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Finding the minimum value that an expression can take.

  1. Jan 11, 2010 #1
    I've recently come across a http://technologyinterface.nmsu.edu/Spring08/" [Broken] for why power is maximised in a component when its resistance [tex]R_L[/tex] equals the internal resistance [tex]R[/tex].
    But in part (5) of method 1, we need to find the minimum non-negative value that the expression [tex]k+k^-^1[/tex] can take. For this they use the inequality [tex](k-1)^2 \geq 0[/tex] and then expand it to [tex]k^2-2k+1 \geq 0 [/tex] which is then in turn divided by [tex]k [/tex] and rearranged to get [tex]k+k^-^1 \geq 2[/tex]
    The problem I have grasping is the part where they come up with the inequality to solve this problem. This is a step I have never come across, and I was wondering if this method had a name.

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 11, 2010 #2


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    Are you asking a physics question, i.e. where did the question about k + 1/k come from?
    The math proof ( ≥ 2) is simple enough (assuming k > 0).
  4. Jan 11, 2010 #3
  5. Jan 11, 2010 #4


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    Use that the arithmetic mean is larger than the geometric mean.
  6. Jan 12, 2010 #5


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    (k-1)2 ≥ 0

    expand quadratic

    k2 -2k + 1 ≥ 0

    divide through by k (need k > 0)

    k - 2 + 1/k ≥ 0

    add 2 to both sides

    k + 1/k ≥ 2
  7. Jan 12, 2010 #6


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    If you're asking where the initial inequality comes from, any real number squared is always greater than or equal to zero. That's the reason for saying [itex](k-1)^2 \geq 0[/itex].
    Last edited by a moderator: May 4, 2017
  8. Jan 14, 2010 #7
    Yes, but is this a matter of just thinking of a quadratic inequality that when manipulated will give desired terms k and 1/k?
  9. Jan 14, 2010 #8


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    The last two replies seem to make it as clear as possible. It is a trivial derivation. What don't you get?
  10. Jan 15, 2010 #9
    I know the inequality can be used to find the minimum and how to manipulate it. My problem is how to actually come up with an inequality that will be of use in the first place, say I have [tex](k-2)^{2}\geq0[/tex] - it's not going to be useful for our purposes when manipulated. Well, obviously one can see that the first (k-1) inequality will be the one to choose. I want to know whether this can be generalised for any quadratic inequality where you want to find not just x is more than/less than a; but x and another non constant term.
    It just seems to me that some steps are being skipped in the paper.
  11. Jan 15, 2010 #10


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    You can generalize with (k-n)2 ≥ 0 to get k + n2/k ≥ 2n.

    Where it might be used I don't know.
  12. Jan 15, 2010 #11
    Thank you, I've gotten what I need to know.
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