Cost & Demand Function

[JoeJ]

I was wondering if anyone could help me on the following problem...

The manager of a 100-unit apartment complex knows from experience that all units will be occupied if the rent is $800 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?

I'm thinking something like this...

revenue = number of units rented * amount of rent per unit

so, revenue would be...

100 * $400 = $40,000

The manager is considering increasing the rent in increments of $5, but for each increment the number of units rented decreases by 1.

Suppose the decision is to increase the rent by x increments. The number of units rented would then be 100 - x and the rent per unit would be $400 + $5x. So, the revenue would be

(100 - x)($400 + $5x)


Can anyone help on this?

 

[soroban]

Hello, JoeJ!

The manager of a 100-unit apartment complex knows from experience
that all units will be occupied if the rent is $800 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?

Youre approach is absolutely correct . . . and the right formula.
. . But you used $400 rent and $5 increases. (?)

The revenue function is: . R(x) . = . (100 - x)(800 + 10x) . = . 80,000 + 200x - 10x²

If you're in a Calculus course, take the derivative, equate to zero, and solve.

Otherwise, note that the revenue function is a quadratic.
. . Its graph is a down-opening parabola, so its vertex is a maximum point.

The formula for the vertex is: . x .= .-b/2a

Our quadratic has: . a = -10, b = 200, c = 80,000.

The vertex is: . x .= .-200/2(-10) .= .10

Therefore, he should raise the rent ten increments (to $900).

He will lose 10 tenants, but his revenue will be at a maximum:
. . . . . . 90 x $900 .= .$81,000

 

[JoeJ]

Ok, when I take the derivative of R(x), I get 200 - 20x, then solving 200-20x = 0, I then get x=10, which I assume is the same as getting the vertex? So I then take $10 * 10, to get $100, and add that to the $800, to get the $900, and 100-10=90, so then the 90 * $900 = $81,000. Just wondering if that's the approach to take by taking the derivative.