All units in a 100-unit apartment building are rented out when the monthly rent is set at r=$580/month. Suppose that one unit becomes vacant with each $10 increase in rent and that each occupied unit costs $80/month in maintenance. Which rent r maximizes monthly profit?
Question
All units in a 100-unit apartment building are rented out when the monthly rent is set at r=10 increase in rent and that each occupied unit costs $80/month in maintenance. Which rent r maximizes monthly profit?
Solution
To solve this problem, we first need to define the variables and the profit function.
Let's denote:
- r as the rent for each apartment,
- x as the number of $10 increases in rent,
- P as the profit.
We know that the initial rent r is 10 increase in rent results in one vacant unit. Therefore, the rent r can be expressed as r = 580 + 10x.
The number of rented apartments is initially 100 and decreases by 1 with each $10 increase in rent. Therefore, the number of rented apartments can be expressed as 100 - x.
The profit P is the total rent income minus the maintenance costs. The total rent income is the number of rented apartments times the rent for each apartment, and the maintenance cost is the number of rented apartments times $80. Therefore, the profit P can be expressed as P = (100 - x)(580 + 10x) - (100 - x)80.
Simplifying the profit function gives P = 500x - 10x^2 + 52000.
To find the rent r that maximizes the profit P, we need to find the maximum of the profit function. The maximum of a quadratic function ax^2 + bx + c occurs at x = -b/2a. In this case, a = -10 and b = 500, so the maximum occurs at x = -500/(2*(-10)) = 25.
Substituting x = 25 into the equation for r gives r = 580 + 10*25 = $830.
Therefore, the rent that maximizes monthly profit is $830.
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