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 36X - X² and profits of the developer are 42Y - 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 profits of the developer. A) H₁ = 18 and H₂ = 20. B) H₁ = 21 and H₂ = 16. C) H₁ = H₂ = 16. D) H₁ = 16 and H₂ = 21. E) H₁ = 20 and H₂ = 24.

Question

Quizplus · Accepted Answer

To maximize profits for a single company owning the airport and housing development, we need to find the value of X and Y that maximizes the combined profit function:
Profit = 36X - XS1U1P12S1S1P0 + 42Y - YS1U1P12S1S1P0 - XY
To maximize profit, we take partial derivatives with respect to X and Y and set them equal to zero:
d(Profit)/dX = 36 - S1U1P12S1S1P0 = 0
d(Profit)/dY = 42 - S1U1P12S1S1P0 - X = 0
Solving these equations gives us:
X = 36 - S1U1P12S1S1P0
Y = (42 - S1U1P12S1S1P0)/2
Substituting these values into the profit function gives us the maximum profit:
Profit = 36(36 - S1U1P12S1S1P0) - (36 - S1U1P12S1S1P0)S1U1P12S1S1P0 + 42(42 - S1U1P12S1S1P0)/2 - (42 - S1U1P12S1S1P0)/2*S1U1P12S1S1P0 - (36 - S1U1P12S1S1P0)(42 - S1U1P12S1S1P0)/2
Simplifying this expression gives us:
Profit = -S1U1P12S1S1P0^2/2 + 1260S1U1P12S1S1P0 - 65520
Now we need to find the values of S1U1B11S1S1P0 and S1U1B12S1S1P0 that maximize the profit function for the two scenarios:
For the first scenario, a single company owns the airport and housing development, so the profit function is given by:
Profit = 36X - XS1U1P12S1S1P0 + 42Y - YS1U1P12S1S1P0 - XY
Substituting the values of X and Y we found earlier gives:
Profit = 18S1U1P12S1S1P0 - S1U1P12S1S1P0^2/2 - 32760
To maximize profit, we take the partial derivative with respect to S1U1P12S1S1P0 and set it equal to zero:
d(Profit)/d(S1U1P12S1S1P0) = 18 - S1U1P12S1S1P0 = 0
Solving for S1U1P12S1S1P0 gives us:
S1U1P12S1S1P0 = 18
Plugging this value back into the profit function gives us:
Profit = 65538
For the second scenario, the airport and housing development are operated independently, so the airport has to pay the developer the total "damages" XY done by the planes to the profits of the developer. The profit function for the developer is given by:
Profit = 42Y - YS1U1P12S1S1P0 - XY
Substituting the value of Y we found earlier gives:
Profit = (42 - S1U1P12S1S1P0/2)Y - YS1U1P12S1S1P0
To maximize profit, we take the partial derivative with respect to S1U1P12S1S1P0 and set it equal to zero:
d(Profit)/d(S1U1P12S1S1P0) = -Y + X = 0
Substituting the value of X we found earlier gives:
Y = (42 - S1U1P12S1S1P0)/2
Substituting this value back into the profit function for the developer gives us:
Profit = (42 - S1U1P12S1S1P0)/2 * (42 - S1U1P12S1S1P0)/2 - (42 - S1U1P12S1S1P0)/2 * S1U1P12S1S1P0
To maximize profit, we take the partial derivative with respect to S1U1P12S1S1P0 and set it equal to zero:
d(Profit)/d(S1U1P12S1S1P0) = 21 - S1U1P12S1S1P0 = 0
Solving for S1U1P12S1S1P0 gives us:
S1U1P12S1S1P0 = 21
Plugging this value back into the profit function for the developer gives us:
Profit = 403.5
Therefore, we have HS1U1B11S1U1B0 = 16 and HS1U1B12S1U1B0 = 21, which is answer choice D.

An Airport Is Located Next to a Housing Development