Lagrange Multipliers to Find Extreme Values of a Multi-Variable Function

[ChrisPls]

Homework Statement

I need to find the extrema of $f(x,y) = 3x^{2} + y^{2}$ given the constraint $x^{2} + y^{2} = 1$

Homework Equations

I'm not sure what goes here. I've been trying to solve it with this:

∇f(x,y) = λ∇g(x,y)

The Attempt at a Solution

$f(x,y) = 3x^{2} + y^{2}$
$g(x,y) = x^{2} + y^{2} = 1$

$∇f(x,y) = <6x, 2y>$
$∇g(x,y) = <2x, 2y>$

With the previous equation, we get a system of equations:

$6x = 2xλ$

$2y = 2yλ$

and the constraint

$x^{2} + y^{2} = 1$

Solving for x and y, I get x=y=0. But that fails the final constraint. I've used the above procedure for other Lagrange problems without much issue, but this one's stumping me.

So I tried finding a solution on Wolfram: Maxes at (-1, 0) and (1, 0), and Mins at (0, -1) and (0, 1). Okay, nice and simple looking solutions. Right?

But I just can't figure this out. What am I missing?

Thanks

 

[Ray Vickson]

Homework Statement

I need to find the extrema of $f(x,y) = 3x^{2} + y^{2}$ given the constraint $x^{2} + y^{2} = 1$

Homework Equations

I'm not sure what goes here. I've been trying to solve it with this:

∇f(x,y) = λ∇g(x,y)

The Attempt at a Solution

$f(x,y) = 3x^{2} + y^{2}$
$g(x,y) = x^{2} + y^{2} = 1$

$∇f(x,y) = <6x, 2y>$
$∇g(x,y) = <2x, 2y>$

With the previous equation, we get a system of equations:

$6x = 2xλ$

$2y = 2yλ$

and the constraint

$x^{2} + y^{2} = 1$

Solving for x and y, I get x=y=0. But that fails the final constraint. I've used the above procedure for other Lagrange problems without much issue, but this one's stumping me.

So I tried finding a solution on Wolfram: Maxes at (-1, 0) and (1, 0), and Mins at (0, -1) and (0, 1). Okay, nice and simple looking solutions. Right?

But I just can't figure this out. What am I missing?

Thanks

$$6x = 2x \lambda \Longrightarrow x = 0 \text{ or } \lambda = 3\\<br /> 2y = 2y \lambda \Longrightarrow y = 0 \text{ or } \lambda = 1.$$
Either x or y must = 0, and then the other one = ± 1. One value of λ corresponds to the constrained min and the other to the constrained max.

RGV

 

[ChrisPls]

Ooh, I see how that works now. Thank you very much!