Question

I have this Lagrange multipliers question which I find difficult to solve. I know the method to go about but I have problems in finding the values of x and y. Here is the question :
f(x,y)=x^3-y^3
subject to the constraint
x^2+y^2=8

Answer 1

Okay, define a function:
\[h(x,y,\phi)=x^3-y^3-\phi [x^2+y^2-8]\]
right?

Answer 2

yes

Answer 3

Okay, now take a partial derivative with respect to x,y, phi. We get:
\[\frac{\partial h}{\partial x}=3x^2-2x \phi\]
\[\frac{\partial h}{\partial y}=-3y^2-2y \phi\]
\[\frac{\partial h}{\partial \phi}=-(x^2+y^2-8)\]

Answer 4

Do you agree?

Answer 5

yes, that's right

Answer 6

Okay. Now we set them to zero. So we get:
\[3x^2-2x \phi=0;-3y^2-2y \phi=0; x^2+y^2=8\]
The last one is the constraint which comes from the dh/dphi set to zero (HAPPENS EVERY TIME! verify yourself)
So solving we get:
\[x(3x-2\phi)=0 \implies x=0,x=\frac{2\phi}{3}\]
\[-y(3y+2\phi)=0 \implies y=0,y=\frac{-2 \phi}{3}\]
We see that the combination (0,0) doesn't solve the last condition so we can throw that out. So we have:
\[(0,\frac{-2 \phi}{3}) (\frac{2 \phi}{3},0)\]
Right?

Answer 7

Well and
\[(\frac{2 \phi}{3},\frac{-2 \phi}{3})\]

Answer 8

Make sense?

Answer 9

yes it does make sense

Answer 10

but i dont understand why I cant solve for \[\phi\]

Answer 11

As in the simultaneuos equations initially. Why do I have to solve for x and y when I can just make \[\phi \]the subject and then find x and y?

Answer 12

We're gonna do it right now.

Answer 13

You DO need to solve for phi. Because you only have your points (x,y) in terms of phi, this is where the constraint is useful. So lets try our points. (2phi/3,-2phi/3) PLUG THIS INTO THE CONSTRAINT!
\[(\frac{2\phi}{3})^2+(\frac{-2 \phi}{3})^2=8 \implies \frac{8\phi^2}{9}=8 \implies \phi=\pm3\]
So x=+/-2,y=+/-2
Plug these into the original function. We have:
\[(2,2),(-2,-2),(-2,2),(2,-2)\]
Get the values. Then do the other two points we had:
(0,-2phi/3)(2phi/3,0) Doing either of those we get phi=+/-3, so we get (0,+/-2) for the other point BUT!!! it does not satisfy the constraint. So that means we only have the "2's" points listed above.

Answer 14

IGNORE THAT LAST PARAGRAPH lol

Answer 15

lol

Answer 16

I haven't slept in 2 days /facepalm/ looking at the constraint if you plug in 0 for x or y you realize that the oppositive has to be +/- sqrt(8) or 2sqrt(2) so test those as well. Those actually turn out being the max and stuff

Answer 17

http://www.wolframalpha.com/input/?i=maximize+x%5E3-y%5E3+subject+to+x%5E2%2By%5E2%3D8
http://www.wolframalpha.com/input/?i=minimize+x%5E3-y%5E3+subject+to+x%5E2%2By%5E2%3D8

Answer 18

These problems require a little thought. I hate how they can be really inconsistent.

Answer 19

yeah they have been giving me a problem lately. The real difficult part of Lagrange isn't understanding the concept, it is trying to find the values of x and y. That sucks..

Answer 20

Haha, true story. All you're really doing is subtracting zero. lol

Answer 21

Thanks a lot for your help man..

Answer 22

Really do appreciate..

Answer 23

No problem