The manager of an 110-unit apartment complex knows from experience that at a rent of $300, all the units will be full. on the average, one additional unit will remain vacant for each $20 increase in rent over $300. furthermore, the manager must keep at least 50 units rented due to other financial considerations. currently, the revenue from the complex is $72 comma 000. how many apartments are rented?

The number of rented apartments is 80

Step-by-step explanation:

Hi there!

The revenue is calculated as the number of units rented times the value of the rent. Let y be the revenue, then:

y = 300 · 110

For each $20 increase, one unit will remain vacant. Then, if  the increase is 20x then the number of vacant apartments will be x. In this case, the revenue will be calculated as:

y = (300 + 20x)(110 - x)

Apply distributive property:

y = 33000 - 300x + 2200x - 20x²

y = -20x² + 1900x + 33000

If the revenue is $72000, then:

72000 = -20x² + 1900x + 33000

0 = -20x² + 1900x + 33000 - 72000

0 = -20x² + 1900x - 39000

Solving the quadratic equation using the quadratic formula:

a = -20

b = 1900

c = -39000

x = [-b±√(b² - 4ac)]/2a

x1 = 30 and x2 = 65

Since at least 50 units have to be rented, the numer of apartments rented is 110 - 30 = 80 (because 110 - 65 = 45<50).

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step-by-step explanation:

my guess would be b and d, but i didn’t actually get out the pen and paper. lol