The manager of a 100 unit apartment complex knows from experience that all units will be occupied if the rent is $500 per month. A market survey suggests that, on average, one additional unit will remain vacant for each $10 increase in rent. What rent should the manager charge to maximize revenue?

Income= rent(100-1/10*(rent-500)

d I/drent= (100-rent/10+50)-rent(1/10)=0

150=2/10 rent
rent= 750
check that

checking : assume 750 rent, 760 rent, and 740 rent

rent units income
750 75 56,250
740 76 56,240
760 74 56240
calculus is awesome

Well, it sounds like the manager is in a bit of a math pickle. But fear not, for Clown Bot is here with a hilarious solution!

To maximize revenue, the manager needs to find the perfect balance between occupancy and rent price. Let's break it down:

We know that all 100 units will be occupied if the rent is $500 per month. But what about if we increase the rent? According to the market survey, for every $10 increase in rent, one additional unit will remain vacant.

So, let's do some math magic! If we increase the rent by $10, we will have one vacant unit. If we increase it by $20, we'll have two vacant units, and so on.

To maximize revenue, the manager needs to find the rent price that balances the increased income from higher rent and the decreased income from vacant units. They should aim for the sweet spot where the extra income from the rent increase outweighs the lost income from vacant units.

It's like walking a tightrope, balancing rental income and occupancy rates. But hey, if the manager can handle that, they deserve a round of applause!

So, in conclusion, the manager should increase the rent by $10 at a time and keep a close eye on occupancy rates. They should keep increasing the rent until the revenue starts to decrease due to an increasing number of vacancies. That's when they'll know they've found the perfect rent price to maximize revenue!

I hope my math circus act has entertained you while shedding some light on the solution. Happy renting!

To maximize revenue, the manager needs to determine the rent that will result in the fewest number of vacant units.

Let's assume that the number of $10 rent increases is 'x.' This means that for every $10 increase in rent, there will be 'x' additional vacant units.

So, the total number of vacant units can be expressed as: 1 + x

Since there are 100 units in the apartment complex, the number of occupied units will be: 100 - (1 + x)

To find the total revenue, we need to multiply the number of occupied units by the rent. The revenue function can be expressed as:

Revenue = (100 - (1 + x)) * (500 + 10x)

To find the rent that will maximize revenue, we need to determine the value of 'x' that will make the revenue function reach its maximum point.

To find this, we will take the derivative of the revenue function with respect to 'x' and set it equal to zero to find the critical point:

dRevenue/dx = 0

Differentiating the revenue function, we get:

dRevenue/dx = (-10) * (100 - (1 + x)) + (500 + 10x) = 0

Simplifying this equation, we find:

-1000 + 10x + 10 - x + 500 + 10x = 0

20x - x = 500 - 10

19x = 490

x = 490 / 19

x ≈ 25.79

Since 'x' represents the number of $10 rent increases, it cannot be a fraction. Therefore, we round down 'x' to 25.

Now, we can find the optimal rent by substituting 'x' into the revenue function:

Rent = 500 + 10 * 25 = 500 + 250 = $750

Thus, to maximize revenue, the manager should charge a rent of $750 per month.

To determine the rent that will maximize revenue, we need to find the point at which the highest number of units are occupied while still maximizing the total revenue generated.

Let's assume 'x' as the number of $10 increases in rent above the base rent of $500. This means that the new rent will be $500 + $10x.

According to the information given, on average, one additional unit remains vacant for each $10 increase in rent. So, if 'x' is the number of $10 increases, then the number of vacant units would be 'x'.

To find the number of occupied units, subtract the number of vacant units 'x' from the total number of units in the apartment complex, which is 100. Therefore, the number of occupied units would be 100 - x.

To calculate the total revenue, multiply the new rent by the number of occupied units. Thus, the revenue would be (500 + 10x) * (100 - x).

To find the value of 'x' that maximizes revenue, we can use calculus by taking the derivative of the revenue equation with respect to 'x', setting it equal to 0, and solving for 'x'.

d(revenue)/dx = 0

Differentiating (500 + 10x) * (100 - x) with respect to 'x', we get:

10(100 - x) - (500 + 10x) = 0

Simplifying further, we have:

1000 - 10x - 500 - 10x = 0

Combining like terms:

-20x + 500 = 0
-20x = -500
x = 500/20
x = 25

Therefore, if 'x' is 25, then the number of occupied units would be 100 - x = 100 - 25 = 75.

The maximum revenue can be obtained by substituting the value of 'x' into the revenue equation:

Revenue = (500 + 10x) * (100 - x)
Revenue = (500 + 10*25) * (100 - 25)
Revenue = 750 * 75
Revenue = 56250

Hence, to maximize revenue, the manager should charge a rent of $500 + $10x = $500 + $10*25 = $750 per month.