The manager of a large apartment complex knows from experience that 100 units will be occupied if the rent is 350 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 7 dollar increase in rent. Similarly, one additional unit will be occupied for each 7 dollar decrease in rent. What rent should the manager charge to maximize revenue?

I'll start you off ....

let the number of $7 increases be n
then number rented = 100 - n
rent charged = 350 + 7n

Revenue = (100-n)(350+7n)

expand, take the derivative, set that equal to zero and solve for n.

bah

25

To find the rent that will maximize revenue, we need to determine the number of occupied units at different rent levels.

Let's assume x represents the number of 7-dollar increments added or subtracted from the initial rent of $350. If x is positive, it means the rent is increased, and if x is negative, it means the rent is decreased.

Based on the information given, we know that for each 7-dollar increase in rent, one additional unit will remain vacant. Therefore, the equation representing the number of occupied units as a result of an increase in rent would be:

Occupied units = 100 - (x/7)

Similarly, for each 7-dollar decrease in rent, one additional unit will be occupied. So, the equation representing the number of occupied units as a result of a decrease in rent would be:

Occupied units = 100 + (x/7)

To maximize revenue, the manager needs to find the rent level where the number of occupied units (and thus the revenue) is the highest.

Revenue can be calculated by multiplying the number of occupied units by the rent:

Revenue = Rent * Occupied units

Substituting the equations for occupied units, we get:

Revenue = Rent * (100 - x/7) for an increase in rent
Revenue = Rent * (100 + x/7) for a decrease in rent

The goal is to find the value of x that maximizes revenue. We will analyze both cases separately to find the optimal rent:

1. For an increase in rent:
Revenue = Rent * (100 - x/7)
To find the value of x that maximizes revenue, we can take the derivative of the revenue equation with respect to x and set it equal to zero:

d(Revenue)/dx = 0
d(Rent * (100 - x/7))/dx = 0
100 - x/7 = 0
x = 700

So, when the rent increases by 700 dollars, the number of occupied units will be 0, resulting in the maximum revenue.

2. For a decrease in rent:
Revenue = Rent * (100 + x/7)
Similarly, we differentiate the revenue equation with respect to x and set it equal to zero:

d(Revenue)/dx = 0
d(Rent * (100 + x/7))/dx = 0
100 + x/7 = 0
x = -700

Therefore, when the rent decreases by 700 dollars, the number of occupied units will be 200, resulting in the maximum revenue.

To calculate the corresponding rents, we add or subtract the value of x from the initial rent:

For an increase in rent: $350 + (700/7) = $450
For a decrease in rent: $350 - (700/7) = $250

Hence, to maximize revenue, the manager should charge a rent of $450 per month.