is H1=12 and H2 =14 . Therefore , option(a) is correct.

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