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Max/min with constraints

  1. Oct 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Find max/min of x^2+y^2+z^2 given x^4+y^4+z^4=3


    2. Relevant equations
    Use of gradient vectors related by LaGrange Multiplier


    3. The attempt at a solution
    [tex]\begin{gathered}
    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 \\
    \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 \\
    \left\langle {2x,2y,2z} \right\rangle = \lambda \left\langle {4{x^3},4{y^3},4{z^3}} \right\rangle \\
    2{x^2} = 2{y^2} = 2{z^2} \to x = \pm y = \pm z \\
    3{x^4} - 3 = 0 \to {x^4} = 1 \to x = \pm 1 \to y = \pm 1,z = \pm 1 \\
    \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) = \\
    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 \\
    \end{gathered}[/tex]

    So I found the maximum but does the minimum exist?
     
  2. jcsd
  3. Oct 27, 2011 #2

    Ray Vickson

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    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
     
  4. Oct 27, 2011 #3
    I'm not quite sure what you mean by compact or Weierstrass' Theorem but I think that the function is continuous
     
  5. Oct 27, 2011 #4

    Ray Vickson

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    Google is your friend.

    RGV
     
  6. Oct 27, 2011 #5
    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...
     
  7. Oct 27, 2011 #6

    Ray Vickson

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    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
     
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