Expert: Ahmed Salami - 4/22/2009

Question

let r be the monthly rent per unit in an apartment building with 100 units. A survey reveals that all units can be rented when r = $900 and that one unit becomes vacant with ech $10 increase in rent. Suppose that the average monthly maintenance per occupied unit is $100 per month.

question

a. show that the functionfor n (number of rente apartments) in terms of r (rent) is n = -1/10r + 190. what is the domain for this function? explain in complete sentences why you know that his is the function?

b. find the function for the net cash intake, C (revenue minus cost) for this complex, in terms of r. What is the domain for this function?

c. determin how many apartments must be rented to maximize the cash intake. what is the maximum intake?

d. What impact would be to answer in part c if the the owner of the apartment comlex has a mortgage payment of $2100?

e. What impact would be to answer in part c if the maintenance costs increased to $110 per month per occupied unit?

Answer
Hi Kai,
a)Suppose that there are n rented apartments, then there are 100 - n vacant ones. Each vacant apartment results in a $10 increase in the initial rent of $900. For (100 - n) vacant apartments there is a total increase of $10(100 - n) in the initial rent. The new rent r is therefore
r = 10(100 - n) + 900
r = 1000 - 10n + 900
r = 1900 - 10n
10n = -r + 1900
n = -r/10 + 190
When all the 100 units are rented, r is $900 and it increases with a decrease in the number of rented units. At least one unit must be rented otherwise there's no rent. At n = 1
r = 10(100 - n) + 900
r = 10(100 - 1) + 900
r = 10(99) + 900
r = 990 + 900
r = 1890
The domain is the set of values that r can take. We see here that r can take values between $900 and $1890 in increases of $10.

b)Revenue R = r.n
           = r(-r/10 + 190)
           = -r^2/10 + 190r
It costs $100 maintenance per unit,
Cost C = 100n
      = 100(-r/10 + 190)
      = -10r + 19000
Net Cash Intake N = R - C
                 = (-r^2/10 + 190r) - (-10r + 19000)
                 = -r^2/10 + 190r + 10r - 19000
                 = -r^2/10 + 200r - 19000
The domain of N is the same as the domain of n as before.

c)The number of apartments n that maximize N is found when the derivative of N with respect to r i.e dN/dr is equal to zero.
From (b),
dN/dr = -r/5 + 200
equating to zero,
-r/5 + 200 = 0
r/5 = 200
r = 1000
From n = -r/10 + 190
n = -1000/10 + 190
 = -100 + 190
 = 90 units
N = R - C
 = r.n - 100n
 = (r - 100)n
 = (1000 - 100)90
 = (900)90
 = $81000

d)If he has to pay $2100 monthly mortgage, only the maximum intake is affected and it becomes
N = $81000 - $2100
 = $78900

e)If C = 110n
      = 110(-r/10 + 190)
      = -11r + 20900
N = R - C
 = (-r^2/10 + 190r) - (-11r + 20900)
 = -r^2/10 + 190r + 11r - 20900
 = -r^2/10 + 201r - 20900
dN/dr = -r/5 + 201
equating to zero,
-r/5 + 201 = 0
r/5 = 201
r = 1005
From n = -r/10 + 190
n = -1005/10 + 190
 = -100.5 + 190
 = 89.5 units
But of course in reality there is nothing like 89.5 apartments and so we have to pick anyone of 89 or 90 to find which one gives us the maximum N. We write N in terms of n.
N = R - C
 = r.n - 110n
 = (r - 110)n
But
r = 1900 - 10n
N = (1900 - 10n - 110)n
 = (1790 - 10n)n
At n = 89
N = (1790 - 10.89)89
 = (1790 - 890)89
 = (900)89
 = $80100
At n = 90
N = (1790 - 10.90)90
 = (1790 - 900)90
 = (890)90
 = $80100
And so both give the same result.

Hope it helps you.

Regards