Application Problem 2 - Solutions (1)
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Washington University in St Louis *
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100
Subject
Mathematics
Date
Apr 3, 2024
Type
Pages
5
Uploaded by DoctorFogTapir37
This application problem is due on Friday 9/15 at 11:59pm. Please see the Specifications
and Sample Solution on Canvas for a guide to what your submission should look like. As
a reminder, it can be hand-written or typed, and you should upload your submission to
Canvas.
Application Problem 2
An
apartment
manager
keeps
careful
records
of
how
the
rent
charged
per
unit
corresponds to the total number of occupied units in a large apartment complex. This
collected data is shown in the table below:
Monthly
Rent
$650
$700
$750
$800
$850
$900
Occupied
Apartmen
ts
203
196
189
182
175
168
1. Why is it reasonable to say that the number of occupied apartments is a linear
function of rent?
a.
Yes, it is reasonable to say that the number of occupied apartments is a linear
function of rent. They are directly interconnected. The amount of occupied
apartments (y-axis) is dependent on the cost of the monthly rent (x axis). The
higher the rent gets, the less tenants there are in the apartment. The lower the
rent is, the more tenants there are in the facility. This function, however, is linear
as well as decreasing with a rate of change of -7/50. As the rent increases by
$50, a group of 7 tenants leaves. This can be calculated through the rate of
change formula which is f(b)-f(a)/b-a. Plugging the values in, I came up with
196-203/700-650 which then simplifies into -7/50. With this answer and
reasoning, we can definitely say that the number of occupied apartments is a
linear function of rent.
2. Let
A
be the number of occupied apartments and
R
the monthly rent charged (in
dollars). What is the slope of the linear function
A
(
R
)? What is the meaning of the
slope in the context of this question? The slope is negative – why does that make
sense in this context?
a.
The slope of the linear function A(R) is equal to -7/50. Once again, this is
calculated through the rate of change formula which is f(b)-f(a)/b-a. Plugging the
values in, I came up with 196-203/700-650 which then simplifies into -7/50.This is
the slope as well as the rate of change for the function. This slope represents the
relationship between the people leaving the apartment and the increase of the
rent. This slope being negative makes sense in this context because the amount
of tenants living in the apartment complex (y-axis) is decreasing as the rent
(x-axis) increases.
3. Determine a formula for the function
A
(
R
). At what monthly rent will there be no
interested renters?
a.
The formula for the function A(R) is A(R)= -7/50R + 294. We have already
established the slope of the function which is -7/50, now we need to determine
the y-intercept to position the line correctly. In order to do this, we can use the
formula A-A
o
=m(R-R
o
). The m stands for the slope, so we can plug the found
slope into the equation to create A-A
o
=-50/7(R-R
o
). For the A
o
and R
o
, we only
need to plug in values from the table. Considering the number of residents is
dependent on the cost of rent, we can assume that the cost (R) represents the x
axis and the number of residents (A) represents the y-axis. Therefore, we can
further this equation by plugging in points. The points I used were (650, 203).
This gives us the equation A-203=-7/50(R-650). Using the distributive property to
multiply -7/50 with R and -650, we are left with A-203=-7/50R+91. We now need
to isolate A, we do this by transferring the number -203 to the other side of the
equation (note: this also makes the number positive instead of negative). Adding
the 230 to the only like number (number without a variable: 91) we are left with
the sum 294. This is what the equation looks like now A=-7/50R+294. This is the
completed function for A(R). I also confirmed this in desmos and all of the points
in the table line up correctly when graphed. The monthly rent cost in which no
buyers would be interested is $2100. We have already found the function of A(R)
and we can use that to find our point where there will be no buyers. Setting our
function equation equal to zero (0=-7/50R+294) we have replaced the number of
tenants (A) with 0 tenants. This will tell us the cost of apartments that no buyer
would be interested in. We now need to subtract 294 from both sides to isolate
the slope (-294=-7/50R). Now we need to divide both sides of the equation by
-7/50 to isolate the variable R (-294/-7/50=-7/50R/-7/50). This leaves us with
2100=R. R stands for the cost of rent, which means $2100 is the cost of rent that
no buyer is interested in.
4. What do you think is a reasonable domain for the function? Why?
a.
A reasonable domain for this function would be [0, 2100]. Considering it is not
realistic for our domain to contain negative numbers in this instance (as it is
impossible to charge negative amounts of rent for each apartment), the lowest
number in our domain must be 0 (lowest price must be 0). If we plug 0 into the R
(meaning the rent is zero, the minimum), we then get the amount of renters that
would be interested in moving in, 294 people. I did this by, again plugging 0 into R
which gave me -7/50(0)+294. 0x-7/50 is equal to zero, and 0+294 is 294. This
means that A(R) (the number of interested renters) is 294, which is the maximum
number of people that can live in the apartment because the price is at the lowest
it can be. I can then find the second number of the domain by finding the
maximum price that the landlord can charge where no buyers are interested by
setting the function A(R)=0 (meaning that the number of renters is 0), this will
give us our second (maximum) price. Plugging in zero to the y-value, we get
0=-7/50R+294. First, I subtracted 294 from both sides to isolate the slope
(-294=-7/50R). Now we need to divide both sides of the equation by -7/50 to
isolate the variable R (-294/-7/50=-7/50R/-7/50). This leaves us with 2100=R.
With this method, I was able to find the amount of money that could be charged
per apartment where there would be no interested buyers. This then solidifies my
domain of the minimum and maximum amount of money that could be charged:
[0, 2100].
5. If the rent were increased to $1000, how many occupied apartments should the
apartment manager expect? How much total profit would the manager collect in a given
month when rent is set at $1000? (Hint: Total profit = number of units sold
×
rent for each
unit.)
a.
The
expected
amount
of
occupied
apartments
is
154.
Using
our
function
determined in earlier questions (A=-7/50R+294), we know that R stands for the
cost of rent. To determine how many occupied apartments the manager should
expect, we need to plug 1000 into R to determine what A (the occupied units) is.
The equation is now A=-7/50(1000)+294. -7/50(1000) is -140, and when added
with 294, the sum is 154. This means that A=154 and the landlord should expect
154 units to be taken. This answer can be confirmed by plugging 154 into the A of
the equation to solve for R. 154=-7/50R+294. Subtracting 294 from both sides,
we get -140=-7/50R. Then, we divide both sides by -7/50, which leaves us with
1000=R. We can confirm that when R=1000, A=154.
6. One of the buildings in the complex is being renovated, and only 140 units are
available for rent. What rent should the apartment manager charge to attract exactly
140 renters?
a.
The price the manager should charge for apartments with 140 units available is
$1100. Using our function determined earlier (A=-7/50R+294), we know that A
stands for the amount of occupied apartments. In order to determine how much
the manager should charge to occupy those apartments, we need to plug 140
into A in the equation. The equation is now 140=-7/50R+294. Once again, we
need to subtract 294 from both sides in order to isolate the slope. This leaves us
with -154=-7/50R. Now, we need to divide both sides by -7/50 in order to isolate
R. Now we are left with 1100=R. This means that the rent that should be charged
to
occupy 140 apartments is $1100. We can confirm this price by working
backwards.
We
can
plug
1100 into the R of the equation to determine A.
A=-7/50(1100)+294. -7/50(1100) is -154. The sum of -154 island 294 is 140,
leaving us with A=140. This confirms our earlier answer.
7. Let
P
be the total profit of the manager.
P
is a function of
R
, the monthly rent, since
the total profit is equal to the rent per apartment times the number of apartments
rented. Find a formula for the function
P
(
R
).
a.
The function of the total profit of the manager is P(R)=A(R)xR. Since P is a
function of R, that means it is dependent on R, which makes it the y-value; hence
why we put it before the R in parentheses. It is stated that P(R) is equal to the
rent per apartment (which is represented by R) times the number of apartments
rented (which is our function A(R)=-7/50R +294, as A(R) is a function of the
number of rented apartments in the unit). When we plug in the function we have
already determined for A(R), we are left with P(R) = (-7/50R+294)R. We know
that A(R) is equal to -7/50R+294, therefore we can clarify P(R) by plugging the
function of A(R) in rather than leaving it as is. We also must account for the
multiplication of R in the function. Since the number of apartments rented (A(R))
is being multiplied by R, we must distribute the R to the entire function of A(R) as
the whole function (number of apartments rented) must be multiplied by rent (R)
to equal to the total profit, P(R). We must distribute the R to both factors in the
parentheses in the function P(R). (-7/50R)R = -7/50R
2
, and R(294)= 294R. With
our distributive multiplication completed, we can combine the factors once more
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and achieve our completed function of P(R)=-7/50R
2
+294R.
8. In Desmos, graph the function
P
(
R
). You will have to write using
y
and
x
as variables
instead, and you will have to change the viewing box in order to see the whole
function. In the top right corner, click the wrench. Change the
x
-axis to
−
200
≤ x ≤
2500 and the
y
-axis to
−
20000
≤ y ≤
160000. It should look like this:
1
Determine by inspecting the graph what the
x
-coordinate of the vertex of the
parabola is. What does the vertex of the parabola represent? Why do you think the
apartment manager would be interested in that information? Somewhere in your
answer, please include the graph of
P
(
R
). You can either sketch it on paper or
include a picture of the Desmos graph.
a.
Plugging in the parabola function P(R)=-7/50R
2
+294R, the x-coordinate of its
vertex is 1050. The apartment manager would be interested in this information
because the vertex is the highest point on a parabola. After the function increases
and reaches the vertex, then it begins to decrease. The R (rent) value that the
landlord can charge without starting to lose tenants is 1050. The vertex of this
graph is at the point where they can no longer gain customers and don’t run the
risk of losing them. Therefore, the most beneficial price they can charge per unit
is $1,050.
2