What are the 8 distinct critical points for the given equations?

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In summary, the optimization problem has 8 critical points at (1,1,-1), (-1,-1,1), (0,1,0), (0,-1,0), (1,0,0), (-1,0,0), (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0), (-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},0).
  • #1
squenshl
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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 - [tex]\lambda[/tex]g = x3 + y3 + z3 - [tex]\lambda[/tex](x2 + y2 + z2 - 1)

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

Now I can't seem to solve these 3 equations to get the critical points & find what [tex]\lambda[/tex] is. Some help please. Thanks.
 
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  • #2


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


Alright. 3x2 = 2[tex]\lambda[/tex]x, 3y2 = 2[tex]\lambda[/tex]y, 3z2 = 2[tex]\lambda[/tex]z
[tex]\Rightarrow[/tex] 3/2x = 3/2y = 3/2z [tex]\Rightarrow[/tex] x = y = z
Putting this into the constraint [tex]\Rightarrow[/tex] x2 = 1 so x = [tex]\pm[/tex]1, y = [tex]\pm[/tex]1, z = [tex]\pm[/tex]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 = [tex]\pm[/tex]1 to find [tex]\lambda[/tex] I get [tex]\lambda[/tex] = 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


squenshl said:
Alright. 3x2 = 2[tex]\lambda[/tex]x, 3y2 = 2[tex]\lambda[/tex]y, 3z2 = 2[tex]\lambda[/tex]z
[tex]\Rightarrow[/tex] 3/2x = 3/2y = 3/2z [tex]\Rightarrow[/tex] x = y = z
Putting this into the constraint [tex]\Rightarrow[/tex] x2 = 1 so x = [tex]\pm[/tex]1, y = [tex]\pm[/tex]1, z = [tex]\pm[/tex]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 = [tex]\pm[/tex]1 to find [tex]\lambda[/tex] I get [tex]\lambda[/tex] = 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).


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


DO NOT DIVIDE WITH POTENTIAL ZEROES!.

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

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:smile:

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:
  • #6


Sorry. The constraint was x2 + y2 - z2 - 1 = 0. My bad. So I think my answers are right.
 
  • #7


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


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


It is x = y = -z
So x = y = -z = [tex]\pm[/tex]1/[tex]\sqrt{3}[/tex]
 
  • #10


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


squenshl said:
Sorry. The constraint was x2 + y2 - z2 - 1 = 0. My bad. So I think my answers are right.
Is your constraint:
a) [tex]x^{2}+y^{2}+z^{2}-1=0[/tex]
OR
b) [tex]x^{2}+y^{2}-z^{2}-1=0[/tex]

Furthermore, it does not seem that you understand that you cannot divide by zero.
 
  • #12


x2 + y2 - z2 - 1 = 0
 
  • #13


squenshl said:
x2 + y2 - z2 - 1 = 0

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

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

Then, we get x=y=-z.

Inserting this into the last equation yields:
[tex]x^{2}-1=0\to{x}=\pm{1}[/tex]
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:
[tex]y^{2}-1=0\to{y}=\pm{1}[/tex]
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:
[tex]2x^{2}-1=0\to{x}=\pm\frac{1}{\sqrt{2}}[/tex]

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



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

1. What is the equation x^2+y^2+z^2=1 used for?

The equation x^2+y^2+z^2=1 is known as the equation of a unit sphere. It is used in mathematics and physics to represent a three-dimensional sphere with a radius of 1 unit.

2. How do you graph the equation x^2+y^2+z^2=1?

To graph the equation x^2+y^2+z^2=1, you can use a three-dimensional graphing software or manually plot points on a three-dimensional coordinate system. The resulting graph will be a perfect sphere with a radius of 1 unit.

3. What does the equation x^2+y^2+z^2=1 represent in terms of points in space?

The equation x^2+y^2+z^2=1 represents all the points in space that are equidistant from the origin (0,0,0) by a distance of 1 unit. These points form a perfect sphere in three-dimensional space.

4. Can the equation x^2+y^2+z^2=1 be solved for a specific variable?

Yes, the equation x^2+y^2+z^2=1 can be solved for a specific variable. For example, if we solve for x, we get x = ±√(1-y^2-z^2). Similarly, we can solve for y and z by rearranging the equation.

5. How is the equation x^2+y^2+z^2=1 related to Pythagorean theorem?

The equation x^2+y^2+z^2=1 is closely related to Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This theorem can be extended to three dimensions, where x^2+y^2+z^2 represents the square of the length of the hypotenuse of a 3D right triangle, with x, y, and z representing the lengths of the three sides.

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