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Equation x^2+y^2+z^2=1by squenshl
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#1
Oct1109, 09:05 PM

P: 328

Optimize f(x,y,z) = x^{3} + y^{3} + z^{3}, subject to the constraint g(x,y,z) = x^{2} + y^{2} + z^{2}  1 = 0
Step 1: I did L = f  [tex]\lambda[/tex]g = x^{3} + y^{3} + z^{3}  [tex]\lambda[/tex](x^{2} + y^{2} + z^{2}  1) Step 2: I got L_{x} = 3x^{2}  2[tex]\lambda[/tex]x = 0, L_{y} = 3y^{2}  2[tex]\lambda[/tex]y = 0, L_{z} = 3z^{2}  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
Oct1209, 03:06 AM

P: 41

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 EL equations, and you need to introduce it back into the system of equations in order to find a solution. 


#3
Oct1209, 04:40 PM

P: 328

Alright. 3x^{2} = 2[tex]\lambda[/tex]x, 3y^{2} = 2[tex]\lambda[/tex]y, 3z^{2} = 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] x^{2} = 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
Oct1209, 04:46 PM

P: 41

Equation x^2+y^2+z^2=1
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
Oct1209, 05:03 PM

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P: 12,016

DO NOT DIVIDE WITH POTENTIAL ZEROES!!.
You have the 4 equations: [tex]x(3x2\lambda)=0[/tex] [tex]y(3y2\lambda)=0[/tex] [tex]z(3z2\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 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 


#6
Oct1209, 06:00 PM

P: 328

Sorry. The constraint was x^{2} + y^{2}  z^{2}  1 = 0. My bad. So I think my answers are right.



#7
Oct1209, 06:05 PM

P: 328

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



#8
Oct1209, 06:09 PM

P: 41

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
Oct1209, 06:14 PM

P: 328

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


#10
Oct1209, 06:16 PM

P: 41

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



#11
Oct1309, 02:35 AM

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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
Oct1309, 02:29 PM

P: 328

x^{2} + y^{2}  z^{2}  1 = 0



#13
Oct1309, 03:44 PM

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[tex]x(3x2\lambda)=0[/tex] [tex]y(3y2\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^21=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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