Recursive Lagrange multipliers

[hiigaranace]

Hey all, this is my first post, so I apologize in advance if data are missing/format is strange/etc.

I'm working with lagrange multipliers, and I can get to the answer about half the time. The problem is, I'm not really sure how to deal with things when the multiplier equation becomes recursive, like in this problem:

Homework Statement

"Find the maximum and minimum values of x²+y² subject to the constraint x²-2x+y²-4y=0"

Homework Equations

$$\grad{}$$f(x, y) = $$\lambda\grad{}$$g(x, y)
...those are supposed to be gradient symbols. They're a little different from what I'm used to seeing, but that's apparently what LaTeX thinks it should be.

The Attempt at a Solution

I let f be the equation to maximize and minimize, and let g the constraint equation. this gave me:

2x$$\vec{i}$$+2y$$\vec{j}$$ = $$\lambda$$(2x-2)$$\vec{i}$$+$$\lambda$$(2y-4)$$\vec{j}$$

Then, you split it to deal with one direction at a time, giving two equations:

2x = $$\lambda$$(2x-2) 2y=$$\lambda$$(2y-4)

Annnnnnnd here's where I get stuck. I need to get everything in terms of one variable so I can put that into the constraint equation and solve for the coordinates, but when I try that, I either get X or Y on both sides of the equation (like above), or I wind up with the lambda isolated, but an expression for X or Y that necessitates their presence in the equation. I suspect I'm making some algebraic misstep (so weird, I can do calc, but algebra always kills me), rather than a calculus one, but I'm not sure. What should I do next?

Thanks!

 

[HallsofIvy]

I don't know what you mean by "recursive" here. There does not appear to me to be anything "recursive" here.

You want to maximize $f(x,y)= x^2+ y^2$ subject to the constraint that $g(x,y)= x^2- 2x+ y^2- 4y= 0$

Okay, $\nabla f= 2x\vec{i}+ 2y\vec{j}$ and $\nabla g= (2x- 2)\vec{i}+ (2x- 4)\vec{j}$. At a max or min with this constraint, we must have $2x\vec{i}+ 2y\vec{j}= \lambda((2x-2)\vec{i}+ (2y-4)\vec{j})$.

That is, we must have $2x= \lambda(2x- 2)$ and $2y= \lambda(2y- 4)$.

That, so far, is exactly what you have. One good way to solve those is to eliminate $\lambda$ by dividing one equation by the other.
x/y= (x-1)/(y- 2) or xy- 2x= xy- y so that xy= -y or y(x+ 1)= 0

I get four distinct points.

 

[hiigaranace]

I follow what you said, and I had a feeling it was something simple like that. I am slightly confused, though, about one of your simplifications of the division. You said that

xy-2x = xy-y gives xy = -y.

Shouldn't that be y = 2x, instead? If not, I'm not quite seeing how you got there.

 

[hiigaranace]

Never mind, I pressed ahead and got the answer. Thanks! Problems like this have been driving me nuts for a couple of days now...

 

[icystrike]

Never mind, I pressed ahead and got the answer. Thanks! Problems like this have been driving me nuts for a couple of days now...

I've gotten y = 2x too ... Is it correct?

 

[hiigaranace]

I've gotten y = 2x too ... Is it correct?

You're doing it right, icy, don't worry.