The manager of a large apartment complex knows from experience that 90 units will be occupied if the rent is 444 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 6 dollar increase in rent. Similarly, one additional unit will be occupied for each 6 dollar decrease in rent. What rent should the manager charge to maximize revenue?
You have any thoughts on any of these?
To determine the rent that will maximize revenue, we need to find the optimal balance between the number of occupied units and the amount of rent charged. To do this, we can create a revenue function and then find its maximum using calculus.
Let's analyze the problem step by step:
1. Let's denote the number of occupied units as x and the rent per month as y. We know that when the rent is $444 per month, 90 units are occupied. Therefore, we have the point (90, 444) on our revenue function.
2. According to the market survey, for every $6 increase in rent, one additional unit will remain vacant. This means that for every $6 decrease in rent, one additional unit will be occupied. So if the rent increases by $6, the number of occupied units will decrease by 1, and if the rent decreases by $6, the number of occupied units will increase by 1.
3. Based on the above information, we can define the relationship between the number of occupied units and the rent with the following equation:
x = 90 - (y - 444)/6
This equation represents the number of occupied units as a function of the rent.
4. To find the revenue function, we need to multiply the number of occupied units (x) by the rent (y):
R = x * y
Substituting the equation for x from step 3 into the revenue function, we get:
R = (90 - (y - 444)/6) * y
5. Now, the goal is to maximize the revenue function R. To do this, we need to differentiate the revenue function with respect to y and set the derivative equal to zero. Solving this equation will give us the value of y where the revenue is maximized.
dR/dy = 0
To differentiate the function, first expand and simplify the equation:
R = (90y - (y^2 - 444y)/6)
dR/dy = 90 - (2y - 444)/6
Setting dR/dy equal to zero and solving for y:
90 - (2y - 444)/6 = 0
90 = (2y - 444)/6
540 = 2y - 444
984 = 2y
y = 492
Therefore, the rent that will maximize revenue is $492 per month.
To summarize, the manager should charge a rent of $492 per month to maximize revenue. This is determined by analyzing the relationship between the number of occupied units and the rent, and finding the y-value where the revenue function reaches its maximum.