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Find an approximate solution to the system of equations x+y+z=2013 x y z = 2013   

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maroalgo | eNotes Newbie

Posted May 20, 2013 at 9:22 PM via web

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Find an approximate solution to the system of equations

x+y+z=2013

x y z = 2013 

 

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mathsworkmusic | (Level 3) Associate Educator

Posted May 21, 2013 at 12:39 AM (Answer #1)

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We have

1) x + y + z = 2013

2) xyz = 2013

It is clear that one of the values, x for example, needs to be very close to 2013.

Using the golden ratio is a natural avenue to try here.

The golden ratio r is such that

1 + 1/r = r,  (r + 1) = r^2 and

(r + 1) + 1/(r+1) = [(r+1)^2+1]/(r+1) = (r^2 + 2r + 2)/(r+1) = 3(r+1)/(r+1) = 3

Since (r + 1)[1/(r+1)] = 1  and 2013/2010 = 1.001493 (approx 1)

we can take

x = 2010,  y = (r + 1) = 2.6180  and z = 2013/[2010(r+1)] = 0.3825

check: x + y + z = 2013.0005

 

 

 

 

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maroalgo | eNotes Newbie

Posted May 21, 2013 at 2:28 AM (Reply #1)

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I found the same results but the question for the fonction is numbers with no decimal like x= 1 , y =2 , z= 3 it couldnt be x=1,8 or like y=2,7 or z=3,4 for exemple, it has to be solid numbers 

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