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Max Profit

mathkid3

New member
Nov 16, 2012
23
A real Estate office handles a 50-unit apt. complex. When the rent is \$580/mo. all units are occupied.

For each \$40 increase in rent, however, an avg of one unit becomes vacant. Each occupied unit requires an avg of \$45 per month for service and repairs.

What rent should be charged to obtain a maximum profit?



I need help setting up the problem

TY!
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The dollar sign character is used for rendering LaTeX here, that is why your post looks that way.

We need to find a function which describes the number of units occupied as a function of the rent charged. Let the rent be $\displaystyle 580+40x$

We also know that for each increase of 1 in x, we get a decrease of 1 in U, the number of units occupied. Hence:

$\displaystyle \frac{dU}{dx}=-1$ where $\displaystyle U(0)=50$

Can you now solve this IVP to find $\displaystyle U(x)$?

You may choose to simply use the point-slope formula to find this linear function.
 

mathkid3

New member
Nov 16, 2012
23
I may need a little further assistance in what to differentiate

first one I have seen like this
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
First, we want to find the number of units occupied as a function of the number of 40 dollar increases in rent.

So, either integrate the ODE I gave, or more simply use the point-slope formula to find the function. You know a point (0,50) and the slope m = -1.
 

mathkid3

New member
Nov 16, 2012
23
im getting y=-x+50


now what coach ? :)
 

MarkFL

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Staff member
Feb 24, 2012
13,775
Okay, so we now have:

$\displaystyle U(x)=50-x$

Now, we know profit is revenue minus cost. Can you state the profit as a function of U and x?

$\displaystyle P(U,x)=?$

Note: don't worry, we will get rid of U to get the profit as a function of one variable in the next step.
 

mathkid3

New member
Nov 16, 2012
23
Im sorry Mark I dont follow
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
For the revenue:

The money coming in is the number of units occupied times the rent charged per unit.

What is this product?

For the cost:

The total cost is the number of units occupied times the cost of upkeep per unit.

What is this product?
 

mathkid3

New member
Nov 16, 2012
23
P = 580(50)-50(45)


?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
That's only true if x = 0. Recall we defined the rent as:

$\displaystyle r(x)=580+40x$

And instead of using 50 for the units occupied, we want to use $\displaystyle U(x)$.

What do you get now?
 

mathkid3

New member
Nov 16, 2012
23
hey is the profile pic u?

P = (580+40z)(50)(45) ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
No, that is one of my intellectual heroes, Dr. Ed Witten, the only physicist to win the Fields Medal in mathematics, and arguably the world's foremost expert in M-theory.

Now, revenue is number of units occupied ($\displaystyle U(x)=50-x$) times rent charged ($\displaystyle r(x)=580+40x$), so we have the revenue:

$\displaystyle R(x)=(50-x)(580+40x)$

And cost is units occupied ($\displaystyle U(x)=50-x$) times average cost per unit (45 dollars), so we have the cost function:

$\displaystyle C(x)=45(50-x)$

So, what is the profit function $\displaystyle P(x)$?
 

mathkid3

New member
Nov 16, 2012
23
P = (50-z)(580+40z)-45(50-z)


?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yep, now I would factor, expand, then optimize. You don't have to expand, you could use the product rule to differentiate. Your choice.

edit: What kind of number do we require x to be?
 
Last edited:

mathkid3

New member
Nov 16, 2012
23
P ' = -40z^2 + 1465z + 26750


?

am I done?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
That's not the derivative, that's the expanded profit function. Now differentiate and equate to zero...but be careful as z must be what kind of number? Hint: look at the function defining the number of units occupied.
 

mathkid3

New member
Nov 16, 2012
23
wait is Z = -1 ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
No, but how did you determine that value?
 

mathkid3

New member
Nov 16, 2012
23
I was just thinking by the way u said it that z = -1


is it that z must be positive ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
First, think about the fact that U must be a non-negative integer in [0,50], so what kind of number must z be and what is the feasible domain?
 

mathkid3

New member
Nov 16, 2012
23
Mark,

sorry I fell asleep on couch last night while chatting with you on the forum. I had to get coffee in me to start to re what we were doing last night :)

ok I have questions. #1 how did I get so lost in the problem from the start?? I think it was from asking your help you introduced so variables in the equation that didn't match with the textbook. Prob just overwhelmed me from the start.

I left u with giving you the expanded profit function instead of taking the derV and setting to zero

let me try again...

P ' (x) = -80x + 14652

solving the org P function for x I get x =
-1387.75 and x = -1452.25 This is for defining the feasable domain (Is that correct Mark?)

anyway... I wish we could start from scratch in this problem and u could walk me through it again. I do not like it when I get lost in a problem. I need to understand why we set it up the way we did and why I am struggling with a concept(s)

Thanks bud!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Okay, we are given:

A real Estate office handles a 50-unit apt. complex. When the rent is \$580/mo. all units are occupied.

For each \$40 increase in rent, however, an avg of one unit becomes vacant. Each occupied unit requires an avg of \$45 per month for service and repairs.

What rent should be charged to obtain a maximum profit?

Here's the way I would look at it:

Let z represent the number of \$40 increases in rent, that is, the monthly rent per unit is:

$\displaystyle r(z)=580+40z$

We are told that for each increase of 1 in z, there is a decrease of 1 in the number of units occupied, which we can state as:

$\displaystyle u(z)=50-z$

The monthly revenue is the total amount of rent collected, which is equal to the number of units occupied times the monthly rent per unit:

$\displaystyle R(z)=u(z)\cdot r(z)=(50-z)(580+40z)=20(50-z)(29+2z)$

The total monthly cost is the average cost per unit times the number of units occupied:

$\displaystyle C(z)=45u(z)=45(50-z)$

Now, the monthly profit is the monthly revenue minus the monthly cost:

$\displaystyle P(z)=R(z)-C(z)=20(50-z)(29+2z)-45(50-z)=5(50-z)(4(29+2z)-9)=5(50-z)(107+8z)$

Now, I am going to let you use differentiation, while I am going to use the fact that we have a parabolic profit function opening downward whose axis of symmetry will be midway between the roots, and thus the vertex (maximum point) will be on this axis.

The roots are:

$\displaystyle 107+8z=0\,\therefore\,z=-\frac{107}{8}$

$\displaystyle 50-z=0\,\therefore\,z=50$

The axis of symmetry is then the line (using the mid-point formula):

$\displaystyle z=\frac{-\frac{107}{8}+50}{2}=\frac{293}{16}$

See if you can get this critical value using the calculus. After this, we'll deal with the fact that we should treat z as a discrete variable rather than a continuous one.