Maximum Revenue Question - Business Calculus

Question

The manager of a 60 unit apartment complex is trying to decide what rent to charge. Experience has shown that at a rent of $900 all the units will be full. On average, one additional unit will remain vacant for each $50 increase in rent. Let x represent the number of $50 increases. What is the maximum revenue the complex will see? What is the rent they should charge to obtain this revenue? Why will this be a maximum and not minimum?

Answer 1

All 60 units are full when the rent is 900. The revenue is (60)(900) = 54000.

 

If the rent is 950, there will be 59 units full. The revenue is (59)(950) = 56050

 

In general, for a rent of 900 + 50n, the occupancy will be 60 - n, if n is an integer. So, the revenue as a function of n is

 

R(n) = (900 + 50n)(60 - n)

 

R(n) = 54000 + 2100n - 50n^2

 

Let's take the derivative of R with respect to n and set that equal to 0, to find the extremum.

 

dR/dn = 2100 - 100n = 0

or n = 21

 

To see whether this is a maximum or minimum, take the second derivative of R with respect to n. This gives -100, which is less than 0, which implies that the curve R(n) is everywhere concave down. So, the extremum at n = 21 is a maximum.

 

The rent to charge to obtain the maximum revenue is 900 + (21)(50) = $1950