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Smart people do this math problem

I know the right answer, because I looked in the back of the book, but I just can't work the problem out. I'm not asking anyone to do my homework because I'll just ask the teacher to do it but here, see if any of you can do it...

Boeing 747s seat 400 passengers and are priced at $200 million each, Boeing 777s seat 300 passengers and are priced at $160 million, and the European Airbus A330's seat 300 passengers and are priced at $120 million. You are the purchasing manager of an airline company and have a spending goal of $2400 million for the purchase of new aircraft to seat a total of 5000 passengers. Your company has a policy of supporting US industries, and you have been instructed to buy twice as many US manufactured aircraft (Boeing) as foreign aircraft (Airbus). Given the selection of three aircraft, how many of each should you order?

Use a system of equations

So I made a system of equations as follows...

400x + 300y + 300z = 5000
200x + 160y + 120z = 2400
x + y - 2z = 0

I don't know if I'v written the system of equations wrong. I'm using the Row Reduction Method of Matrices and I know for a fact I'm doing the operation correctly, so it must be my system of equations. I'v worked and reworked this problem for about an hour, and I'v just given up...

use substitution

quote:

Originally posted by Sunsnail use substitution

for 3 equations? How?

http://en.wikipedia.org/wiki/Montante's_method

i think that's the easiest method to solve your problem. theres also a method called, gausse? i guess

hope it helped

Re: Smart people do this math problem

Good ol linear programming. Simplex method!!!!

quote:

Originally posted by Krypton 400x + 300y + 300z = 5000 200x + 160y + 120z = 2400 x + y - 2z = 0

I checked what you got[5,5,5] That makes sense because
747:QTY 5 + 777:QTY 5 = 10 which is double of the A330:QTY 5

Only question I have is why did you do it 1x+1y-2z = 0
I see the problem as 2x+2y+z=0 because of the doubling with respect to the european plane? The answer is already there but I just want to know the reasoning for that.

Re: Re: Smart people do this math problem

quote:

Originally posted by Trancealot Only question I have is why did you do it 1x+1y-2z = 0 I see the problem as 2x+2y+z=0 because of the doubling with respect to the european plane? The answer is already there but I just want to know the reasoning for that.

That threw me off for a bit too, but I'm pretty sure its right.

Re: Re: Smart people do this math problem

quote:

Originally posted by Trancealot Good ol linear programming. Simplex method!!!! I checked what you got[5,5,5] That makes sense because 747:QTY 5 + 777:QTY 5 = 10 which is double of the A330:QTY 5 Only question I have is why did you do it 1x+1y-2z = 0 I see the problem as 2x+2y+z=0 because of the doubling with respect to the european plane? The answer is already there but I just want to know the reasoning for that.

2(5) + 2(5) + 5 = 25

Maybe you meant...

2(5) + 2(5) - 5 but it still equals 15 which is not zero. That is an invalid equation. Whereas...

5 + 5 - 2(5) = 0 is true. So I think it must be x + y - 2z = 0 but I still can't find the answer (5, 5, 5) doing the operation. My head is about to explode...The next question is a hybrid of this question, and so I can't do either one!

KRYPTON, THE SYSTEM DOES NOT WANT YOU TO SUCCEED. THEY ARE WEEDING YOU OUT, YOU ARE FIGHTING AGAINST YOUR NATURE, YOU WERE MEANT TO BE A CLERK, I KNOW YOU DON'T WANT TO ADMIT THAT SO YOU'LL WORK YOUR ASS OFF TRYING TO PROVE YOURSELF THAT YOU CAN 'GO THE MILE'

I'LL TELL YOU WHAT, WHY DON'T YOU SHOVE A BIT OF HOPE AND CHANGE UP YOUR ASS, GRAB YOUR LAME ASS GLOWSTICKS AND GO DANCE WITH YOUR REDNECK MOM. YOU REDNECK ASSHOLE, I HOPE YOU FAIL MISERABLY

quote:

Originally posted by winston KRYPTON, THE SYSTEM DOES NOT WANT YOU TO SUCCEED. THEY ARE WEEDING YOU OUT, YOU ARE FIGHTING AGAINST YOUR NATURE, YOU WERE MEANT TO BE A CLERK, I KNOW YOU DON'T WANT TO ADMIT THAT SO YOU'LL WORK YOUR ASS OFF TRYING TO PROVE YOURSELF THAT YOU CAN 'GO THE MILE' I'LL TELL YOU WHAT, WHY DON'T YOU SHOVE A BIT OF HOPE AND CHANGE UP YOUR ASS, GRAB YOUR LAME ASS GLOWSTICKS AND GO DANCE WITH YOUR REDNECK MOM. YOU REDNECK ASSHOLE, I HOPE YOU FAIL MISERABLY

AFGA F AEUFIA UIFH A FHAEIUR AIUOE AWIOHF EF AHWOIEFH AWIOHWIF EWI FWAHFWIAOFH OIEWFH IOWAFEI FHIEWAIEO OIWAEHFOI WAHFOIWAHFIOW FHWAIOEF YOU TOO!

Re: Smart people do this math problem

quote:

Originally posted by Krypton 400x + 300y + 300z = 5000 200x + 160y + 120z = 2400 x + y - 2z = 0

substitution is the only way I know but it takes alot of work.. Just pick one equation and eliminate a variable.

Lets use the third one:

x + y - 2z = 0
so..
x + y = 2z

divide both sides by 2..

(x + y)/2 = z

There is your substitution. Replace z in the first two equations with (x + y)/2. Your answer will be in all x's and y's.

Let's take the first equation and substitute:
400x + 300y + 300[(x+y)/2] = 5000
= 400x + 300y + (300x)/2 + (300y)/2 = 5000

eliminate the fraction by multiplying everything by 2..

= 800x + 600y + 300x + 300y = 10000
= 1100x + 900y = 10000

now get Y by itself..

900y = 10000 - 1100x
900y/900 = (10000/900 - 1100x/900)
Y = (10000/900 - 1100x/900)

You now have the values for Y and Z priced in terms of X. ***We've eliminated the Y and Z variables entirely***, so you can rewrite your original equations using nothing but X. If you do this, you should be able to get a numeric value for X..and can then plug it back in and solve for Y and Z.

My brain just exploded.

If you need to show work, Gau�ian elimination seems like the least annoying method -- it will work, the only annoying thing is that it's easy to make mistakes in arithmetic.

If you don't need to show work, then matlab is your friend

Re: Re: Smart people do this math problem

quote:

Originally posted by Capitalizt substitution is the only way I know but it takes alot of work.. Just pick one equation and eliminate a variable. Lets use the third one: x + y - 2z = 0 so.. x + y = 2z divide both sides by 2.. (x + y)/2 = z There is your substitution. Replace z in the first two equations with (x + y)/2. Your answer will be in all x's and y's. Let's take the first equation and substitute: 400x + 300y + 300[(x+y)/2] = 5000 = 400x + 300y + (300x)/2 + (300y)/2 = 5000 eliminate the fraction by multiplying everything by 2.. = 800x + 600y + 300x + 300y = 10000 = 1100x + 900y = 10000 now get Y by itself.. 900y = 10000 - 1100x 900y/900 = (10000/900 - 1100x/900) Y = (10000/900 - 1100x/900) You now have the values for Y and Z priced in terms of X. ***We've eliminated the Y and Z variables entirely***, so you can rewrite your original equations using nothing but X. If you do this, you should be able to get a numeric value for X..and can then plug it back in and solve for Y and Z. My brain just exploded.

Unfortunately, I'm not allowed to use substitution, so this would be pointless...

And, for everyone who suggested the Gaussian Method, that's the method I'm using, I just called Row Reduction Method. They'r the same thing. I'm doing the operation correctly but I am still getting off the wall answers like negatives.

----------------------------------------------------------

"You have been instructed to buy twice as many American planes (x & y) as European planes (z)."

I would think this is the equation for that...

x + y = 2z -----> x + y - 2z = 0

I can't see how my system of equations is wrong, or how my operation using the Gausse Row Reduction Method is wrong...

http://forums.anandtech.com/categor...id=50&forumid=1

Post your question there..lots of smart people.

Since I'm bored..

| 400 300 300 | | 5000 |
| 200 160 120 | | 2400 |
| 1 1 -2 | | 0 |

| 4 3 3 | | 50 |
| 5 4 3 | | 60 |
| 1 1 -2 | | 0 |

| 4 3 3 | | 50 |
| 0 4-(5/4)*3 3-(5/4)*3 | | 60-(5/4)*50 |
| 1 1 -2 | | 0 |

| 4 3 3 | | 50 |
| 0 1/4 -3/4 | | -5/2 |
| 0 1-(1/4)*3 -2-(1/4)*3 | | 0-(1/4)*50 |

| 4 3 3 | | 50 |
| 0 1/4 -3/4 | | -5/2 |
| 0 1/4 -11/4 | | -50/4 |

| 4 3 3 | | 50 |
| 0 1/4 -3/4 | | -5/2 |
| 0 0 -11/4+3/4 | | -50/4+5/2 |

| 4 3 3 | | 50 |
| 0 1/4 -3/4 | | -5/2 |
| 0 0 -2 | | -10 |

| 4 3 3 | | 50 |
| 0 1/4 -3/4 | | -5/2 |
| 0 0 1 | | 5 |

| 4 3 3 | | 50 |
| 0 1/4 0 | | -5/2+(3/4)*5 |
| 0 0 1 | | 5 |


| 4 3 3 | | 50 |
| 0 1/4 0 | | 5/4 |
| 0 0 1 | | 5 |

| 4 0 0 | | 50-5*3-5*3 |
| 0 1 0 | | 5 |
| 0 0 1 | | 5 |

| 4 0 0 | | 20 |
| 0 1 0 | | 5 |
| 0 0 1 | | 5 |

| 1 0 0 | | 5 |
| 0 1 0 | | 5 |
| 0 0 1 | | 5 |

quote:

Originally posted by piggy Since I'm bored..

Nice, I'm gonna have a hard look at this tomorrow since my brain is already fried, thanks a lot.

Whatever you're studying, you suck at it. Go and become a trash collector.

quote:

Originally posted by Meat187 Whatever you're studying, you suck at it. Go and become a trash collector.

Got something stuck up ur ass I see. Get a life...

quote:

Originally posted by Krypton Got something stuck up ur ass I see. Get a life...

you get a life! don't you see that you annoy half of the population inside this forum? yet you stay to prove a point. to prove that you simply don't care, and that comes off as obnoxious, and kong. the thread title just proves us that you need affection from internet buddies, just get out!

It's a highschool level problem dude. http://en.wikipedia.org/wiki/Gaussian_elimination

gaussian elimination makes that shit so much easier.

Re: Re: Re: Smart people do this math problem

quote:

Originally posted by Krypton I can't see how my system of equations is wrong, or how my operation using the Gausse Row Reduction Method is wrong...

Maybe you're doing row operations on the matrix but not on applying them on the right hand side?

add the first two equations and get rid of a variable

400x + 300y + 300z = 5000
200x + 160y + 120z = 2400


400x + 300y + 300z = 5000
-400x - 320y - 240z = -4800

ad0d two equations and solve for one of the variables (y, or z) then substitute

I've finished linear algebra few days ago and have absolutely no intention of solving those problems again.

Oh, and my username is perfect for this thread.

that's way way below the
__ABOVE THIS LINE IS SMART__
group.

But eh, America. When I graduated in California, all that was required (in mathematics) was a passing grade in GEOMETRY. Take into account, the last mathematics class in middle school is Pre-Algebra, which is the pre-requisite for Geometry.

Are you required to learn Matrices? Both times I've come across the method in my life, the instructor's have passed it on saying it's some archaic compton shit. The substitution from above(shadow_419) is what I was taught and it's very easy.

quote:

Originally posted by piggy Since I'm bored.. | 400 300 300 | | 5000 | | 200 160 120 | | 2400 | | 1 1 -2 | | 0 | | 4 3 3 | | 50 | | 5 4 3 | | 60 | | 1 1 -2 | | 0 | | 4 3 3 | | 50 | | 0 4-(5/4)*3 3-(5/4)*3 | | 60-(5/4)*50 | | 1 1 -2 | | 0 | | 4 3 3 | | 50 | | 0 1/4 -3/4 | | -5/2 | | 0 1-(1/4)*3 -2-(1/4)*3 | | 0-(1/4)*50 | | 4 3 3 | | 50 | | 0 1/4 -3/4 | | -5/2 | | 0 1/4 -11/4 | | -50/4 | | 4 3 3 | | 50 | | 0 1/4 -3/4 | | -5/2 | | 0 0 -11/4+3/4 | | -50/4+5/2 | | 4 3 3 | | 50 | | 0 1/4 -3/4 | | -5/2 | | 0 0 -2 | | -10 | | 4 3 3 | | 50 | | 0 1/4 -3/4 | | -5/2 | | 0 0 1 | | 5 | | 4 3 3 | | 50 | | 0 1/4 0 | | -5/2+(3/4)*5 | | 0 0 1 | | 5 | | 4 3 3 | | 50 | | 0 1/4 0 | | 5/4 | | 0 0 1 | | 5 | | 4 0 0 | | 50-5*3-5*3 | | 0 1 0 | | 5 | | 0 0 1 | | 5 | | 4 0 0 | | 20 | | 0 1 0 | | 5 | | 0 0 1 | | 5 | | 1 0 0 | | 5 | | 0 1 0 | | 5 | | 0 0 1 | | 5 |

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