The manager of a 100-unit apartment complex knows from experience that all units will be occupied if the rent is $800 per month. A market survey suggests that, on average, one additional unit will remain vacant for each $10 increase in rent. What rent should the manager charge to maximize revenue?

Answer 1

To determine the rent that maximizes revenue, we first need to express the number of occupied units and the revenue as functions of the rent.

Let \( x \) be the number of $10 increases in rent. Therefore, the rent can be expressed as:

\[ R(x) = 800 + 10x \]

The number of occupied units decreases by 1 for each $10 increase in rent. Thus, the number of occupied units \( N \) can be expressed as:

\[ N(x) = 100 - x \]

The revenue \( P \) can be defined as the product of the rent per unit and the number of occupied units:

\[ P(x) = R(x) \times N(x) \]

Substituting the expressions for rent and number of occupied units into the revenue formula, we have:

\[ P(x) = (800 + 10x)(100 - x) \]

Next, we will expand this equation:

\[ P(x) = 800 \times 100 - 800x + 10 \times 100x - 10x^2 \] \[ P(x) = 80000 - 800x + 1000x - 10x^2 \] \[ P(x) = 80000 + 200x - 10x^2 \]

This is a quadratic equation in the standard form \( P(x) = -10x^2 + 200x + 80000 \).

To find the value of \( x \) that maximizes revenue, we use the vertex formula for a quadratic equation \( ax^2 + bx + c \). The x-coordinate of the vertex is given by:

\[ x = -\frac{b}{2a} \]

In our equation, \( a = -10 \) and \( b = 200 \). Thus:

\[ x = -\frac{200}{2 \times -10} = \frac{200}{20} = 10 \]

Now, substituting \( x = 10 \) back to find the optimal rent:

\[ R(10) = 800 + 10 \times 10 = 800 + 100 = 900 \]

Therefore, the rent that the manager should charge to maximize revenue is:

\[ \boxed{900} \text{ dollars per month.} \]