# System of equations

1. Jul 2, 2008

### snoggerT

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

3. The attempt at a solution

2. Jul 2, 2008

### Defennder

Is $$\lambda$$ 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. Jul 2, 2008

### snoggerT

- 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. Jul 2, 2008

### Defennder

Yes that's right.

5. Jul 2, 2008

### snoggerT

- 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. Jul 2, 2008

### rocomath

Wait wait! You're solving for lambda?

$$\lambda=\frac{3}{2x}$$

$$\lambda=\frac{1}{2y}$$

$$\lambda=\frac{1}{3z}$$

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: Jul 2, 2008
7. Jul 2, 2008

### snoggerT

- so I get +/- sqrt(27/41) for x, is that right?

8. Jul 2, 2008

### rocomath

Hmm, what did you do?

9. Jul 2, 2008

### snoggerT

- 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. Jul 2, 2008

### rocomath

$$y=\frac x 3$$

$$z=\frac 2 9 x$$

11. Jul 2, 2008

### snoggerT

- 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. Jul 2, 2008

### snoggerT

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. Jul 2, 2008

### Dick

This is a lagrange multiplier problem isn't it? Can you state the original problem?

14. Jul 2, 2008

### snoggerT

- 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. Jul 3, 2008

### Dick

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.

Last edited: Jul 3, 2008