Need help here guys :S

romeroroma

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Hard word problem for me :S

Postby romeroroma » Fri May 04, 2012 6:41 pm
hey all. here is the word problem am having troubles in. they wana me to use gaus elimination method to solve it u now equations matrices and this stuff. and am rly bad at word problems so i wish for u all to help me please.

A chemical manufacturer wants to lease a fleet of 24 railroad tank cars with a combined carrying capacity of 420.000 gallons.. Tank cars with three different carrying capacities are available : 7.000 gallons, 14.000 gallons, and 28.000 gallons. How many each type of tank car should be leased ?.
he also said let X1 be the number of cars with a 7.000 gallon capacity, X2 be the number of cars with 14.000 gallon capacity. and X3 be the number of cars with a 28.000 gallon
 
I solved the system of equations and got 12 cars carrying 7000 gallons and 12 cars carrying 28000 gallons and no cars carrying 14000 gallons.

If there are other solutions how would I generate them?
 
Hard word problem for me :S

Postby romeroroma » Fri May 04, 2012 6:41 pm
hey all. here is the word problem am having troubles in. they wana me to use gaus elimination method to solve it u now equations matrices and this stuff. and am rly bad at word problems so i wish for u all to help me please.

A chemical manufacturer wants to lease a fleet of 24 railroad tank cars with a combined carrying capacity of 420.000 gallons.. Tank cars with three different carrying capacities are available : 7.000 gallons, 14.000 gallons, and 28.000 gallons. How many each type of tank car should be leased ?.
he also said let X1 be the number of cars with a 7.000 gallon capacity, X2 be the number of cars with 14.000 gallon capacity. and X3 be the number of cars with a 28.000 gallon

You need another boundary condition such as minimum cost, or minimum number of cars, or some distinguishing characteristic between the cars.

Without any, you have the following:

1--7000a + 14000b + 28000c = 420,000.
2--a + b + c = 24.
3--Multiplying (2) by 7 yields 7000a + 7000b + 7000c = 168,000.
4--Subtracting from (1) yields 7000y + 21000z = 252,000.
5--Dividing through by 7000 yields y + 3z = 36 or z = (12 - y)/3 = 12 - y/3
6--With y being divisible by 3, we have
y...3....6...9..12
z..11..10..9...8
x..10...8..6...4
 
i wanted it to be solved in this way guys.
"A chemical manufacturer wants to lease a fleet of 24 railroad tank cars"
X1+ X2+ X3= 24

"he also said let X1 be the number of cars with a 7.000 gallon capacity, X2 be the number of cars with 14.000 gallon capacity. and X3 be the number of cars with a 28.000 gallon capacity."

So total capacity must be 7000X1+ 14000X2+ 28000X3= 420000

That could be written in matrix form as
png.latex


Unfortunately, you have only two conditions with three unknown values. that is not enough to be able to solve the system.



but ty anyway. i`ll post another word problem now. and wana it to be solved in the same way :D
 
Why did you post it then?
Because he had no clue how to solve it himself! He posted this same question on another board and what he has in post #5 is my response on that board (without attribution).
What I missed in my response was what people here caught- that X1, X2, X3, being the "number of cars" must be integers. The matrix form of the equation lets us set limits on the possible value for X1, X2, and X3 and then integers with those "possible" values can be tested.
 
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