E. A real estate firm owns an apartment complex which consists of 96 apartments. At $550 per month every apartment can be rented. However, for each $25 increase per apartment there will be three vacancies with no possibility of filling them. The firm wants $54600 per month from rent. What amount should be charged for each apartment.

To solve this problem, we need to create an equation that models the situation and then solve for the rent amount that meets the firm's target revenue of $54,600 per month.

### Step 1: Define Variables
Let:
- $ x $ be the number of $25 increases in the rent.
- The rent per apartment after $ x $ increases will be $ 550 + 25x $.
- The number of apartments rented after $ x $ increases will be $ 96 - 3x $, since for every $25 increase, 3 apartments become vacant.

### Step 2: Set Up the Revenue Equation
The total revenue is the rent per apartment times the number of rented apartments. This gives the revenue as:

\[
\text{Revenue} = (\text{Rent per apartment}) \times (\text{Number of apartments rented})
\]
Substituting in the expressions for rent and number of apartments:

\[
\text{Revenue} = (550 + 25x)(96 - 3x)
\]

### Step 3: Set the Revenue Equal to the Desired Amount
The firm wants a total revenue of $54,600 per month, so we set up the equation:

\[
(550 + 25x)(96 - 3x) = 54600
\]

### Step 4: Solve the Equation
Now, expand and simplify the equation:

\[
(550 \times 96) + (550 \times -3x) + (25x \times 96) + (25x \times -3x) = 54600
\]

\[
52800 - 1650x + 2400x - 75x^2 = 54600
\]

Simplify further:

\[
-75x^2 + 750x + 52800 = 54600
\]

Now, move all terms to one side:

\[
-75x^2 + 750x + 52800 - 54600 = 0
\]

\[
-75x^2 + 750x - 1800 = 0
\]

### Step 5: Solve the Quadratic Equation
To make calculations simpler, divide the entire equation by -75:

\[
x^2 - 10x + 24 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

\[
x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(24)}}{2(1)}
\]

\[
x = \frac{10 \pm \sqrt{100 - 96}}{2}
\]

\[
x = \frac{10 \pm \sqrt{4}}{2}
\]

\[
x = \frac{10 \pm 2}{2}
\]

Thus, $ x = \frac{10 + 2}{2} = 6 $ or $ x = \frac{10 - 2}{2} = 4 $.

### Step 6: Find the Rent for Each Apartment
Now that we have $ x = 6 $ or $ x = 4 $, let's calculate the rent for each apartment in both cases:

- If $ x = 6 $, the rent per apartment is:
  \[
  550 + 25(6) = 550 + 150 = 700
  \]

- If $ x = 4 $, the rent per apartment is:
  \[
  550 + 25(4) = 550 + 100 = 650
  \]

### Step 7: Check Which Solution Meets the Desired Revenue
- For $ x = 6 $, the number of rented apartments is:
  \[
  96 - 3(6) = 96 - 18 = 78
  \]
  The total revenue is:
  \[
  78 \times 700 = 54600
  \]
  This satisfies the firm's target revenue.

- For $ x = 4 $, the number of rented apartments is:
  \[
  96 - 3(4) = 96 - 12 = 84
  \]
  The total revenue is:
  \[
  84 \times 650 = 54600
  \]
  This also satisfies the firm's target revenue.

### Conclusion:
The rent per apartment can be **$650** or **$700** to meet the target revenue of $54,600.

Would you like more details, or do you have any questions?

### Related Questions:
1. What would happen if the number of vacancies per rent increase was reduced to two?
2. How does the target revenue change if the rent increase per apartment is reduced to $20?
3. Can you model the total revenue as a quadratic function of $ x $ and determine its maximum?
4. What would the rent per apartment be if the firm wanted $60,000 per month instead?
5. How would the solution change if the number of apartments started at 100 instead of 96?

### Tip:
When solving real-world problems like this, always double-check solutions by plugging them back into the original scenario to verify they meet all conditions.

A real estate firm owns an apartment complex which consists of 96 apartments. At $550 per month, every apartment can be rented. However, for each $25 increase per apartment, there will be three vacancies with no possibility of filling them. The firm wants $54,600 per month from rent. What amount should be charged for each apartment?

Discover how to use quadratic equations to solve a real-world revenue problem. This detailed example shows how a real estate firm can adjust apartment rents to meet a target revenue of $54,600 per month by balancing rent and vacancy rates. Suitable for Grades 9-11 algebra students.

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