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    tocloz2me's Avatar
    tocloz2me Posts: 3, Reputation: 1
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    #1

    Nov 10, 2009, 11:12 AM
    System of 3 Equations and 3 Unknowns
    I have this problem and have no clue how to set up the three equations. I thought it would have been
    200x+400y+1700z=4100
    160x+300y+1940z=4200
    30x+200y+2040z=4300

    Here's the equation.. I placed a order totaling $2100 mil to purchase a new plane seating a total of 4,500 passengers. Some of the aircraft were 747s for $200 mil seating 400, 777's for $160 mil seating 300 and some A321's for $60 mil seating 200. At the time I was instrusted to by twice as many U.S. planes as foreign aircraft.
    How many of each did I order.

    I just need to find out how to set up the equation. The solving part I know how to do.
    donf's Avatar
    donf Posts: 5,679, Reputation: 582
    Printers & Electronics Expert
     
    #2

    Nov 10, 2009, 11:54 AM
    Forgive me for being obtuse, but your equasions don't make sense to me.

    Shouldn't you first determine how many planes are needed to accommodate 4500 passengers?

    Also, what is the top dollar you are allowed to spend?

    Are you to purchase one plane to accommodate all of the passengers or an assortment of planes?
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #3

    Nov 10, 2009, 12:13 PM
    If you let x=# of 747, y=# of 777, z=# of A321's

    The cost is: 200x+160y+60z=2100

    The number of passengers is 400x+300y+200z=4500

    A constraint is twice as many domestic as foreign: x+y=2z

    Solve the system.
    ArcSine's Avatar
    ArcSine Posts: 969, Reputation: 106
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    #4

    Nov 10, 2009, 12:13 PM
    Staying with your notation of x = # 747s; y = # 777s; and z = # Airbus units; then

    the cost constraint sets up as 200x + 160y + 60z = 2,100

    The seating requirement models as 400x + 300y + 200z = 4,500

    And finally the ratio requirement sets up as x + y = 2z
    ... which can be rendered as x + y - 2z = 0.
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #5

    Nov 10, 2009, 12:14 PM
    Wow, Arcsine, talk about thinking alike:)
    ArcSine's Avatar
    ArcSine Posts: 969, Reputation: 106
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    #6

    Nov 10, 2009, 12:18 PM
    Right down to the minute... :)
    tocloz2me's Avatar
    tocloz2me Posts: 3, Reputation: 1
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    #7

    Nov 10, 2009, 03:02 PM

    Ok thank you all for your help. The second part of the equation is even worse.

    The revised budge for the new aircraft is now $1810mil for 800 fewer total passenger seats. At this point in production Boeing charges $40 mil for each 747 aircraft cancelled and $35 million for each 777 aircraft cancelled. In similar fashion Airbus charges $20 million for each A321 aircraft cancelled.
    Given that all other criteria remain the same, how many of each typed of aircraft did you cancel?
    donf's Avatar
    donf Posts: 5,679, Reputation: 582
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    #8

    Nov 11, 2009, 04:00 PM

    I'm really curious, is this a real planning exercise or a theoretical class project.

    I ask this because if your goal is to use one plane capable of carrying 4500 passengers, crew and luggage, then you are basically trying to put wings on an Aircraft Carrier.

    I surely do not want to be under that beast as it rumbles down the runway.

    Remember, there is no airport runway capable of carrying that much weight without crumbling.

    Simple math says it would take 11 plus 747s. 11 747s would carry a passenger load of 4400 passengers.
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #9

    Nov 11, 2009, 04:31 PM
    I believe the problem is poorly stated. All f tha planes will carry a total of 4500 passengers. Upon solving the system in the first problem, there are 15 planes. 5 of each.
    tocloz2me's Avatar
    tocloz2me Posts: 3, Reputation: 1
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    #10

    Nov 13, 2009, 09:32 AM

    It is a theoretical class problem that is ridiculous.

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