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Thread: Differentiation Problem!

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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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    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.

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

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

    Therefore $\displaystyle u + v = 1$ and $\displaystyle uv = 20$.

    Solving these equations simultaneously gives...

    $\displaystyle u = 1 - v$

    $\displaystyle (1 - v)v = 20$

    $\displaystyle v - v^2 = 20$

    $\displaystyle 0 = v^2 - v + 20$

    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
    $\displaystyle \frac{1}{u} + \frac{1}{v} = \frac{1}{20}$

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

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

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

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

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

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

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

    Solving these equations simultaneously gives...

    $\displaystyle u = 1 - v$

    $\displaystyle (1 - v)v = 20$

    $\displaystyle v - v^2 = 20$

    $\displaystyle 0 = v^2 - v + 20$

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