Finding Minimum with Lagrange Multipliers

[smashyash]

Homework Statement

Find the minimum of f(x,y) = x^2 + y^2 subject to the constraint g(x,y) = xy-3 = 0

Homework Equations

delF = lambda * delG

The Attempt at a Solution

Okay, after lecture, reviewing the chapter and looking at some online information, this is what I have so far:
(using l for lambda)

x^2 + y^2 -(l)(xy-3)
x^2 + y^2-xyl-3l

find critical points:

Fx = 2x-ly = 0 --> 2x = ly --> x = ly/2
Fy = 2y-lx = 0 --> 2y = lx --> y = lx/2
Fl = -xy-3 = 0

so then substituting x and y values into the Fl equation,

(ly/2)(lx/2) - 3 = 0

so in all the examples I've seen, there's never an x or y in this equation, it should all be in terms of lambda. So what do I do know? I'm stuck!

 

[Dick]

Try, for example, putting x=ly/2 into y=lx/2.

 

[smashyash]

Try, for example, putting x=ly/2 into y=lx/2.

Ok, I've done that and through the magic of algebra...

l^3*y^2 = 12

Isn't this still a problem because I'm trying to solve for l to get my critical points... right??

 

[Dick]

Ok, I've done that and through the magic of algebra...

l^3*y^2 = 12

Isn't this still a problem because I'm trying to solve for l to get my critical points... right??

My magic of algebra tells me that y=l^2*y/4. So I can find the possible values of l or that y=0 (which isn't true from the constraint). Try that again.

 

[smashyash]

My magic of algebra tells me that y=l^2*y/4. So I can find the possible values of l or that y=0 (which isn't true from the constraint). Try that again.

Oops! Yes, you're right. I was getting ahead of myself. This problem seems very different than others I've done..

so y = l^2*y/4

So now do I plug this y value into my constraint xy-3=0? Along with my x value gives me..

-[l^3*y^2]/8 + 3 = 0

I'm not sure where to go from here..

 

[Dick]

Oops! Yes, you're right. I was getting ahead of myself. This problem seems very different than others I've done..

so y = l^2*y/4

So now do I plug this y value into my constraint xy-3=0? Along with my x value gives me..

-[l^3*y^2]/8 + 3 = 0

I'm not sure where to go from here..

Figure out what the possible values of l are first. l^2/4=1, right?

 

[smashyash]

Figure out what the possible values of l are first. l^2/4=1, right?

I values to satisfy the equation y = l^2*y/4 ??

wouldn't that just be 2 or -2?

I'm sorry, this is all very new to me.

 

[Dick]

I values to satisfy the equation y = l^2*y/4 ??

wouldn't that just be 2 or -2?

I'm sorry, this is all very new to me.

That's ok. It takes a while to get used to. Sure l=2 or -2. (Since y can't be zero - otherwise you need to keep track of that possibility). So what's the relation between x and y?

 

[smashyash]

That's ok. It takes a while to get used to. Sure l=2 or -2. (Since y can't be zero - otherwise you need to keep track of that possibility). So what's the relation between x and y?

well, since x = ly/2 and y = lx/2, does this mean that y = x??

 

[Dick]

well, since x = ly/2 and y = lx/2, does this mean that y = x??

What about x=(-y)? 'l' can be -2 also, yes?

 

[smashyash]

What about x=(-y)? 'l' can be -2 also, yes?

oh! yes, I suppose that is also true... how do we incorporate both a + and - value?

 

[Dick]

oh! yes, I suppose that is also true... how do we incorporate both a + and - value?

That's only from what you know so far given l=2 or l=(-2). x=(-y) might not work in the constraint. Does it?

 

[smashyash]

That's only from what you know so far given l=2 or l=(-2). x=(-y) might not work in the constraint. Does it?

I may be off here but my constraint is g(x,y)= x*y -3 =0

If we plug in -y for x, we get (-y)*y - 3 = 0 or -y^2 - 3 = 0

so the sign doesn't matter as it's canceled out by the square??

 

[Dick]

I may be off here but my constraint is g(x,y)= x*y -3 =0

If we plug in -y for x, we get (-y)*y - 3 = 0 or -y^2 - 3 = 0

so the sign doesn't matter as it's canceled out by the square??

Hmm. -y^2-3 is never zero. Is it? I didn't mean to say x=(-y) IS a solution. I just wanted you to tell me why it wasn't. So yes, that makes the only possibility x=y.

 

[smashyash]

Hmm. -y^2-3 is never zero. Is it? I didn't mean to say x=(-y) IS a solution. I just wanted you to tell me why it wasn't. So yes, that makes the only possibility x=y.

Oh! I'm sorry.. this is confusing. So x and y have to be sqrt(3) to satisfy that equation, so these are my critical points??

 

[Dick]

Oh! I'm sorry.. this is confusing. So x and y have to be sqrt(3) to satisfy that equation, so these are my critical points??

When you are doing Lagrange multipliers you get bits of information one at a time as you solve parts of the problem. It's sort of a detective job. When you figured out y=l^2*y/4 that meant either y=0 or l=2 or l=(-2). You can eliminate y=0 by looking at the constraint. You can also eliminate l=(-2) the same way. So 'l' must be 2 and x=y. It might be a little confusing because applying the method isn't totally cookbook. You have to think. Yes, x=sqrt(3) and y=sqrt(3) is one critical point. There's another one. Can you find it?

 

[smashyash]

When you are doing Lagrange multipliers you get bits of information one at a time as you solve parts of the problem. It's sort of a detective job. When you figured out y=l^2*y/4 that meant either y=0 or l=2 or l=(-2). You can eliminate y=0 by looking at the constraint. You can also eliminate l=(-2) the same way. So 'l' must be 2 and x=y. It might be a little confusing because applying the method isn't totally cookbook. You have to think. Yes, x=sqrt(3) and y=sqrt(3) is one critical point. There's another one. Can you find it?

The only other numeric to satisfy this would be -(sqrt(3)). That's just from looking at the equation. Is this right?

 

[Dick]

The only other numeric to satisfy this would be -(sqrt(3)). That's just from looking at the equation. Is this right?

Sure. That's right.