# Lagrange multipliers.

• squenshl
The constraint says that both x and y must be non-zero. The z equation has a solution only if z=0. When z=0, x=2y and y=2x. But the constraint says that x and y must be non-zero, not just 2x and 2y.f

## 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.

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.

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.

(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.

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.

Yup.

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

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.

Why?