Let r be the monthly rent per unit in an apartment building with 100 units. All the units are rented at r= $900. One unit becomes vacant for each $10 increase in rent. Maintenance cost are $100 for occupied units and nothing for unoccupied units..... so A) show that the number of units rented is n= 190 - (r/10) for 900 is less than or equal to r and 1900 is greater than or equal to r B) net cash intake= revenue - maintenance costs. determine the rent r that maximizes net cash intake Note: the answer to part A is 0 is less than or equal to r and 100 is greater than or equal to r .. i just need help with part B
A)
n=floor(190-(r/10)) for 900≤r≤1900
Since the number of units has to be integral, and the question says vacancy increases for an increase of $10.
900≤r≤1900 can also be expressed as an interval [900,1900].
B)
Revenue for each unit = r-100
number of units = n = floor(190-(r/10))
Profit, P(r)
=Revenue - maintenance
= (190-(r/10))(r-100)
= (r^2-2000*r+190000)/10
= (2000r-r²-190000)/10
For a maximum profit, set marginal profit to zero
P'(r) = 200-2r/10 = 0
r=1000
if r=1000 is at the maximum, P"(r)<0
P"(r) = -2/10, therefore r=1000 is a maximum.
Also check that n is an integral number:
n=(190-(r/10))=190-1000/10=90
Profit = P(1000)=$81,000
B) To determine the rent r that maximizes net cash intake (revenue - maintenance costs), we need to find the value of r that yields the highest net cash intake.
The revenue is given by the formula: revenue = number of units rented * rent per unit.
In this case, the number of units rented is represented by n = 190 - (r/10) as derived in part A.
So, the revenue can be expressed as: revenue = (190 - (r/10)) * r
The maintenance costs remain constant at $100 per occupied unit.
The net cash intake is calculated by subtracting the maintenance costs from the revenue:
net cash intake = revenue - maintenance costs
Plugging in the revenue formula, we get:
net cash intake = (190 - (r/10)) * r - 100 * (190 - (r/10))
Now, to find the value of r that maximizes net cash intake, we can take the derivative of the net cash intake formula with respect to r and set it equal to zero to find the critical point:
d(net cash intake)/dr = 0
Simplifying the derivative and setting it to zero, we get:
190 - (r/5) - 100/10 = 0
Multiplying through by 10 to get rid of the denominators:
1900 - 2r - 100 = 0
1900 - 100 = 2r
1800 = 2r
r = 900
Therefore, the rent r that maximizes net cash intake is $900.
To determine the rent, r, that maximizes net cash intake, we need to find the point where the revenue is maximized and the maintenance costs are minimized. The revenue is given by the formula:
Revenue = number of units rented * monthly rent per unit
And the maintenance costs are given by the formula:
Maintenance costs = number of occupied units * cost per occupied unit
Given that we have 100 units, let's determine the number of occupied units based on the rent, r:
number of occupied units = 190 - (r/10)
Substituting this into the maintenance costs formula, we have:
Maintenance costs = (190 - (r/10)) * 100
Now, let's substitute the number of occupied units and the monthly rent per unit into the revenue formula:
Revenue = [(190 - (r/10)) * r]
The net cash intake is the revenue minus the maintenance costs:
Net cash intake = Revenue - Maintenance costs
Net cash intake = [(190 - (r/10)) * r] - [(190 - (r/10)) * 100]
Simplifying this equation, we get:
Net cash intake = (190 - (r/10)) * (r - 100)
To find the value of r that maximizes the net cash intake, we can take the derivative of the net cash intake with respect to r and set it equal to zero:
d(Net cash intake)/dr = 0
Differentiating the equation (190 - (r/10)) * (r - 100) with respect to r, we get:
(-1/10) * (r - 100) + (190 - (r/10))
Expanding and rearranging terms, we have:
(r/10) - 10 - (r/10) + 190 = 0
This simplifies to:
(1/10) * r + 180 = 0
Solving for r, we find:
r = -1800
However, since rent cannot be negative, this solution is not valid.
Therefore, there is no value of r that maximizes net cash intake within the given constraints.
To determine the rent value that maximizes the net cash intake, we need to calculate the revenue and maintenance costs as functions of the rent value.
Let's break it down:
1. Revenue:
Since there are 100 units in total, and each unit is rented at $900, the revenue generated per month can be calculated as the product of the rent per unit and the number of rented units. Let's represent the revenue function as R(r).
R(r) = r * (100 - (r/10))
2. Maintenance Costs:
The maintenance costs depend on the number of occupied units. Given that the maintenance cost for each occupied unit is $100, and the number of occupied units is (100 - (r/10)), the total maintenance cost per month can be calculated by multiplying these two values. Let's represent the maintenance cost function as M(r).
M(r) = 100 * (100 - (r/10))
3. Net Cash Intake:
The net cash intake can be calculated as the difference between the revenue and the maintenance costs. Let's represent the net cash intake function as N(r).
N(r) = R(r) - M(r)
To find the rent value that maximizes the net cash intake, we need to find the value of r that maximizes N(r).
To do this, we can take the derivative of N(r) with respect to r and set it equal to zero. Then solve for r. Let's denote the derivative of N(r) with N'(r).
N'(r) = d(N(r))/dr
Now, let's calculate the derivative:
N(r) = r * (100 - (r/10)) - 100 * (100 - (r/10))
To simplify the calculations, we can expand and rearrange this function:
N(r) = 900r - r^2/10 - 10000 + r^2/10
N(r) = 900r - 10000
Next, we differentiate N(r) with respect to r:
N'(r) = 900
Since N'(r) is a constant (900), it implies that N(r) is a linear function. Therefore, there is no maximum or minimum value for N(r). The net cash intake will increase linearly as the rent increases.
So, there is no specific rent value that maximizes the net cash intake. The higher the rent, the higher the net cash intake.