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

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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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