The manager of an 80-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 30
units rented due to other financial considerations.
(a) Suppose that x represents the number of $20 increases over $300. Represent
the number of apartment units that will be rented in terms of x .
(b) Represent the rent per unit in terms of x .
(c) Use the answers in part (a) and (b) to write an equation that defines the
revenue generated when there are x increases of $20 over $300.
(d) The revenue currently generated is $35,000. How many apartment units are
rented?

We will go through each part of the problem step by step:

### Part (a) - Number of Apartment Units Rented in Terms of $ x $:

Let $ x $ represent the number of $20 increases over the initial $300 rent. 

- At $300 rent, all 80 units are rented.
- For each $20 increase in rent, one additional unit will become vacant.

Thus, the number of units rented as a function of $ x $ is:
\[
\text{Units rented} = 80 - x
\]

### Part (b) - Rent per Unit in Terms of $ x $:

The rent starts at $300, and for each increase of $ x $, the rent rises by $20. 

So the rent per unit in terms of $ x $ is:
\[
\text{Rent per unit} = 300 + 20x
\]

### Part (c) - Revenue as a Function of $ x $:

Revenue is calculated as:
\[
\text{Revenue} = \text{(Number of units rented)} \times \text{(Rent per unit)}
\]

Substituting the expressions from part (a) and part (b):

\[
\text{Revenue} = (80 - x) \times (300 + 20x)
\]

Now, expanding this expression:

\[
\text{Revenue} = 80(300 + 20x) - x(300 + 20x)
\]
\[
\text{Revenue} = 80(300) + 80(20x) - x(300) - x(20x)
\]
\[
\text{Revenue} = 24000 + 1600x - 300x - 20x^2
\]
\[
\text{Revenue} = 24000 + 1300x - 20x^2
\]

So the revenue as a function of $ x $ is:
\[
R(x) = 24000 + 1300x - 20x^2
\]

### Part (d) - Finding How Many Units Are Rented When Revenue is $35,000:

We are given that the revenue is $35,000. So, we set $ R(x) = 35000 $ and solve for $ x $:

\[
24000 + 1300x - 20x^2 = 35000
\]
Subtract 35000 from both sides:
\[
24000 + 1300x - 20x^2 - 35000 = 0
\]
\[
-11000 + 1300x - 20x^2 = 0
\]
Multiply the equation by -1 to simplify:
\[
11000 - 1300x + 20x^2 = 0
\]
Rearranging it in standard quadratic form:
\[
20x^2 - 1300x + 11000 = 0
\]

Now, solve this quadratic equation using the quadratic formula:
\[
x = \frac{-(-1300) \pm \sqrt{(-1300)^2 - 4(20)(11000)}}{2(20)}
\]
\[
x = \frac{1300 \pm \sqrt{1690000 - 880000}}{40}
\]
\[
x = \frac{1300 \pm \sqrt{810000}}{40}
\]
\[
x = \frac{1300 \pm 900}{40}
\]

Now, compute the two possible values for $ x $:

1. $ x = \frac{1300 + 900}{40} = \frac{2200}{40} = 55 $
2. $ x = \frac{1300 - 900}{40} = \frac{400}{40} = 10 $

Since the manager must keep at least 30 units rented, and the number of units rented is $ 80 - x $, we can only accept $ x = 10 $ because $ 80 - 55 = 25 $ units is below the required 30 units. Thus, $ x = 10 $ is the valid solution.

Now, the number of units rented when $ x = 10 $ is:
\[
80 - 10 = 70 \text{ units}
\]

### Final Answer:
When the revenue is $35,000, 70 units are rented.

---

Would you like more details on any part? Here are some related questions:

1. What is the maximum revenue the manager can generate?
2. How would the equation change if rent increased by $30 instead of $20?
3. How many units are rented when the rent is $400?
4. How does the vacancy rate change as rent increases?
5. Can you graph the revenue function to visualize the relationship between rent and units?

**Tip:** Always check that your solutions satisfy all problem constraints, such as minimum or maximum values.

The manager of an 80-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 30 units rented due to other financial considerations. (a) Suppose that x represents the number of $20 increases over $300. Represent the number of apartment units that will be rented in terms of x. (b) Represent the rent per unit in terms of x. (c) Use the answers in part (a) and (b) to write an equation that defines the revenue generated when there are x increases of $20 over $300. (d) The revenue currently generated is $35,000. How many apartment units are rented?

Solve an apartment manager's revenue problem using algebra and quadratic equations. Understand how rent changes affect occupancy, calculate optimal revenue, and determine the number of units rented at $35,000 revenue. This math problem applies algebraic concepts like quadratic equations and revenue models, ideal for Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation