Does a Minimum Exist for x^2+y^2+z^2 Given x^4+y^4+z^4=3?

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Homework Statement


Find max/min of x^2+y^2+z^2 given x^4+y^4+z^4=3

Homework Equations


Use of gradient vectors related by LaGrange Multiplier

The Attempt at a Solution


\begin{gathered}<br /> f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2};g\left( {x,y,z} \right) = {x^4} + {y^4} + {z^4} - 3 = 0 \\<br /> \vec \nabla f = \left\langle {2x,2y,2z} \right\rangle ;\vec \nabla g = \left\langle {4{x^3},4{y^3},4{z^3}} \right\rangle \\<br /> \left\langle {2x,2y,2z} \right\rangle = \lambda \left\langle {4{x^3},4{y^3},4{z^3}} \right\rangle \\<br /> 2{x^2} = 2{y^2} = 2{z^2} \to x = \pm y = \pm z \\<br /> 3{x^4} - 3 = 0 \to {x^4} = 1 \to x = \pm 1 \to y = \pm 1,z = \pm 1 \\<br /> \max = f\left( {1,1,1} \right) = f\left( {1,1, - 1} \right) = f\left( {1, - 1,1} \right) = f\left( {1, - 1, - 1} \right) = \\<br /> f\left( { - 1,1,1} \right) = f\left( { - 1,1, - 1} \right) = f\left( { - 1, - 1,1} \right) = f\left( { - 1, - 1, - 1} \right) = 3 \\ <br /> \end{gathered}

So I found the maximum but does the minimum exist?
 
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arcyqwerty said:

Homework Statement


Find max/min of x^2+y^2+z^2 given x^4+y^4+z^4=3


Homework Equations


Use of gradient vectors related by LaGrange Multiplier


The Attempt at a Solution


\begin{gathered}<br /> f\left( {x,y,z} \right) = {x^2} + {y^2} + {z^2};g\left( {x,y,z} \right) = {x^4} + {y^4} + {z^4} - 3 = 0 \\<br /> \vec \nabla f = \left\langle {2x,2y,2z} \right\rangle ;\vec \nabla g = \left\langle {4{x^3},4{y^3},4{z^3}} \right\rangle \\<br /> \left\langle {2x,2y,2z} \right\rangle = \lambda \left\langle {4{x^3},4{y^3},4{z^3}} \right\rangle \\<br /> 2{x^2} = 2{y^2} = 2{z^2} \to x = \pm y = \pm z \\<br /> 3{x^4} - 3 = 0 \to {x^4} = 1 \to x = \pm 1 \to y = \pm 1,z = \pm 1 \\<br /> \max = f\left( {1,1,1} \right) = f\left( {1,1, - 1} \right) = f\left( {1, - 1,1} \right) = f\left( {1, - 1, - 1} \right) = \\<br /> f\left( { - 1,1,1} \right) = f\left( { - 1,1, - 1} \right) = f\left( { - 1, - 1,1} \right) = f\left( { - 1, - 1, - 1} \right) = 3 \\ <br /> \end{gathered}

So I found the maximum but does the minimum exist?

Is the feasible set S = {(x,y,z): x^4 + y^4 + z^4 = 3} compact? Is the function f(x,y,z) = x^2 + y^2 + z^2 continuous on S? Have you heard of Weierstrass' Theorem?

RGV
 
Ray Vickson said:
Is the feasible set S = {(x,y,z): x^4 + y^4 + z^4 = 3} compact? Is the function f(x,y,z) = x^2 + y^2 + z^2 continuous on S? Have you heard of Weierstrass' Theorem?

RGV

I'm not quite sure what you mean by compact or Weierstrass' Theorem but I think that the function is continuous
 
arcyqwerty said:
I'm not quite sure what you mean by compact or Weierstrass' Theorem but I think that the function is continuous

Google is your friend.

RGV
 
Ray Vickson said:
Google is your friend.

RGV

So...
"A subset S of a topological space X is compact if for every open cover of S there exists a finite subcover of S."

Not quite sure what that means exactly, but perhaps its compact if there can be a finite subset of the points defined by the function?

And...
There seems to be two different Theorems, one about estimating functions with polynomials and another about sequence convergence...
 
arcyqwerty said:
So...
"A subset S of a topological space X is compact if for every open cover of S there exists a finite subcover of S."

Not quite sure what that means exactly, but perhaps its compact if there can be a finite subset of the points defined by the function?

And...
There seems to be two different Theorems, one about estimating functions with polynomials and another about sequence convergence...

If you keep searching you will eventually find a document in which all this is put into the context of ordinary 3-D space with the usual distance measure. In that case there is a theorem saying that a set is compact if and only if it is closed and bounded. So, is the set S closed (i.e., contains all its limit points)? Is it bounded? Then there is a theorem of Weierstrass saying that a continuous function on a compact set assumes both its maximum and its minimum. (These are theorems that are proven in advanced Calculus classes, well before 'topology'.) So, in your case the answer is YES: S is compact, and f has a minimum on S, as well as a maximum. None of this helps you *find* the minimum, but it does tell you that the search makes sense.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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