Math please help

Find the smallest value on function (3/x)+(4/1-x) on interval (0,1)

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asked by Sarah
  1. let y = 3/x + 4(1-x) = 3/x + 4(1-x)^-1
    dy/dx = -3/x^2 -4(1-x)^-2 (-1)
    = -3/x^2 + 4/(1-x)^2
    = 0 for a max/min of y

    4/(1-x)^2 = 3/x^2
    4x^2 = 3(1-x)^2
    2x = ± √3(1-x)

    2x = √3 - √3x
    2x + √3x = √3
    x = √3/(2 + √3) = appr .4641
    or
    2x = -√3 + √3x
    x = -√3/(2 - √3) = appr -6.464 , outside of your domain

    for x = √3/(2+√3) = 2√3 - 3 after rationalizing
    y = 2/(2√3-3) + 4/(1 - 2√3 + 3)
    = 2/(2√3-3) + 4/(4 - 2√3)
    = appr 13.93

    confirmed by Wolfram
    http://www.wolframalpha.com/input/?i=minimum+y+%3D+3%2Fx+%2B+4%2F%281-x%29

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    posted by Reiny

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