Finding Minimum Values with Lagrange Multipliers

[squenshl]

Homework Statement

Minimise = x² + y² subject to C(x,y) = 4x² + 3y² = 12.

Homework Equations

The Attempt at a Solution

I let h(x,y) = x² + y² + \lambda(4x² + 3y² - 12).
I got h_x = 2x + 8\lambdax = 0, h_y = 2y + 6\lambday = 0, but here I get 2 values of \lambda, \lambda = -1/4 & -1/3. So I don't get a min/max value.

 

[fzero]

When you solve each equation for \lambda, you're making an assumption about the values of x and/or y. Perhaps one of these assumptions is wrong.

 

[Dick]

How about, for example, lambda=(-1/4) and y=0. Doesn't that satisfy both equations?

 

[squenshl]

So a min/max point is (0,0). How do I get the other 3.

 

[vela]

(0,0) doesn't satisfy the constraint. Try again.

 

[squenshl]

I think I got it. 4 points: (0,2),(0,-2),(\sqrt{3},0),(-\sqrt{3},0). When it asks to minimise do I just find the 2 min values.

 

[vela]

Yup.

 

[squenshl]

Cheers.
I have anther question.
How do I find the point(s) on the surface z² + xy = 1 that lie closest to the origin.
Is it let h(x,y,z) = x² + y² + z² + \lambda(z² + xy - 1) then find the partial derivatives & equate to 0 & the rest.

 

[vela]

Yup.

 

[squenshl]

I get \lambda = 0 so the point is (0,0,0) which is the origin but that can't be right?

 

[vela]

No, that point doesn't satisfy the constraint, so the multiplier can't be 0.

Look at the z equation. Assume z isn't 0. What does λ have to be? Then use that value to solve for x and y. Then solve for z.

Now assume z=0. To satisfy the constraint, neither x nor y can vanish. For what values of λ will the x and y equations have non-zero solutions?

 

[squenshl]

Oops. I got my h_z wrong. \lambda = -1. So y = 2x & x = 2y. So we get z² + 2x³ = 1 therefore z = \pm (1-2x²)^1/2. I get (0,0,1) & (0,0 -1). I actually get 4 more points making 6 in total.

 

[squenshl]

I think this is wrong.

 

[vela]

Why?