Calculus Rent Problem?

Question

The manager of a large apartment complex knows from experience that 100 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 5 dollar increase in rent. Similarly, one additional unit will be occupied for each 5 dollar decrease in rent. What rent should the manager charge to maximize revenue?

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The answer I got was 460 but that doesn't seem right at all.
Any help? Suggestions?
Thanks!

Answer 1

Initial Revenue, R= 100x320 =32000
Let x be no $5 increases in rent
One additional unit will remain vacant for each 5 dollar increase in rent: so
R= (100-x)(320+5x)
R= 32000+180x-5x^2
dR/dx = 180 -10x
dR/dx = 0 if x = 18

So maximum revenue if x=18,
ie rent = 320+5x18 = 410
Then no of apartments occupied = 100-18=82

So revenue = 410x82 =$ 33620

Answer 2

Let x be the number of apartments rented at a particular price p. The gross rent G will be px. The constraint is that x = 100 - ((p-320)/5) = 100 -p/5 +64 = -p/5+164. Plugging in, we have G = p(-p/5+164) = -p^2/5 +164p. We wish to maximize G with respect to p. Then dG/dp = -2p/5 +164 = 0, and p = 410. At that price, there will be 82 tenants and the gross rent roll will be $33,620 as opposed to the present roll of $33,000.

Answer 3

n(320) = 100
n(325) = 99
(n - 100)/(r - 320) = - 1/5
5n - 500 = 320 - r
5n = 820 - r
n = 160 - (1/5)r
R = nr
R = 160r - (1/5)r^2
dn/dr = 160 - (2/5)r = 0
(2/5)r = 160
r = $400

check:
R = r(160 - r/5)
R(395) = 395(160 - 395/5)
R(395) = $31,995
R(400) = 400(160 - 400/5)
R(400) = $32,000
R(405) = 405(160 - 405/5)
R(405) = $31,995

Answer 4

p = hire u = instruments rented (p - 800)/(u - one hundred) = 10/-a million -p/10 + 80 = u - one hundred u = -p/10 + one hundred eighty r = u*p r = -p^2/10 + 180p dr/dp = -2p/10 + one hundred eighty -p/5 + one hundred eighty = 0 p = 5*one hundred eighty p = $900