Maximize REVENUE PROBLEM??

Question

The manager of a large apartment complex knows from experience that 120 units will be occupied if the rent is 448 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 8 dollar increase in rent. Similarly, one additional unit will be occupied for each 8 dollar decrease in rent. What rent should the manager charge to maximize revenue?

Answer 1

To solve with simple calculus:

Method one - solve via price:

revenue = price * number rented

R = p * n

n = 120 + (448-p)/8

n = 120 + 56 - p/8 = 176 - p/8

R = p (176 - p/8) = -1/8p^2 + 176p

The maximum will be where the derivative is zero

derivative => -1/4p + 176

solve for zero derivative

p/4 = 176

p = 704

at p=704, n=88 and R=61,952

Method two - solve via unrented apartments :

n = number of unrented apartments (below 120)

loss = n *448

gain = (120-n) * 8n = 960n - 8n^2

change = (960-448)n - 8n^2 = 512n - 8n^2

You want to find value of n which maximizes change. It won't be affected by dividing by 8.

changeX = 64n - n^2

The maximum will be where the derivative is zero

derivative => 64-2n

solve for zero derivative:

n=32

and your rent is 448 + (32*8) = 704

(nice they come out the same)

To solve without calculus, you have to be a bit more clever:

Use the second version, so you still want to find the maximum value of 64n - n^2

The first part increases in value by 64 each time n increases by one.  The second part decreases the value by the difference between n^2 and (n-1)^2.  As long as that's less than 64, you want to keep increasing n.  So where does the difference equal 64?

(n-1)^2 = n^2 - 2n + 1

Difference = n^2 - (n^2 - 2n + 1) = 2n-1

Solve for diff=64

2n -1 = 64

2n = 65

n = 32.5

So n=32 is the last number before the difference exceeds 64.

n=32

and your rent is 448 + (32*8) = 704

(and it comes out the same all 3 ways)