The manager of a large apartment complex knows from experience that 100 units will be occupied if the rent is 425 dollars per month. A market survey suggests that, on average, one additional unit will remain vacant for each 9 dollar increase in rent. Similarly, one additional unit will be occupied for each 9 dollar decrease in rent.
Let the rent on an apartment be x dollars per month, and let N be the number of apartments rented each month, and let R be the revenue (the gross income) brought in each month by the apartment manager.
N(x)=______apartments
R(x)=______dollars
What rent should the manager charge to maximize revenue?
N(x) = 100-(x-425)/9
R(x) = x*N(x) = x(100-(x-425)/9) = 1/9 (1325x-x^2)
dR/dx = 1/9 (1325-2x)
dR/dx=0 at x=1325/2 = $662.50
To find the rent that will maximize revenue, let's first define the variables:
N(x) represents the number of apartments rented each month.
R(x) represents the revenue (gross income) brought in each month.
Given the information provided, we know that when the rent is $425 per month, 100 units are occupied. Based on the market survey, for every $9 increase in rent, one additional unit will remain vacant, and for every $9 decrease in rent, one additional unit will be occupied.
Let's start by determining the equation for N(x), the number of apartments rented each month.
If the rent is $425 per month, 100 units are occupied. Therefore, we have:
N(425) = 100
According to the market survey, for every $9 increase in rent, one additional unit will remain vacant. This means that for each $x increase in rent, we have (x/9) additional vacancies. Therefore, the equation for N(x) would be:
N(x) = 100 - (x - 425)/9
Now, let's determine the equation for R(x), the gross income brought in each month.
To calculate the revenue, we multiply the rent (x) by the number of apartments rented (N(x)). Hence, the equation for R(x) would be:
R(x) = x * N(x)
= x * (100 - (x - 425)/9)
To find the rent that maximizes revenue, we need to find the maximum value of R(x). We can do this by finding the critical points or the turning points of R(x) using calculus. By taking the derivative of R(x) with respect to x and setting it equal to zero, we can find the value of x that maximizes R(x).
Let's differentiate R(x) with respect to x:
R'(x) = (100 - (x - 425)/9) + x * (-(1/9))
= 100 - (x - 425)/9 - x/9
= 100 - x/9 + 425/9 - x/9
= 525/9 - x/9
Setting R'(x) equal to zero and solving for x:
525/9 - x/9 = 0
525 - x = 0
x = 525
Therefore, the value of x that maximizes revenue is $525 per month.
To calculate N(x) and R(x) at x = 525, substitute x = 525 into the respective equations:
N(525) = 100 - (525 - 425)/9
= 100 - 100/9
= 100 - 11.11
= 88.89 (approximately)
R(525) = 525 * N(525)
= 525 * 88.89
= 46,428.75 (approximately)
Therefore, to maximize revenue, the manager should charge a rent of $525 per month. At this rent, approximately 88.89 apartments will be rented each month, and the gross income will be approximately $46,428.75.
To find the rent that will maximize revenue, we need to determine the number of apartments rented each month (N(x)) and the revenue brought in each month (R(x)) as functions of the rent amount (x).
We are given that at $425 rent, 100 units will be occupied, so we can start by writing the equation for N(x) in terms of the rent:
N(x) = 100 + (x - 425)/9
This equation represents the number of apartments occupied, where (x - 425)/9 represents the additional units occupied for each $9 decrease in rent.
To find the revenue function (R(x)), we multiply the number of apartments occupied (N(x)) by the rent amount (x):
R(x) = N(x) * x
Substituting the equation for N(x) into R(x), we get:
R(x) = (100 + (x - 425)/9) * x
Now, to find the rent that maximizes revenue, we need to find the value of x that maximizes the revenue function R(x). This can be done by taking the derivative of R(x) with respect to x and setting it equal to zero:
dR(x)/dx = 0
Differentiating R(x) with respect to x:
dR(x)/dx = (100 + (x - 425)/9) + x/9
Setting dR(x)/dx equal to zero:
0 = (100 + (x - 425)/9) + x/9
Simplifying the equation:
0 = (9(100 + (x - 425)) + x) / 9
0 = 9(100 + (x - 425)) + x
Now, solve for x:
0 = 900 + 9x - 3825 + x
0 = 10x - 2925
10x = 2925
x = 292.5
Therefore, the rent that the manager should charge to maximize revenue is $292.50 per month.