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Let's break down **Problem 1.1** step-by-step:

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### Given:
- **80 units** in the apartment complex.
- At **$300 rent**, all units are occupied (80 rented).
- **For every $20 increase**, one more unit becomes vacant.
- At least **30 units must be rented** for financial stability.
  
---

### Now, let's solve part-by-part:

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**(a) Represent the number of rented units in terms of $x$**:  
- $x$ is the number of **$20 increases** in rent beyond $300.
- Each increase makes **1 unit vacant**.

So, the number of rented units becomes:
\[
\text{Units rented} = 80 - x
\]

---

**(b) Represent the rent per unit in terms of $x$**:
- Each increase adds $20 to the base rent of $300.
- So, the rent per unit is:
\[
\text{Rent per unit} = 300 + 20x
\]

---

**(c) Write an equation for revenue**:
- **Revenue** = (Number of rented units) × (Rent per unit).
\[
R(x) = (80 - x)(300 + 20x)
\]
Now, expand the expression:
\[
R(x) = 80(300 + 20x) - x(300 + 20x)
\]
\[
R(x) = 24000 + 1600x - 300x - 20x^2
\]
\[
R(x) = -20x^2 + 1300x + 24000
\]

This is the **revenue function**.

---

**(d) Find how many units are rented when revenue is $35,000**:
We need to solve:
\[
R(x) = 35000
\]
Using the revenue function:
\[
-20x^2 + 1300x + 24000 = 35000
\]
Simplify:
\[
-20x^2 + 1300x - 11000 = 0
\]
Divide the whole equation by -20:
\[
x^2 - 65x + 550 = 0
\]
Now, solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, $a = 1$, $b = -65$, and $c = 550$:
\[
x = \frac{-(-65) \pm \sqrt{(-65)^2 - 4(1)(550)}}{2(1)}
\]
\[
x = \frac{65 \pm \sqrt{4225 - 2200}}{2}
\]
\[
x = \frac{65 \pm \sqrt{2025}}{2}
\]
\[
x = \frac{65 \pm 45}{2}
\]
So:
\[
x_1 = \frac{65 + 45}{2} = 55 \quad \text{(Not feasible, since too few units would be rented)}
\]
\[
x_2 = \frac{65 - 45}{2} = 10
\]

Thus, $x = 10$.

---

Now, substitute $x = 10$ into the equation for **rented units**:
\[
80 - x = 80 - 10 = 70
\]

So, **70 units** are rented.

---

### Final Answers:
- **(a)**: $80 - x$ units are rented.
- **(b)**: Rent per unit is $300 + 20x$.
- **(c)**: Revenue function: $R(x) = -20x^2 + 1300x + 24000$.
- **(d)**: **70 units** are rented when revenue is $35,000.

---

Would you like a further breakdown of any part?

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### 5 Related Questions:
1. What would the rent per unit be if 60 units were rented?
2. How many $20 increases will yield the maximum possible revenue?
3. What is the minimum revenue the manager can generate if at least 30 units must be rented?
4. How does the revenue change if the rent increases by $10 increments instead of $20?
5. What happens to the revenue if more than 80 units are available?

---

### Tip:
When working with quadratic equations in revenue models, the maximum or minimum value can be found using the **vertex formula**: $x = -\frac{b}{2a}$.

The manager of an 80-unit apartment complex knows from experience that at a rent of $300, all the units will be full. On 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?

This problem explores revenue modeling in an 80-unit apartment complex. Learn how rent increases and unit vacancy affect total revenue using algebraic equations and quadratic formula concepts. Suitable for Grades 8-10.

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