A real estate company owns 120 apartments which are fully occupied when the rent is $300 per month.

Question

A real estate company owns 120 apartments which are fully occupied when the rent is $300 per month. The company estimates that for each additional $10 a month increase in rent, live apartments will become vacant. What rent should be charged in order for the company to obtain the largest gross income?

Answer 1

Let x = number of $10 increases in rent.

G(x) = gross monthly ncome = (number of apartments rented)(rent per apartment)

If x = 0, then all apartments are rented and if x = 24, 120 - 5x = 0, so no apartments are rented.

G(x) = (120 - 5x)(300 + 10x) = -50x² - 300x + 36000, where 0 ≤ x ≤ 24

G'(x) = -100x - 300 = 0 when x = -3.

So, there are no critical points in the interval [0,24]. Therefore, G(x) is maximized at an endpoint.

G(0) = $36,000 and G(24) = 0

Gross monthly income is maximized when x = 0. In other words, the current rent of $300 per month maximizes gross monthly income, so don't raise the rent.

Answer 2

If you mean this to say "for each additional $10 a month increase in rent, five apartments will become vacant" then here you go:

Let "n" represent the number of apartments that are rented. This value must range between 0 and 120. And let "r" represent the rent charged. You could make a table like this:

This is a linear relationship so fits y = mx + b (in this case r = mn + b)

The slope is (120 - 115)/(300 - 310) = 5/-10 or -0.5

You could use the point-slope form to get the equation of the line:

y - y₁ = m(x - x₁) or, in this case:

n - n₁ = m(r - r₁)

then plugging in m = -0.5 and (r₁, n₁) = (300, 120) we get:

n - 120 = -0.5(r - 300)

n = -0.5r + 150 + 120

n = -0.5r + 270

If we plug in zero for the number of apartments rented, we see that the rent that would be so high as to result in zero apartments being rented is:

0 = -0.5r + 270

0.5r = 270

r - 540

Therefore the rent must range between $300 and $540

To calculate income, multiply the number of apartments rented and the rent:

Income (I) = (number of apartments rented)(rent) = (n)(r) but n = -0.5r + 270 so:

I = (-0.5r + 270)(r) = -0.5r² + 270r

I(r) = -0.5r² + 270r

We are trying to maximize this so we take the derivative and set it equal to zero:

I'(r) = (2)(-0.5)r + 270 = -r +270 = 0 or r = 270

But we see that this is not a possible answer as rent must be between $300 and $540 therefore the lowest rent of $300 produces the maximum income.