How Do You Solve a Lagrange Multiplier Problem with a Circle Constraint?

[MrCreamer]

Homework Statement

Find the extrema of f(x, y) = x²−2xy+ 2y², subject to the
constraint x² +y² = 1.

Homework Equations

∇f(x,y) = λg(x,y)

The Attempt at a Solution

This is the work I have thus far:

Letting g(x,y) = x²+y²-1,

We obtain the following three equations from the Lagrange Multiplier equation:

2x-2y = 2λx
4y-2x = 2λy
x²+y²-1=0

I know that x and y cannot both be zero due to the constraint equation.

If x = 0:

4y = 2λy

--> λ = 2

If y = 0:

2x = 2λx

--> λ = 1

I have no idea what else to do

 

[Ray Vickson]

Homework Statement

Find the extrema of f(x, y) = x²−2xy+ 2y², subject to the
constraint x² +y² = 1.

Homework Equations

∇f(x,y) = λg(x,y)

The Attempt at a Solution

This is the work I have thus far:

Letting g(x,y) = x²+y²-1,

We obtain the following three equations from the Lagrange Multiplier equation:

2x-2y = 2λx
4y-2x = 2λy
x²+y²-1=0

I know that x and y cannot both be zero due to the constraint equation.

If x = 0:

4y = 2λy

--> λ = 2

If y = 0:

2x = 2λx

--> λ = 1

I have no idea what else to do

Using the symbol u instead of λ, you have:
$(1-u)x = y$ and $(2-u)y = x$, so $0 = x(u^2 - 3u + 1)$.

 

[vela]

We obtain the following three equations from the Lagrange Multiplier equation:

2x-2y = 2λx
4y-2x = 2λy
x²+y²-1=0

I know that x and y cannot both be zero due to the constraint equation.

If x = 0:

4y = 2λy

--> λ = 2

If y = 0:

2x = 2λx

--> λ = 1

I have no idea what else to do

At this point, there's really no reason to consider the cases x=0 and y=0. Nothing in the equations obviously suggests that either of those cases have anything to do with solving the system of equations. However, after you combine the equations the way Ray suggests, you find x=0 could be part of a possible solution. You found this implies that $y \ne 0$ and $\lambda=2$. The first equation, however, then requires that $y=0$, which is a contradiction. You can therefore eliminate x=0 as a possibility.