The landlord of a 40-unit apartment building is planning to increase the rent.

Question

Currently, residents pay $700/month. 4 units are vacant. Every $25 increase in monthly rent results in one more vacant unit. What rent results in max. revenue?

Answer 1

Let A = # of occupied apts and R = rent

A = (800-R)/25

Revenue = A * R = (800R - R²)/25

dRevenue/dR = 32 - 2R/25

Max revenue when dRevenue/dR = 0 when R=400

Answer 2

Let n = the number of vacant apartments. Then the number of rented apartments = 40 - n.

Let x = the monthly rent. So the total revenue (R) is x(40 - n)

Now, let's use the information given to write a relationship between the number of vacant apartments and the rent. It's a linear relationship. $700 monthly rent results in 4 vacant apartments, $725 results in 5, etc. like this:

x | n

------ |------

700 | 4

725 | 5

750 | 6

So we can use y = mx + b, or in our case n = mx + b with m = Δn/Δx = (5-4)/(725 - 700) = 1/25 so n = 1/25x + b. Plugging in a data point (700, 4) we can calculate b: 4 = 1/25(700) + b or b = -24 so:

n = 1/25x - 24

Taking our revenue equation (R = x(40 - n) and replacing n with "1/25x - 24", we get R = x[40 - (1/25x - 24)] and simplifying, we get:

R(x) = x[40 - 1/25x + 24]

R(x) = x(64 - 1/25x)

R(x) = 64x - 1/25x²

To find the maximum revenue, we take the derivative and set it equal to zero.

R'(x) = 64 - 2/25x

64 - 2/25x = 0

x = 64*25/2 = $800/month

As a sanity check, $800 will result in 8 vacant apartments leaving 32 filled and 32*800 = $25,600 in revenue. $775 results in 7 vacant apartments or 33 filled and 33*775 = $25,575 (lower). $825 in rent results in 9 vacant apartments or 31 filled and 31*825 = $25,575 (lower). Looks good.

Answer 3

Let n = the number of vacant apartments. Then the number of rented apartments = 40 - n.

Let x = the monthly rent. So the total revenue (R) is x(40 - n)

Now, let's use the information given to write a relationship between the number of vacant apartments and the rent. It's a linear relationship. $700 monthly rent results in 4 vacant apartments, $725 results in 5, etc. like this:

x | n

------ |------

700 | 4

725 | 5

750 | 6

So we can use y = mx + b, or in our case n = mx + b with m = Δn/Δx = (5-4)/(725 - 700) = 1/25 so n = 1/25x + b. Plugging in a data point (700, 4) we can calculate b: 4 = 1/25(700) + b or b = -24 so:

n = 1/25x - 24

Taking our revenue equation (R = x(40 - n) and replacing n with "1/25x - 24", we get R = x[40 - (1/25x - 24)] and simplifying, we get:

R(x) = x[40 - 1/25x + 24]

R(x) = x(64 - 1/25x)

R(x) = 64x - 1/25x²

To find the maximum revenue, we take the derivative and set it equal to zero.

R'(x) = 64 - 2/25x

64 - 2/25x = 0

x = 64*25/2 = $800/month

As a sanity check, $800 will result in 8 vacant apartments leaving 32 filled and 32*800 = $25,600 in revenue. $775 results in 7 vacant apartments or 33 filled and 33*775 = $25,575 (lower). $825 in rent results in 9 vacant apartments or 31 filled and 31*825 = $25,575 (lower). Looks good.