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  1. #1
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    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.
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  2. #2
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    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.
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  3. #3
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    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
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  4. #4
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    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...
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  5. #5
    Grand Panjandrum
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    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
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  6. #6
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    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?
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  7. #7
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    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
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  8. #8
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    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.
    Last edited by ticbol; February 23rd 2006 at 10:48 AM. Reason: for, not of
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  9. #9
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    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
    Attached Thumbnails Attached Thumbnails Help Me Guys-hotel.gif  
    Last edited by earboth; February 24th 2006 at 09:06 PM.
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