Lagrange Multipliers

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer does not exist, enter DNE.)

f(x, y) = 5x^2 + 5y^2; xy = 1

I'm confused. Normal way isn't working?

F(x,y,λ = f(x,y) - λ(g(x,y)) - k
F(x,y,λ = 5x^2 + 5y^2 - λxy + λ

Fx = 10x - λy
Fy = 10y - λx
Fλ = -xy + 1

Setting derivatives = 0 and solving:

y = 10x/λ; x = 10y/λ; xy = 1

Sub y and x into xy = 1:

(10x/λ(10y/λ = 1

λ = 10sqrt(xy)

Sub λ back into fx = 0, fy = 0:

y = 10x/10sqrt(xy)
y^2 = x/y
y^3 = x

x = 10y/10sqrt(xy)
x^2 = y/x
x^3 = y

What do I do from here? I feel like I messed up somewhere in the process, because I don't know what to do with a variable.

5x^2 + 5y^2 - λ(xy-1)

d/dx = 10x - λy
d/dy = 10y - λx
d/dλ = 1 - xy

Set all to 0 and rearrange

10x = λy
10y = λx

10x/10y = λy/λx <== This is the step where you get rid of λ, it shouldn't get a value assigned to it.

x/y = y/x

x = y

1 - x^2 = 0

x = +/- 1
y = +/- 1

And remember if x = -1, y must = -1 and if x = 1, y must = 1

Okay.
That means both f(1,1) and f(-1,-1) are both minimums because fxx = 10 (which is greater than 0), and there are no maximums.

I guess I was confused because I had just done another problem by solving for x and y, which had λ in its value, and substituting that into the constraint and solving for λ, then substituting that value into the x and y equations.

Can anyone explain what a lagrange multiplier is and how/why it works? I don't quite understand the textbook/wikipedia/google explanations.

Okay.
That means both f(1,1) and f(-1,-1) are both minimums because fxx = 10 (which is greater than 0), and there are no maximums.

If you think about it this makes sense just by thinking about the equation. Both X and Y terms are squared so you will have identical positive and negative values. Ergo two minimum values.

Since there are no deducting terms (no subtraction signs), the values will only continue increasing. Ergo no maximum.

Can anyone explain what a lagrange multiplier is and how/why it works? I don't quite understand the textbook/wikipedia/google explanations.

Can't help you here unfortunately. Physics major. So long as the math works, I just go with it. They're used all over the place in matrix algebra...

If you think about it this makes sense just by thinking about the equation. Both X and Y terms are squared so you will have identical positive and negative values. Ergo two minimum values.

Since there are no deducting terms (no subtraction signs), the values will only continue increasing. Ergo no maximum.

Yeah, I figured that much - just like to check with equations.

Can't help you here unfortunately. Physics major. So long as the math works, I just go with it. They're used all over the place in matrix algebra...

I'm a physics major too! First year

I understand math equations more if I know how they are derived and how/why they work (e.g. derivative is the instantaneous slope).

You can also kinda think about it this way.

If xy = 1, that makes y a function of x, ie. y = 1/x

f(x) = 5x^2 + 5/x^2
d/dx = 10x - 10/x^3
10x = 10/x^3
x^4 = 1
x = +/- 1 which is either a maximum or a minimum

d^2/dx^2 = 2 + 6/x^4

Since x^4 is always positive, the second derivative is always positive so you know there is no maximum. It follows that x = +/- 1 is always a minimum, and since y = 1/x there are minima at (1,1) and (-1,-1)

As for why it works, I see it as a way of taking one equation with two unknowns and playing with it until you have three equations with three unknowns. As it shows in my working above, you can then reduce it in to two equations with two unknowns and solve for happy fun times.

I'm a physics major too! First year

I understand math equations more if I know how they are derived and how/why they work (e.g. derivative is the instantaneous slope).

Oh the blissful ignorance of innocence. I'd forgotten... I'm a senior and Higg's Boson willing I will graduate this year!

You will soon learn to ignore what isn't on the test. You just plain have to take a lot of the math for granted or you will never finish. Take Newton to heart and feel free to stand on the shoulders of others.

In fairly simple words, the method of Lagrange multipliers is exactly what you have said: find the turning points given a set of constraints. So in the first step, you introduce lambda(xy-1) into your function that leaves it unchanged because it equals to 0, and you WANT the function to be governed by this fixed set of x-y values that it can only take. You don't actually need to find lambda, because it's a dummy variable that you introduce with the aim of removing it later without finding out what it is i.e. its an auxillary function.

Just like that. Lambda comes in useful later at some other points, but let someone more versed in this come and give a better explanation. I'm late for lecture.

Hadriel

I think your question is answered, so if you don't me hijacking your topic. How do you exactly determine if the critical points you find are a maximum or a minimum. I now have a solution to a problem, by using the langrange method, expressed through a few other symbols like this for example: x* = 3m / (3px + 2py), y* = 2m / (3px + 2py), where m is income, px is price of good x and py is price of good y. The question was to find the maximum, but how can I determine if this solution is the maximum?

I think your question is answered, so if you don't me hijacking your topic. How do you exactly determine if the critical points you find are a maximum or a minimum. I now have a solution to a problem, by using the langrange method, expressed through a few other symbols like this for example: x* = 3m / (3px + 2py), y* = 2m / (3px + 2py), where m is income, px is price of good x and py is price of good y. The question was to find the maximum, but how can I determine if this solution is the maximum?

I use the second derivative test. You'll have to translate the variables to work for that economics problem, but that shouldn't be difficult.

D = fxx(a,b)*fyy(a,b)-(fxy(a,b))^2 [Same thing as fxx cross fyy]

if D > 0 and fxx(a,b) > 0; f(a,b) is a local minimum
if D > 0 and fxx(a,b) < 0; f(a,b) is a local maximum
if D < 0 then test fails (saddle point I believe)

if D < 0 and fxx(a,b) > 0; f(a,b) is a local minimum
if D > 0 and fxx(a,b) < 0; f(a,b) is a local maximum
if D = 0 then test fails (saddle point I believe)

Fixed for ya

Or even more simply, just plug the numbers in

On second thought it's probably outside of what we need to know so far, considering this is supposed to be an introduction to the lagrange method and second partial derivative test for functions with three variables were nowhere near being mentioned. During the lecture I caught that a condition of using the lagrange method is that both the constraint function and the optimization function need to be differentiable. Is there any way to proof that, or would just actually taking the derivative be enough? Functions in question are: u(x,y) = x^3y^2, g(x,y) = pxX +pyY - m; u(x,y,z) = x^5y^2z^3, g(x,y,z)= pxX + pyY + pzZ - m. Thank you for your replies

just plug the numbers in

This works well.

I am pretty sure that it's :
if D > 0 and fxx(a,b) > 0; f(a,b) is a local minimum
if D > 0 and fxx(a,b) < 0; f(a,b) is a local maximum
if D < 0 then saddle point
if D = 0 then test fails

I looked up my notes and that's what I have written, and Wikipedia also says this.

This works well.

I am pretty sure that it's :
if D > 0 and fxx(a,b) > 0; f(a,b) is a local minimum
if D > 0 and fxx(a,b) < 0; f(a,b) is a local maximum
if D < 0 then saddle point
if D = 0 then test fails

I looked up my notes and that's what I have written, and Wikipedia also says this.

Oh whoops, misread it. You're completely correct.