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.

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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.