Lagrange multipliers.

Homework Statement

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

The Attempt at a Solution

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

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

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

vela
Staff Emeritus
Homework Helper
(0,0) doesn't satisfy the constraint. Try again.

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
Staff Emeritus
Homework Helper
Yup.

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
Staff Emeritus
Homework Helper
Yup.

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

vela
Staff Emeritus
Homework Helper
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?

Oops. I got my hz 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.

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I think this is wrong.

vela
Staff Emeritus