Calculus
posted by Chris .
Let r be the monthly rent per unit in an apartment building with 100 units. All the units are rented at r= $900. One unit becomes vacant for each $10 increase in rent. Maintenance cost are $100 for occupied units and nothing for unoccupied units..... so A) show that the number of units rented is n= 190  (r/10) for 900 is less than or equal to r and 1900 is greater than or equal to r B) net cash intake= revenue  maintenance costs. determine the rent r that maximizes net cash intake Note: the answer to part A is 0 is less than or equal to r and 100 is greater than or equal to r .. i just need help with part B

A)
n=floor(190(r/10)) for 900≤r≤1900
Since the number of units has to be integral, and the question says vacancy increases for an increase of $10.
900≤r≤1900 can also be expressed as an interval [900,1900].
B)
Revenue for each unit = r100
number of units = n = floor(190(r/10))
Profit, P(r)
=Revenue  maintenance
= (190(r/10))(r100)
= (r^22000*r+190000)/10
= (2000rr²190000)/10
For a maximum profit, set marginal profit to zero
P'(r) = 2002r/10 = 0
r=1000
if r=1000 is at the maximum, P"(r)<0
P"(r) = 2/10, therefore r=1000 is a maximum.
Also check that n is an integral number:
n=(190(r/10))=1901000/10=90
Profit = P(1000)=$81,000
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