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Math Help - Differentiation Problem!

  1. #1
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    Differentiation Problem!

    The variables u and v take positive values such that 1/u + 1/v = 1/20. If s = u + v, find the least value of s as u varies. Thanks
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  2. #2
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    Quote Originally Posted by puggie View Post
    The variables u and v take positive values such that 1/u + 1/v = 1/20. If s = u + v, find the least value of s as u varies. Thanks
    I don't really see how this is a differentiation problem.

    \frac{1}{u} + \frac{1}{v} = \frac{u + v}{uv}.

    So \frac{u + v}{uv} = \frac{1}{20}.

    Therefore u + v = 1 and uv = 20.

    Solving these equations simultaneously gives...

    u = 1 - v

    (1 - v)v = 20

    v - v^2 = 20

    0 = v^2 - v + 20

    This value does not have a solution... (check it's discriminant).
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  3. #3
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    Quote Originally Posted by puggie View Post
    The variables u and v take positive values such that 1/u + 1/v = 1/20. If s = u + v, find the least value of s as u varies. Thanks
    \frac{1}{u} + \frac{1}{v} = \frac{1}{20}

    \frac{uv}{u+v} = 20

    \frac{u(s-u)}{s} = 20

    s = \frac{u^2}{u - 20}

    find \frac{ds}{du} and minimize
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  4. #4
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    Quote Originally Posted by Prove It View Post
    I don't really see how this is a differentiation problem.

    \frac{1}{u} + \frac{1}{v} = \frac{u + v}{uv}.

    So \frac{u + v}{uv} = \frac{1}{20}.

    Therefore u + v = 1 and uv = 20.
    You are arguing that because a/b= c/d, "therefore" a= c and b= d. That is not true! 2/4= 1/2 but 2\ne 1 and 4\ne 2.

    Solving these equations simultaneously gives...

    u = 1 - v

    (1 - v)v = 20

    v - v^2 = 20

    0 = v^2 - v + 20

    This value does not have a solution... (check it's discriminant).
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