# Optimization/ Max Revenue.

• Dec 11th 2013, 12:58 PM
skg94
Optimization/ Max Revenue.
Suppose you own an apartment building containing 100 units. If you charge \$400 per
month for each unit, then all units can be rented out. For every \$20 increase in monthly rent, you
will lose one customer. What monthly rent should you charge to maximize your revenue? Show that
the corresponding revenue is an absolute maximum.

Attempt: (400+20x)(100-x)
40,000+1600x-20x^2
16000-40x
x=400

completing the square method:
(x+40)^2+400

both yield 400 as answer which would make max rent 8400 which is incorrect. IS my formula wrong? Where did i go wrong
• Dec 11th 2013, 01:59 PM
HallsofIvy
Re: Optimization/ Max Revenue.
Quote:

Originally Posted by skg94
Suppose you own an apartment building containing 100 units. If you charge \$400 per
month for each unit, then all units can be rented out. For every \$20 increase in monthly rent, you
will lose one customer. What monthly rent should you charge to maximize your revenue? Show that
the corresponding revenue is an absolute maximum.

Attempt: (400+20x)(100-x)

It is hard to say what you might be doing wrong when you don't say what you are doing!
I see "20x" so is "x" then is the number of "\$20 increases"? Then the revenue from each unit is 400+ 20x and the number of apartments rented is 100- x.
So, yes, the total revenue is (400+ 2x)(100- x)= 40000- 1600x+ x^2[/tex]

Quote:

40,000+1600x-20x^2
16000-40x
You have an extra "0" here. Although you do not say it, you have taken the derivative of the revenue function. That would be 1600- 40x, not 16000- 40x.

Quote:

x=400
Setting the derivative equal to 0, and solving for x gives x= 40, not 400. Maximum revenue comes when there are 40 "\$20 increases in monthly rent" which would mean each apartment rents for 400+ 20(40)= \$1200 so that you rent 100- 40= 60 apartments for total revenue 60(1200)= \$72000.

Quote:

completing the square method:
(x+40)^2+400
No, it doesn't. $(x+ 40)^2+ 400= x^2+ 80x+ 1600+ 400= x^2+ 80x+ 2000$.
Completing the square on $-(20x^2- 1600x+ 4000)= -20(x^2- 80x+ 2000)= -20(x^2- 80x+ 1600)+ 400= (x- 40)^2+ 400$.
The graph of that is a parabola which has vertex at x= 40, y= 400. But it is the "x= 40" that is the number of "\$20 increases", NOT the "y= 400".

both yield 400 as answer which would make max rent 8400 which is incorrect. IS my formula wrong? Where did i go wrong[/QUOTE]
No, both yield 40, not 400, as the answer which makes the rent on each apartment \$1200
• Dec 11th 2013, 06:01 PM
skg94
Re: Optimization/ Max Revenue.
omg, so i had it right just a simple mistake.

And for completing the square thank you for clearing that up