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 = r-100
number of units = n = floor(190-(r/10))
Profit, P(r)
=Revenue - maintenance
= (190-(r/10))(r-100)
= (r^2-2000*r+190000)/10
= (2000r-r²-190000)/10
For a maximum profit, set marginal profit to zero
P'(r) = 200-2r/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))=190-1000/10=90
Profit = P(1000)=$81,000
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
1 answer
1 answer