Find the maximized rent in $.

Question

The manager of a 100-unit apartment complex knows from experience that all units will be occupied if the rent is $600 per month. A market survey suggests that, on average, one additional unit will remain vacant for each $10 increase in rent. What rent should the manager charge to maximize revenue?

Find the maximized rent in $.

Answer 1

Hello, Sofia,

Ari R. has the same answer I came up with (I won't say the correct answer, because I'm wrong sometimes). He solved it graphically, and I had the same result. We both used the number of $10 rent increases, x, as the variable on the x axis and the total revenue as the y.

The equation that is plotted is a hyperbola, with a vertx at 20 (A rent of $600 + 20*(10) = $800/month) to produce a total revenue of $64,000/month. Not bad, but please remember that you now have 20 empty units since each $10 increase means one less renter [what;s the definition of slumlord?].

In addition to plotting and graphically finding the vertex, we could also differentiate and set the derivative to zero, since the apex has a slope of zero. I'll do that here:

Let R be the rent ($/month)

x is the number of $10 increments in rent

N is the number of rented units (paying rent)

We can say the following:

R = 600 + 10x [e.g., if we raise the rent in 5 increments, the toal rent per month becomes $650/month)]

N = 100 - x [The number of renters decrease by x for x increases in rental rate at $10 each]

Revenue = N*R

Revenue = (600+10x)(100-x) [total monthly revenue is tne number of rented units times the rental rate per unit]

Revenue = -10x² +400x + 60000 [The 60000 is the revenue at the base price of $600/month, which would be x = 0, or no increase]

Plot that equation to find a vertex at (20, 64,000).

Or, find the first derivative and set it equal to zero, the point on the hyperbola that slope is zero (the apex).

d(Revenue)/dx = -20x - 400

0 = -20x - 400

x = 20 [20 increments of the rent would produce a maximum monthly income.

Rent = 600 + 20*10 = $800/month

Total revenue would then be the number of rented units (100 - 20, or 80) times the new rental rate ($800/month).

That equals $64,000/month, confirming the graphical analysis.

I hope this helps. It was fun.

Bob

Answer 2

Very similar to a recent SAT problem, and can be figured out (mostly) with Algebra.

We want to maximize the price, while figuring out the ideal number of units difference from the original amount (100) Number of units changed is our x variable and price is our Y. As there are two factors that change with more units. (units and price/per) it's actually going to be a quadratic, and we are going to be looking for the maximum value of Y; in this case the vertex. (cool, right!)

For n, which equals the total amount of unit "shifts "

The total profit can be written as follows:

(100 - n) * ( 600 + 10n)

(This is the total amount of units) ( price per unit) = Total rent roll.

We need to find the maximum of this quadratic. N² is negative, so the maximum is just the vertex.

This is in intercept form, so we take the average of the zeros.

The zeros are -60 and 100.

The average of that is 20. So the vertex is 20 shifts from the original price.

The problem asks for rent price. 20 shifts = 600+ 10(20) = $800

The problem does not ask for the total price, but that would just be (100-20) (600+200) = $64,000

Hope this helps!