Setting up the equations of a word problem

[tarynt1]

1) The manager of a 100-unit apartment complex knows from experience that all units will be occupied if the rent is $400 per month. A market survey suggests that, on average, one additional unit will remain vacant for each $5 increase in rent. What rent should the manager charge to maximize revenue?

I know that revenue = (# of units rented)(rent), or if you substitute in random values, R = xc.

I need a primary and a secondary equation to set up this problem, and I'm not exactly sure how the revenue equation fits in. The issue is finding those equations; I know how to solve beyond that.

I have one more problem that I'm having trouble on:

2) A manufacturer has been selling 1000 televisions a week at $450 each. A market survey indicates that for each $10 rebate offered to the buyer, the number of sets sold will increase by 100 per week.

How large should the rebate be in order to maximize the revenue?

Same issue; I'm having difficulty setting up the primary and secondary equations.

 

[galactus]

Here's the apartment problem. The 2nd one is done the same way.

\(\displaystyle R(x)=(400+5x)(100-x)\)

Now, differentiate, set to 0 and solve for x.

 

[tarynt1]

I'm having some difficulty with the equation; after differentiating, setting to 0, and solving for x, I get an answer of x = 10. I'm not sure how to apply that to my answer if the question is asking what the rent should be in order to maximize revenue.

When I take the second derivative of the R(x) equation, I get R'' = -10, which would imply downward concavity and a minimum rather than a maximum. Am I doing something wrong?

 

[galactus]

You're misinterpreting your concavity. Concave down implies a maximum.

Remember, it concave DOWN. That means it's a 'hill', not a 'valley'.


Since x=10, use it in your function. 400+5(10)=450.

$450 is what the rent should be. That means they will rent 100-10=90 apartments.

(450)(90)=40500

 

[soroban]

Hello, tarynt1!

Since your difficulty is in the set-up,
giving you the revenue function was no help, right?

1) The manager of a 100-unit apartment complex knows from experience
that all units will be occupied if the rent is $400 per month.
On average, one additional unit will remain vacant for each $5 increase in rent.
What rent should the manager charge to maximize revenue?

I know that revenue = (# of units rented)(rent) . . . yes

I need a primary and a secondary equation to set up this problem . . . no

This problem does not require two equations.


You're correct: \(\displaystyle \:\begin{pmatrix}\text{total}\\\text{Revenue }\end{pmatrix}\;=\;\begin{pmatrix}\text{no. of}\\ \text{Units}\end{pmatrix}\:\times \:\begin{pmatrix}\text{monthly}\\ \text{Charge}\end{pmatrix}\;\;\Rightarrow\;\;R\:=\:U\cdot\,C\)

Presently: \(\displaystyle \:\text{Revenue}\;=\;\text{(100 units)}\,\times\,\text{(\$400)}\;=\;\$40,000\)

Let \(\displaystyle x\) = number of five-dollar increases in the monthly charge.
The $400 charge will be increased by \(\displaystyle 5x\) dollars.
. . Then: \(\displaystyle \,C\,=\,400\,+\,5x\)

For each \(\displaystyle x\), one unit will be vacant.
There will be only \(\displaystyle 100\,-\,x\) occupied units.
. . Then: \(\displaystyle \,U\,=\,100\,-\,x\)

Hence: \(\displaystyle \,R\;=\;(100\,-\,x)(400\,+\,5x)\)


You found that \(\displaystyle x\,=\,10.\) . . . What does that mean?

Well, what is \(\displaystyle x\)?
\(\displaystyle x\) is the number of five-dollar increases, remember?

So the $400 charge has ten five-dollar increases: \(\displaystyle \,400\,+\,10\cdot5\,=\,450\)
The new rent will be $450 per month.

As a result, 10 units will be vacant; only 90 will be occupied.

Since \(\displaystyle 90\,\times\,\$450 \:= \:\$40,500\), there is more revenue,
. . and this is the maximum possible revenue.