# Help Me Guys

• Feb 22nd 2006, 04:16 PM
usm_67
Help Me Guys
In a motal there are 30 rooms. Motal owner charges 20$per day. The owner is thinking to increase the rent by 1 dollar but if he increses the price by 1 dollar one room will be vacant(emptied). Daily service and maintanance to clean the room will cost c=x^2-60x+900 find the break even point and find out how many rooms will be vacant and how many rooms will be filled. • Feb 23rd 2006, 02:32 AM ticbol Quote: Originally Posted by usm_67 In a motal there are 30 rooms. Motal owner charges 20$ per day. The owner is thinking to increase the rent by 1 dollar but if he increses the price by 1 dollar one room will be vacant(emptied).
Daily service and maintanance to clean the room will cost c=x^2-60x+900 find the break even point and find out how many rooms will be vacant and how many rooms will be filled.

The problem as posted is not very clear. I think it is your interpretation of the original problem as shown in a book or somewhere else.
----What is x? Is that number of rooms filled? So if price of rent per room remains at $20, then x=30? If price is$21, x=29?
----To clean the room? To clean one room only? Or should that be "to clean the rooms, the rooms that are filled? The x-rooms?

Let me solve the problem based on the following, per day:
---x = number of rooms filled.
---C = x^2 -60x +900 is cost for all filled rooms, for the x-rooms.

Here is one way.

Break even. Meaning, revenue = cost. No loss, no profit.

Revenue = x*(rent of a room) ---***

Rent of a room = $20 +$1*(30-x) = 20 +30 -x = (50-x) dollars.
where (30-x) is the number of rooms emptied. [See what the "x" did, too much confusion.]
See what $(50-x) means: If x=30, or no empty room, (50-x)=(50-30)=$20 per room.
If x=29, (50-x)=(50-29)=$21 per room. It is what the problem say. So, Revenue, R = x*(50-x) = (50x -x^2) dollars. Equate that to the C = x^2 -60x +900 for break even, x^2 -60x +900 = 50x -x^2 x^2 -60x +900 -50x +x^2 = 0 2x^2 -110x +900 = 0 x^2 -55x +450 = 0 Use the Quadratic Formula, x = {-(-55) +,-sqrt[(55)^2 -4(1)(450)]} /(2*1) x = (55 +,-35)/2 x = 45 or 10 rooms filled. Of course, 45 rooms filled is impossible---the motel has max 30 rooms only. Hence, x = 10 rooms filled. Therefore, for break even, 10 rooms must be filled up, or 20 rooms must be vacant. -----answer. • Feb 23rd 2006, 02:43 AM CaptainBlack Quote: Originally Posted by ticbol The problem as posted is not very clear. I think it is your interpretation of the original problem as shown in a book or somewhere else. ----What is x? Is that number of rooms filled? So if price of rent per room remains at$20, then x=30? If price is $21, x=29? ----To clean the room? To clean one room only? Or should that be "to clean the rooms, the rooms that are filled? The x-rooms? Let me solve the problem based on the following, per day: ---x = number of rooms filled. ---C = x^2 -60x +900 is cost for all filled rooms, for the x-rooms. Here is one way. Break even. Meaning, revenue = cost. No loss, no profit. Revenue = x*(rent of a room) ---*** Rent of a room =$20 +$1*(30-x) = 20 +30 -x = (50-x) dollars. I think the intended meaning is that the rent (per room) is increased to$21,
not that the rent for a room be made dependant on the number of occupied
rooms.

RonL
• Feb 23rd 2006, 02:52 AM
ticbol
Quote:

Originally Posted by CaptainBlack
I think the intended meaning is that the rent (per room) is increased to $21, not that the rent for a room be made dependant on the number of occupied rooms. RonL So if there are 29 rooms filled, the rent is$21?
There are 28 rooms filled, the rent is $21 also? There are 15 rooms filled, the rent is$21 still?
Etc...
• Feb 23rd 2006, 10:02 AM
CaptainBlack
Quote:

Originally Posted by ticbol
So if there are 29 rooms filled, the rent is $21? There are 28 rooms filled, the rent is$21 also?
There are 15 rooms filled, the rent is $21 still? Etc... The rent when 28 rooms are filled is$21 per room for a total revenue
of $21 x 28 =$588.

RonL
• Feb 23rd 2006, 10:41 AM
ticbol
Quote:

Originally Posted by CaptainBlack
The rent when 28 rooms are filled is $21 per room for a total revenue of$21 x 28 = $588. RonL Total revenue when 29 rooms are filled = 29*21 =$609?
Total revenue when 20 rooms are filled = 20*21 = $420? Total revenue when 10 rooms are filled = 10*21 =$210?
And when is the break even?
• Feb 23rd 2006, 11:34 AM
CaptainBlack
Quote:

Originally Posted by ticbol
Total revenue when 29 rooms are filled = 29*21 = $609? Total revenue when 20 rooms are filled = 20*21 =$420?
Total revenue when 10 rooms are filled = 10*21 = $210? And when is the break even? Costs = x^2 -60x +900 When revenue equals cost which is somewhere between 13 and 14 rooms. RonL • Feb 23rd 2006, 11:46 AM ticbol Quote: Originally Posted by CaptainBlack Costs = x^2 -60x +900 When revenue equals cost which is somewhere between 13 and 14 rooms. RonL So, using C = x^2 -60x +900, C(13) = 13^2 -60*13 +900 =$289
C(14) = $256 R(13) = 13*21 =$273
R(14) = 14*21 = $294 Nothing even matches. And if break even is between 13 rooms and 14 rooms occupied---cannot occupy part of a room and pay part of the rent for that partial room. • Feb 24th 2006, 12:04 PM earboth Quote: Originally Posted by usm_67 In a motal there are 30 rooms. Motal owner charges 20$ per day. The owner is thinking to increase the rent by 1 dollar but if he increses the price by 1 dollar one room will be vacant(emptied).
Daily service and maintanance to clean the room will cost c=x^2-60x+900 find the break even point and find out how many rooms will be vacant and how many rooms will be filled.

Hello,

if I understand you right, you describe the following situation:

There 30 rooms for rent.
If the price increases by 1 \$ then 1 room will not be rented.
The function c describes how much it will cost the owner to clean the empty rooms.

Let x be the emty, not rented rooms, then you get:

Rented rooms: (30-x)
Daily revenue: (30-x)(20+x)
Daily costs: ((30-x)-30)^2=x^2

If revenues have the same value as the cost, you've got the even point(?):
(here I'm not quite certain, if I understood the problem properly!)
$(30-x)(20+x)=x^2$

Solve this equation for x and you'llget x= -15 or x=20. (-15 not rented rooms are not very realistic)

Rented rooms - daily income
$30 \to 600$
$29 \to 608 (=29*21-(1)^2)$
$28 \to 612$
$27 \to 612$
$26 \to 608$
...
$10 \to 0$ which isn't so much.

I've attached a graph to demonstrate what I've calculated.

Greetings

EB