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

  1. 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.
     
  2. jcsd
  3. 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.
     
  4. Re: Lagrange

    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).
     
  5. 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
     
  6. arildno

    arildno 12,015
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    Re: Lagrange

    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: Oct 12, 2009
  7. Re: Lagrange

    Sorry. The constraint was x2 + y2 - z2 - 1 = 0. My bad. So I think my answers are right.
     
  8. 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
  9. 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.
     
  10. Re: Lagrange

    It is x = y = -z
    So x = y = -z = [tex]\pm[/tex]1/[tex]\sqrt{3}[/tex]
     
  11. 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.
     
  12. arildno

    arildno 12,015
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    Re: Lagrange

    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.
     
  13. Re: Lagrange

    x2 + y2 - z2 - 1 = 0
     
  14. arildno

    arildno 12,015
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    Re: Lagrange

    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).
     
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