Rent Problem! (Math)?

Question

Can anyone do this problem? I need help please.

The manager of a large apartment complex knows from experience that 80 units will be occupied if the rent is 320 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?

Answer 1

Occupancy as a function of Rent is
O(r) = 80 - (r-320)/10

Revenue(r) = rent x Occupancy
= r O(r)
= 80 r - r(r-320)/10
= 80 r - (r^2)/10 + 32 r
= 112 r - (r^2)/10

To maximize, set derivative = 0:

d Revenue/dr = 112 - r/5 ; this = 0 when r = 560

So the optimal rent is r = 560

Answer 2

im assuming that you know calculus. Here is the solution:

Revenue = number of units rented X rent per unit
In the case when rent is 320:
Revenue = 320 X 80
If he decdes to increse the rent by a, then this is what happens:
Revenue = (320+a)(80-(a/10))

Now you have Revenue as a function in terms of a.
You should know the rest. To find at which a, the revenue is max, you gotta differentiate:
[(320+a)(80-(a/10))]' = 0
a = 240 <-This is what i got might be wrong
So the final rent should be 320 + 240 = 560

If you wanna check whether you answer is making sense:
(320+240)(80-(240/10)) = 31360
(320+230)(80-(230/10)) = 31350 <-Ur making less if you decrease the price
(320+250)(80-(250/10)) = 31350<-Ur making less if you increase the price

So the answer is right.

Answer 3

okay, wow, i'm a calculus teacher, and I have to think it thru but here goes

at $320, he rents out 80, so takes in 80*320 = 25600

if he increases rent by 10, he loses one unit ... decrease 80 by 1, and increase 320 by 10 each time... see when the rent is highest...

79*330=26070
78*340=26520
77*350=26950
76*360=27360
75*370=27750
74*380=28120
73*390=28470
72*400=28800
71*410=29110
70*420=29400
69*430=29670
68*440=29920
67*450=30150
66*460=30360
65*470=30550
64*480=30720
63*490=30870
62*500=31000
61*510=31110
60*520=31200
59*530=31270
58*540=31320
57*550=31350
56*560=31360
55*570=31350

(so with rent increase, highest income is at the 560 rent level that is 31360)

go back to original figures of

80*320 = 25600
and now go with rent decrease... increase 80 by 1, and decrease 320 by 10 each time...

81*310=25110
82*300=24600
83*290=24070
84*280=23520

so , by the pattern, the highest income is at the 560 rent level that is 31360)

There is a way to do this with "limits" but this is the long way to get to the answer...

Answer 4

Let x = the number of people who leave with an increase in rent.
The income is then (80 - x)(320 + 10x) = -10x^2+480x+25600
The graph of this is a parabola which has a max at -b/2a =
-480/-20 = 24. The revenue is maximized when 24 people leave and the rent is 560.

Answer 5

If he charges 320 and 80 are full he gets 25600 per month
If he charges 330 and 79 are full he gets 26070
If he charges 400 and 72 are full he gets 28800

The more he charges the more he makes. Find where he can make the most by charging the most, no matter how many are vacant. Can you figure it out from here?

Answer 6

$320

Answer 7

560

Answer 8

charge 23 dollars more for everyone a month and blame it on a new few by repairs,goverment policies, inflation, gas prices employee benifit health cost increases or what ever get rich...

Answer 9

$560 dollars... results in 56 tenants and $31360 total... hope this helps

Answer 10

Sorry, you need to do your own homework.