let the number of $10 increases be n
new rent = 460+10n
number rented= 80-n
Revenue = (80-n)(460+10n)
= 36800 + 340n - 10n^2
d(Rev)/dn = 340 - 20n
= 0 for a max/min
20n=340
n = 17
So there should be an increase of 17(10) or 170
and the new rent should be 630
and 63 units would be rented
check:
let n = 16
rent = 620
units rented = 64
Rev = 620(64) = 39680
let n = 17
rent = 630
units rented = 63
Rev = 39690 , a bit more , our maximum
let n = 18
rent = 640
units rented = 62
Rev = 39680 , not as much as the 39690 at n = 17
The manager of a large apartment complex knows from experience that 80 units will be occupied if the rent is 460 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 10 dollar increase in rent. Similarly, one additional unit will be occupied for each 10 dollar decrease in rent. What rent should the manager charge to maximize revenue?
2 answers
jnjkn