# Equation x^2+y^2+z^2=1

1. Oct 11, 2009

### squenshl

Optimize f(x,y,z) = x3 + y3 + z3, subject to the constraint g(x,y,z) = x2 + y2 + z2 - 1 = 0

Step 1:
I did L = f - $$\lambda$$g = x3 + y3 + z3 - $$\lambda$$(x2 + y2 + z2 - 1)

Step 2:
I got Lx = 3x2 - 2$$\lambda$$x = 0, Ly = 3y2 - 2$$\lambda$$y = 0, Lz = 3z2 - 2$$\lambda$$z = 0

Now I can't seem to solve these 3 equations to get the critical points & find what $$\lambda$$ is. Some help please. Thanks.

2. Oct 12, 2009

### lstellyl

Re: Lagrange

Remember that you also need to use the equation x^2+y^2+z^2=1 to solve for lambda...

even though you used this equation in your lagrangian, the "=1" part kind of gets washed away when applying the E-L equations, and you need to introduce it back into the system of equations in order to find a solution.

3. Oct 12, 2009

### squenshl

Re: Lagrange

Alright. 3x2 = 2$$\lambda$$x, 3y2 = 2$$\lambda$$y, 3z2 = 2$$\lambda$$z
$$\Rightarrow$$ 3/2x = 3/2y = 3/2z $$\Rightarrow$$ x = y = z
Putting this into the constraint $$\Rightarrow$$ x2 = 1 so x = $$\pm$$1, y = $$\pm$$1, z = $$\pm$$1, I got 8 critical points (1,1,1), (1,-1,1),(1,1,-1),(1,-1,-1),(-1,1,1),(-1,-1,1,),(-1,1,1) & (-1,-1,-1). Putting x = y = z = $$\pm$$1 to find $$\lambda$$ I get $$\lambda$$ = 3/2 for x, y & z. Is this right? I'm pretty sure it is. Then I just use the reduced hessian to find the nature of the critical points (Max, min or saddle).

4. Oct 12, 2009

### lstellyl

Re: Lagrange

I think you made a mistake when putting the equations back into the constraint...

with x=y=z, you should get 3x^2=1 (not x^2=1) which will give you factors of 1/Sqrt(3) in your answer.

with the equations you have written out i think lambda ends up being Sqrt(3)/2

5. Oct 12, 2009

### arildno

Re: Lagrange

DO NOT DIVIDE WITH POTENTIAL ZEROES!!.

You have the 4 equations:
$$x(3x-2\lambda)=0$$
$$y(3y-2\lambda)=0$$
$$z(3z-2\lambda)=0$$
$$x^{2}+y^{2}+z^{2}-1=0$$

Consider the case x=0:

This gives you an additional 3 cases:
a) y=0, z something else
b) z=0, y something else
c) Neither z or y 0.

Thus, since this can be copied for the two other variables as well, you'll have a lot more critical points to consider

EDIT:

Furthermore, since those points you found REQUIRED x=y=z, then evidently a point like (-1,1,1) couldn't possibly work, right?

FURTHERMORE:

Don't bother about the Hessian; too much trouble!
You ought to be able to see which of the solutions represent optimizations of f

Last edited: Oct 12, 2009
6. Oct 12, 2009

### squenshl

Re: Lagrange

Sorry. The constraint was x2 + y2 - z2 - 1 = 0. My bad. So I think my answers are right.

7. Oct 12, 2009

### squenshl

Re: Lagrange

Nope, you are right. It is x = y = z = 1/sqrt(3). Am I right to say there are 8 critical points.

Last edited: Oct 12, 2009
8. Oct 12, 2009

### lstellyl

Re: Lagrange

well if the constraint is the one you gave us the second time, then indeed you should not get x=y=z=1/Sqrt(3). Check your L_z equation again...

regardless, arildno is correct in that you must cover your bases and watch your divisions by 0 here.

9. Oct 12, 2009

### squenshl

Re: Lagrange

It is x = y = -z
So x = y = -z = $$\pm$$1/$$\sqrt{3}$$

10. Oct 12, 2009

### lstellyl

Re: Lagrange

NO, here you have x=y=-z and x^2+y^2-z^2=1 which leads to 2x^2-x^2=1 and thus x=1,x=-1. HOWEVER, notice that this is in the case that x,y,z<>0.

11. Oct 13, 2009

### arildno

Re: Lagrange

a) $$x^{2}+y^{2}+z^{2}-1=0$$
OR
b) $$x^{2}+y^{2}-z^{2}-1=0$$

Furthermore, it does not seem that you understand that you cannot divide by zero.

12. Oct 13, 2009

### squenshl

Re: Lagrange

x2 + y2 - z2 - 1 = 0

13. Oct 13, 2009

### arildno

Re: Lagrange

In that case, you have the 4 equations:
$$x(3x-2\lambda)=0$$
$$y(3y-2\lambda)=0$$
$$z(3z+2\lambda)=0$$
$$x^{2}+y^{2}-z^{2}-1=0$$

A) Suppose that neither x, y or z are 0.

Then, we get x=y=-z.

Inserting this into the last equation yields:
$$x^{2}-1=0\to{x}=\pm{1}$$
Thus, we get TWO critical points, namely (1,1,-1) and (-1,-1,1)

B) Suppose that x=z=0
Then, we get from the last equation:
$$y^{2}-1=0\to{y}=\pm{1}$$
Thus, we get TWO critical points, namely (0,1,0) and (0,-1,0)

C) Suppose y=z=0.
Then, in analogy with B, we get the critical points (1,0,0), (-1,0,0)

D) Impossible cases:
Assume that x=y=0. Then the last equation reads -z^2-1=0, which has no solutions.

Furthermore, assume x=0, neither y or z 0. Then y=-z, and the last equation reduces to -1=0, which is false.

Similarly for y=0, neither x or z 0.

E) z=0, neither x or y 0.
Then, x=y, and the last equation becoms:
$$2x^{2}-1=0\to{x}=\pm\frac{1}{\sqrt{2}}$$

This yields the last two critical points:
$$(\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}},0)$$

Thus, in total, we get 8 distinct critical points, to be found in A), B), C) and E).