# Lagrange multipliers (yes, again)

1. May 5, 2014

### Feodalherren

1. The problem statement, all variables and given/known data

f=xy^2

C: x^2 + y^2 = 3

2. Relevant equations

3. The attempt at a solution

I don't understand how he can say that x=0 is a solution in this one. Looking at the contours, there are no solutions for f if x=0.

2. May 6, 2014

### Ray Vickson

I also don't understand it. In fact, it is false. Points of the form $(0 \pm \sqrt{3})$ are neither maxima nor minima of f, because as $x$ increases through 0, f increases from negative to positive if $y^2 = 3$: it looks like $f = 3x$ for fixed $y = \pm \sqrt{3}$---that is, if we go along a line parallel to the x-axis. Since such a line is tangent to the constraint circle, the same type of behavior will happen if we go a small distance along the circle instead of its tangent line. In other words, points with x = 0 are neither local maxima nor minima in the constrained problem.

Anyway, points with x = 0 are not among those we get from the Lagrange multiplier method.

3. May 6, 2014

### Feodalherren

I thought it seemed suspicious, though I have to admit I didn't fully understand your explanation.

4. May 6, 2014

### Ray Vickson

Think about it, take your time, draw pictures, plug numbers into formulas, etc.

5. May 6, 2014

### BvU

Function is not f = xy^2 but f = yx^2. Print is barely readable but teachers handwriting confirms.
I agree with teacher!

6. May 6, 2014

### HallsofIvy

Staff Emeritus
Langrange's method gives the two equations $2xy= 2\lambda x$ and $x^2= 2\lambda y$. What I usually do in such a case is eliminate $\lambda$ by dividing one equation by the other, say dividing the first equation by the second, to get $2xy/x^2= x/y$ but since we cannot divide by 0, that is only valid if x and y are not 0. We need to check those separately. If x= 0, the first equation, $2xy= 2\lambda x$, is satisfied for all y. The second equation, $x^2= 2\lambda y$ is satisfied if y= 0 or $\lambda= 0$. If both x and y are 0, the equation $x^2+ y^2= 3$ cannot be satisfied so that is NOT a solution. If x and $\lambda$ are 0, the equation $x^2+ y^2= 3$ becomes $y^2= 3$ so $(0, \sqrt{3})$ and $(0, -\sqrt{3})$ are valid solutions.

If y= 0, the equation $x^2= \lambda y= 0$ is satisfied only for x= 0 and we have already seen that (0, 0) does not satisfy $x^2+ y^2= 3$.

7. May 6, 2014

### Ray Vickson

I use the OP's (wrong) description that f = xy^2. I will not bother to look at the OP's submitted messy and illegible photocopies, and since the OP cannot manage to submit correct problem descriptions (on more than one occasion) I will no longer respond to his postings.

Last edited: May 6, 2014
8. May 7, 2014

### Feodalherren

Well excuse me for being human and making mistakes. Burn me at the the stake, right?

Thanks HallsOfIvy that's a great explanation!

9. May 7, 2014

### BvU

No need to get burnt (both). I do think that if you manage to antagonize Ray by apparently repeating human errors, there is something there for you to pick up: at least you could promise to try to correctly formulate problem statements (*1) and check if you don't make it difficult for potential helpers (*2). I was on the wrong track for a while too.

(*1) it is human to make that kind of mistake and then overseeing subsequent pointers that it's the other way around. But you can learn from it.
(*2) different matter, more about attitude, meticulousness etc. From that one you can learn that investing a little time to do it completely right saves time and goodwill in the long run

All in good, constructive spirit

Personally I am a slow learner and I do repeat mistakes quite often. Hence the (moderate amount of) understanding sympathy.