System of equations

  • Thread starter snoggerT
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  • #1
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[lambda]=3/(2x) = 1/(2y) = 1/(3z) with constraint eq. x^2+2y^2+6z^2=1





The Attempt at a Solution



- I can't figure out how to solve this system of equations. Can somebody please help.
 

Answers and Replies

  • #2
Defennder
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Is [tex]\lambda[/tex] supposed to be constant? If so, then it appears simple. Just Express x,y,z in terms of lambda, substitute it into the constraint equation and solve for lambda.
 
  • #3
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Is [tex]\lambda[/tex] supposed to be constant? If so, then it appears simple. Just Express x,y,z in terms of lambda, substitute it into the constraint equation and solve for lambda.

- yeah, thats way simpler than I was making it. So I would take lambda and plug back in to find my x, y, z, right?
 
  • #4
Defennder
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Yes that's right.
 
  • #5
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Yes that's right.

- so I get lambda to be +/- sqrt(41/12), which can be reduced to +/- 1/2*sqrt(41/3), but that gets somewhat ugly when plugging back in to find my x,y,z. I'm trying to find the min/max on this problem, so I have to find my x,y,z to plug back into the f(x,y,z) function. Is there any way to simplify things further?
 
  • #6
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Wait wait! You're solving for lambda?

[tex]\lambda=\frac{3}{2x}[/tex]

[tex]\lambda=\frac{1}{2y}[/tex]

[tex]\lambda=\frac{1}{3z}[/tex]

Set them equal to each other so that you have y = ... in terms of x and z = ... in terms of x so you can plug it back into your equation to solve for x.
 
Last edited:
  • #7
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Wait wait! You're solving for lambda?

[tex]\lambda=\frac{3}{2x}[/tex]

[tex]\lambda=\frac{1}{2y}[/tex]

[tex]\lambda=\frac{1}{3z}[/tex]

Set them equal to each other so that you have y = ... in terms of x and z = ... in terms of x so you can plug it back into your equation to solve for x.

- so I get +/- sqrt(27/41) for x, is that right?
 
  • #8
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- so I get +/- sqrt(27/41) for x, is that right?
Hmm, what did you do?
 
  • #9
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Hmm, what did you do?

- I set 3/(2x)= 1/(2y) and got y in terms of x and did the same for z. then plugged into the constraint equation and solved for x. I'm guessing I did it wrong though.
 
  • #10
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[tex]y=\frac x 3[/tex]

[tex]z=\frac 2 9 x[/tex]
 
  • #11
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[tex]y=\frac x 3[/tex]

[tex]z=\frac 2 9 x[/tex]

- thats what I got for y and z, so I must of done something wrong with my algebra in the constraint equation. Let me work it again.
 
  • #12
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I still get x=+/- sqrt(27/41). So I'm not sure what I'm doing wrong. This is how I'm setting the problem up:

x^2+2(x/3)^2+6(2x/9)^2=1
 
  • #13
Dick
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This is a lagrange multiplier problem isn't it? Can you state the original problem?
 
  • #14
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This is a lagrange multiplier problem isn't it? Can you state the original problem?

- sure. I probably need to start over anyways because I think I'm just confusing myself.

find the min/max:

f(x,y,z)=3x+2y+4z with constraint equation x^2+2y^2+6z^2=1
 
  • #15
Dick
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Ok, I think you are on completely the right track. lambda^2=12/41 is what I get too. And yes, x=+/-sqrt(27/41). I'll admit the numbers aren't that pretty. But I think you are doing exactly the right thing. Just keep on doing it.
 
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