Math Problem?

Find a positive number x such that the sum of 16x and 1/x is as small as 

possible

Does this require an open or closed interval?

3 Answers

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  • atsuo
    Lv 6
    4 months ago

    x is positive, so x = (0,inf) ← open interval

    16x + 1/x

    = (4√x - 1/√x)^2 + 8

    Clearly (4√x - 1/√x)^2 has its minimum value 0 when

    4√x - 1/√x = 0

    4x - 1 = 0

    x = 1/4

    So 16x + 1/x has its minimum value 8 when x = 1/4.

  • sepia
    Lv 7
    4 months ago

    Find a positive number x such that 

    the sum of 16x and 1/x is as small as possible.

    y = 16x + 1/x

    dy/dx = 16 - 1/x²

    dy/dx = 0 when 16 -1/x² = 0 that's when x² = 1/16, i.e. when x = 1/4.

    When x = 1/4, the sum of 16 x and 1/x is 8.

    That's the minimum value over the open interval x > 0.

  • Bryce
    Lv 7
    4 months ago

    y'= 16 - 1/x²= 0

    x= 1/4

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