The manager of a large apartment complex knows from experience that 80 units will be occupied if the rent is 460 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 10 dollar increase in rent. Similarly, one additional unit will be occupied for each 10 dollar decrease in rent. What rent should the manager charge to maximize revenue?
The manager of a large apartment complex knows from experience that 80 units will be occupied if the rent is 360 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 10 dollar increase in rent. Similarly, one additional unit will be occupied for each 10 dollar decrease in rent. What rent should the manager charge to maximize revenue?
To maximize revenue, the manager needs to find the rent that will maximize the number of occupied units.
Let's start by determining the relationship between the number of occupied units and the rent.
From the given information, we know:
- At $460 rent, there are 80 occupied units.
- For every $10 increase in rent, there will be one additional unit unoccupied.
- For every $10 decrease in rent, there will be one additional unit occupied.
To find the optimal rent, we need to analyze the impact of increasing or decreasing the rent on the number of occupied units.
First, let's examine the impact of a $10 increase in rent:
- For every $10 increase, one unit will become unoccupied.
- As a result, the number of occupied units will decrease by one.
Next, let's look at the impact of a $10 decrease in rent:
- For every $10 decrease, one unit will become occupied.
- As a result, the number of occupied units will increase by one.
Based on this information, we can conclude that there is a linear relationship between the number of occupied units and the rent, with a negative slope of -1 (for every $10 increase in rent, one unit becomes unoccupied).
To find the rent that maximizes revenue, we need to determine the rent at which the number of occupied units is highest.
Let's denote the rent as R and the number of occupied units as U.
We know that:
- At $460 rent, there are 80 occupied units (R = 460, U = 80).
From the linear relationship, we can write the equation as:
U = (-1/10)(R - 460) + 80
To find the rent that maximizes revenue, we need to find the maximum value of U. This occurs at the vertex of the quadratic equation.
The formula for the x-coordinate of the vertex of a quadratic equation in the form y = ax^2 + bx + c is given by:
x = -b / (2a)
In our case, the equation is in the form U = (-1/10)(R - 460) + 80, so a = -1/10, b = 460, and c = 80.
Substituting these values into the formula for the x-coordinate of the vertex, we get:
R_vertex = -b / (2a)
R_vertex = -460 / (2 * (-1/10))
R_vertex = -460 / (-1/5)
R_vertex = -460 * (-5)
R_vertex = 2300
The rent that maximizes revenue is $2300.
Therefore, the manager should charge $2300 per month to maximize revenue.
To find the rent that will maximize revenue, we need to determine the number of units that will be occupied at different rent prices and calculate the corresponding revenue for each scenario. Then, we can analyze the data to identify the rent value that yields the highest revenue. Let's break down the problem step-by-step:
1. Determine the number of units occupied at different rent prices:
- From the given information, we know that the manager expects 80 units to be occupied if the rent is $460 per month.
- According to the market survey, one additional unit will remain vacant for each $10 increase in rent. Similarly, one additional unit will be occupied for each $10 decrease in rent.
- Based on this information, we can form a linear relationship between the number of units occupied and the rent price:
- For each $10 increase in rent, the number of occupied units decreases by 1.
- For each $10 decrease in rent, the number of occupied units increases by 1.
2. Calculate the revenue for each scenario:
- Revenue is determined by multiplying the number of occupied units by the rent price.
- Let's calculate the revenue at different rent prices. We will assume that the number of units can not be negative, so we will set a lower limit for the number of units at zero.
Rent: $460
Number of Occupied Units: 80
Revenue: $460 * 80 = $36,800
Rent: $470
Number of Occupied Units: 79
Revenue: $470 * 79 = $37,130
Rent: $480
Number of Occupied Units: 78
Revenue: $480 * 78 = $37,440
...
Rent: $550
Number of Occupied Units: 72
Revenue: $550 * 72 = $39,600
3. Analyze the revenue data and identify the maximum value:
- By examining the revenue values calculated for different rent prices, we can observe that the revenue initially increases as the rent price increases, but then starts to decrease.
- We are interested in finding the rent value that yields the highest revenue.
- From the calculations above, the highest revenue value is reached at a rent price of $550 per month, where 72 units are occupied and the revenue is $39,600.
Therefore, to maximize revenue, the manager should charge a rent of $550 per month.
Let n be the number of $10 increases in rent. Total revenue = (# units)(rent)
=(80-n)(460+10n)
=36800 + 340n - 10n^2
has a max at 17
so, rent = 460-170 = 290