The manager of a large apartment complex knows from experience that 120 units will be occupied if the rent is 294 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?

To determine the rent that should be charged to maximize revenue, we need to find the point where the number of occupied units and the rent per unit will generate the highest total revenue.

Let's assume x represents the number of $7 decreases in rent from the initial $294 rent. Therefore, the new rent can be represented as 294 - 7x.

According to the market survey, for each $7 decrease in rent, one additional unit will be occupied. Therefore, the number of occupied units can be represented as 120 + x.

The revenue generated will be the product of the number of occupied units and the rent per unit. So, the revenue function can be written as:

Revenue = (294 - 7x) * (120 + x)

To maximize the revenue, we need to differentiate the revenue function with respect to x and set it equal to zero to find the critical points.

dRevenue/dx = 0

Simplifying this equation will give us:

(294 - 7x) * 1 + (120 + x) * -7 = 0

Simplify further:

294 - 7x - 840 - 7x = 0

Combine like terms:

-14x - 546 = 0

-14x = 546

x ≈ -39

Since we cannot have a negative number of $7 decreases in rent, we discard this solution.

Now, we need to evaluate the revenue at the endpoints of a reasonable range of x values. Since the manager wants to maximize revenue, we will check the revenue at x = 0 (no change in rent) and x = 120 (maximum decrease in rent).

Revenue1 = (294 - 7 * 0) * (120 + 0) = 294 * 120 = $35,280

Revenue2 = (294 - 7 * 120) * (120 + 120) = (294 - 840) * 240 = $14,160

Comparing the revenues, it is clear that Revenue1 is higher than Revenue2. Therefore, the maximum revenue will be generated when there is no change in rent.

Hence, the manager should charge $294 per month to maximize revenue.

To maximize revenue, the manager needs to find the rent that will result in the highest number of occupied units.

Let's break the problem down step by step:

Step 1: Determine the additional occupied units for each $7 decrease or increase in rent.

Based on the market survey information:
- One additional unit will be occupied for each $7 decrease in rent.
- One additional unit will remain vacant for each $7 increase in rent.

Step 2: Calculate the number of additional units for each $7 change in rent.

With a decrease of $7 in rent, one additional unit will be occupied.
With an increase of $7 in rent, one additional unit will remain vacant.

Step 3: Determine the change in units when the rent changes.

Since the manager already knows that 120 units will be occupied at $294 rent, we can use this information as a baseline.
- Decreasing the rent by $7 will result in one more unit being occupied. So, if the rent decreases by $7, the number of occupied units will be 120 + 1 = 121.
- Increasing the rent by $7 will result in one less unit being occupied. So, if the rent increases by $7, the number of occupied units will be 120 - 1 = 119.

Step 4: Find the optimal rent for maximizing revenue.

To maximize revenue, the manager needs to find the rent that yields the maximum number of occupied units.
To do this, compare the number of occupied units when the rent decreases by $7 with the number of occupied units when the rent increases by $7.

If the rent is decreased by $7, the number of occupied units is 121.
If the rent is increased by $7, the number of occupied units is 119.

Therefore, the manager should decrease the rent by $7 to maximize revenue.

The optimal rent would be $294 - $7 = $287 per month.