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Calculus - How to minimize x^2 + (1/x^2)

  1. May 16, 2014 #1
    1. The problem statement, all variables and given/known data

    Find a positive number such that the sum of the square of the number and its reciprocal is a minimum.

    2. Relevant equations



    3. The attempt at a solution
    y=X^2+(1/X^2)
    Derivative=2x-2x^(-3)
    0=2x-2x^3
    2/x^3=2x
    2=2x^4
    1=x^4
    x=1

    The answer I got was 1 but I'm not sure if this is right. Can someone please tell me if I did it right or wrong. If wrong, can you point out my mistake. Thanks :)
     
  2. jcsd
  3. May 16, 2014 #2

    vela

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    [strike]Looks good except that you missed a possible solution. What other value of x satisfies x^4=1?[/strike]

    Oops, never mind. I didn't notice you were looking for a positive number.

    You just now need to verify that the critical point corresponds to a minimum.
     
  4. May 16, 2014 #3
    Is it possible for the minimum to be a smaller value or would the smallest value be 1?
     
  5. May 16, 2014 #4

    LCKurtz

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    If the graph is concave up for ##x>0## your relative minimum will be an absolute minimum on that interval. Have you checked that?
     
  6. May 16, 2014 #5

    Ray Vickson

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    Your function is strictly convex in the convex region {x > 0}, so any stationary point is automatically a global minimum within the region.
     
  7. May 16, 2014 #6
    Oh ok so since it is absolute it would be 1 right?
     
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