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April 20, 2014

April 20, 2014

Posted by **julie** on Sunday, November 11, 2007 at 7:57pm.

this is my work so far:

So i want maximize the revenue so I need an expression for the revenue.

revenue = (number of units rented) (rent per unit)

100 * $400 = $40 000

let increments be x:

The number of units rented would then be 100 - x and the rent per unit would be $400 + $5x. Hence the revenue would be

(100 - x)($400 + $5x)

Then i took the derivative of that function above to maximize it and i got x=30.

then i plugged in 30 in the revenue function and i got 4900 dollars. i don't think that right because increasing the rent from 800 to 4900 is too much.

would equation would i substitute x=30 in then?

- please chech my answer optimization -
**Ms. Sue**, Sunday, November 11, 2007 at 8:10pmI'm sorry -- but I don't know how to figure this out algebraically -- but I made a chart showing the revenue for rents up to $450. With 90 apartments occupied, the monthly revenue is $40,500. At this point the revenue keeps increasing. However, at rents of $500 and 80 units occupied, the total revenue is back at $40,000.

- my answer #2 -
**julie**, Sunday, November 11, 2007 at 8:38pmas my total rent, i got 850 dollars.

the man will increase the rent by a maxium of 50 dollars.

i plugged in 30 into the revenue function and i got 4900.

4900/100 is 49 dollars per unit.

i rounded 49 to 50.

- my answer #2 -
- please chech my answer optimization -
**Ms. Sue**, Sunday, November 11, 2007 at 8:54pmI think you mean that the highest rent would be $450 per unit (not $850).

I think there's something wrong with your numbers. However, you've found the right answer. The maximum rent should be $450 for 90 occupied apartments -- bringing in $40,500 a month to the apartment complex.

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