- #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.

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- Thread starter snoggerT
- Start date

- #1

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

- #2

Defennder

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- #3

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- 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.

[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.

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- #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

- 1,752

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

- #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]

[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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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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