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#1 2013-12-05 09:11:49

mom
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Finding vertexes

I have the following problem. I do not know why I am stuck but I am can someone please help?

1.    An airline company wants to fly 1400 members of a ski club to Colorado.  The airline owns two types of planes. Type A can carry 50 passengers, requires 3 flight attendants, and costs $14,000 for the trip.  Type B can carry 300 passengers, requires 4 flight attendants, and costs $90,000 for the trip.  If the airline must use at least as many type A planes as type B and has available only 42 flight attendants, how many of each type should be used to minimize the cost for the trip.

For the constraints I got:

50x + 300y > 1400
3x + 4y < 42
x > y
x > 0, y > 0

I got standard form to be this:

y > -1/6x + 14/3
y < -3/4x + 21/2
y < x
x > 0, y > 0

With only two standard real constraints, how do I get three vertexes? Also I have big fractions which creates a challenge. Am I missing something?
Please help.

#2 2013-12-05 12:25:42

mom
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Re: Finding vertexes

I couldn't figure out how to find the vertexes so I tried a different approach. I think I got it.

Step 1
Row Operation    X           Y          U    V    W    M
Type A         -50          -300           1    0    0    -1400
Type B           3           4           0    1    0    42
                 14000      90000      0    0    1    0

Pivot on -50


Row Operation    X    Y          U    V    W    M
Type A            1    6        -1/50    0    0    28
Type B           0    -14         3/50    1    0    -42
                   0    6000      280    0    1    0

Pivot on -14

Row Operation    X    Y    U           V          W    M
Type A            1    0    1/175    3/7           0    10
Type B            0    1    -3/700    -1/14       0    3
                    0    0    2410/7    3000/7     1    -410000



10x+3y
To minimize the cost of the trip, the airline should fly 10 type A planes and 3 type B planes.

Please advise

#3 2013-12-05 12:52:35

anonimnystefy
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Re: Finding vertexes

That solution is correct.

What you can do to get it is get the intersections of each two lines. For each intersection you plug in the coordinates into the expression 14000x+90000y and see which of the three gets you the lowest value.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#4 2013-12-05 19:17:53

bob bundy
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Re: Finding vertexes

hi mom,

There are three constraints and I'm getting a triangular solution space.  The graph below has the minimum cost line shown in red.

The way I get this, is to plot any cost line          (eg.             1.4x + 9y + C      and then consider which parallel line to this will give the minimum.  For linear constraints this will always be a vertex and you can see from the graph which one.

The graph shows (10,3) as the minimum.

Hope that helps.

Bob

btw The inequalities are all this type ≤ ≥ rather than < >


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