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Math Help - Maxima problem

  1. #1
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    Maxima problem

    The variables u and v take positive values such that \frac{1}{u}+\frac{1}{v}=\frac{1}{20}. If s=u+v, find the least value of s as u varies.

    Since they want the least value of s, my guess is that differentiating s and equating that equation to zero will get me the answer. but i cant figure of anything to differentiate s with respect to
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  2. #2
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    Well, you will have to clearly make use of the first equation. Notice that

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

    Now you want to find the least value of s as u varies. Stated differently, I would interpret that as finding the least value of s while v is held constant (partial differentiation?). We already have s as a function of u and v. So I think this translation is sound, no? See if that helps you figure out the problem.
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  3. #3
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    Quote Originally Posted by Punch View Post
    The variables u and v take positive values such that \frac{1}{u}+\frac{1}{v}=\frac{1}{20}. If s=u+v, find the least value of s as u varies.

    Since they want the least value of s, my guess is that differentiating s and equating that equation to zero will get me the answer. but i cant figure of anything to differentiate s with respect to
    I would set up this problem by solving the first equation for v.

    \frac{1}{u}+\frac{1}{v}=\frac{1}{20} \iff 20v+20u=uv \iff v=\frac{20u}{u-20}

    Now plug this into the equation for s to get

    s=u+\frac{20u}{u-20}

    Now you can use single variable calculus to minimize the function.
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  4. #4
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    On second thought, I would think about solving the first equation (the restrictions on u and v) in terms of v. This should help a lot more!

    Late Edit: Yeah, like he said lol
    Last edited by bryangoodrich; June 3rd 2011 at 08:39 PM. Reason: Reviewed this question a little late
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  5. #5
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    so with that equation, all i have to do now is ds/du=0, right?
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  6. #6
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    Quote Originally Posted by Punch View Post
    so with that equation, all i have to do now is ds/du=0, right?
    Yes that is correct, but watch out for extraneous solutions.
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  7. #7
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    I managed to form the equation s=u+\frac{20u}{20-u} with your help.

    Then, \frac{ds}{du}=1+\frac{(20-u)(20)-(20u)(-1)}{(20-u)^2}

    =1+\frac{20(20-u)+20u}{(20-u)^2}

    \frac{ds}{du}=0,    1+\frac{20(20-u)+20u}{(20-u)^2}=0

    \frac{400-20u+20u}{(20-u)^2}=-1

    400=-(20-u)^2

    400=-400+40u-u^2

    u^2-40u+800=0

    but i couldn't find u because u^2-40u+800 has no roots.
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    So you have

    s=u+\frac{20u}{u-20}=\frac{u^2}{u-20} \right)

    Now if you use the product rule you get or quotient rule

    s'=\frac{2u}{(u-20)}-\frac{u^2}{(u-20)^2}=\frac{u}{u-20}\left(2-\frac{u}{u-20} \right)

    Can you finish from here?
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  9. #9
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    This is how I would do it

    Plug in values, find the minimum s
    s = u + v
    1/u + 1/v = 1/20

    1/40 + 1/40 = 1/20
    s=80

    1/30 + 1/60 = 1/20
    s=90

    S keeps growing, the v will be > 80 (your smallest value of s so far)

    1/25 + 1/100 = 1/20
    s=125

    1/24 + 1/120 = 1/20
    s=144

    1/22 + 1/220 = 1/20
    s=242

    Etc. Your min s is where u=v so 80.
    Last edited by Stro; June 3rd 2011 at 09:46 PM.
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  10. #10
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    Quote Originally Posted by TheEmptySet View Post
    So you have

    s=u+\frac{20u}{u-20}=\frac{u^2}{u-20} \right)

    Now if you use the product rule you get or quotient rule

    s'=\frac{2u}{(u-20)}-\frac{u^2}{(u-20)^2}=\frac{u}{u-20}\left(2-\frac{u}{u-20} \right)

    Can you finish from here?
    \frac{ds}{du}=0,    1+\frac{20(20-u)+20u}{(20-u)^2}=0

    \frac{400-20u+20u}{(20-u)^2}=-1

    400=-(20-u)^2

    400=-400+40u-u^2

    u^2-40u+800=0

    But was what i did wrong?
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  11. #11
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    Quote Originally Posted by Punch View Post
    \frac{ds}{du}=0,    1+\frac{20(20-u)+20u}{(20-u)^2}=0

    \frac{400-20u+20u}{(20-u)^2}=-1

    400=-(20-u)^2

    400=-400+40u-u^2

    u^2-40u+800=0

    But was what i did wrong?
    20-u \ne u-20

    1+\frac{20(u-20)-20u}{(u-20)^2}=0

    \frac{(u-20)^2-400}{(u-20)^2}=\frac{u(u-40)}{(u-20)^2}
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  12. #12
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    Starting with s=\frac{u^2}{u-20}

    s'=\frac{(u-20)(2u)-u^2}{(u-20)^2}

    =\frac{u^2-40u}{(u-20)^2}

    s'=0, \frac{u^2-40u}{(u-20)^2}=0

    u(u-40)=0

    u=0(reject) or u=40

    but i dont have the value of v, so how am i to find s?
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  13. #13
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    Quote Originally Posted by Punch View Post
    Starting with s=\frac{u^2}{u-20}

    s'=\frac{(u-20)(2u)-u^2}{(u-20)^2}

    =\frac{u^2-40u}{(u-20)^2}

    s'=0, \frac{u^2-40u}{(u-20)^2}=0

    u(u-40)=0

    u=0(reject) or u=40

    but i dont have the value of v, so how am i to find s?
    Use the constraint equation.

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

    \frac{1}{40}+\frac{1}{v}=\frac{1}{20}

    Now solve for v.
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  14. #14
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    For a completed proof, we were given

    s = u + v,\ u>0,\ v>0,\  \frac{1}{u} + \frac{1}{v} = \frac{1}{20}

    By solving the constraint for v we get

    Spoiler:
    u + v = \frac{uv}{20} \iff 20u + 20v = uv \iff v = \frac{20u}{u - 20}


    We can then restate s with this substitution:

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

    Let us take the derivative of this with respect to u. This is a straight application of the product rule with two functions:

    f(x) = x^2\ \mathrm{and}\ g(x) = (x-20)^{-1}

    Each of which has derivatives:

    f'(x) = 2x\ \mathrm{and}\ g'(x) = -(x-2)^{-2}\times 1

    Using the product rule we get,

    Spoiler:
    s'= f' (u)g(u) + f(u)g' (u) = 2u(u-20)^{-1} - u^2(u-20)^{-2}


    Did you get this result? Now let us put it back into a more manageable form.

    s' = \frac{2u}{u-20} - \frac{u^2}{(u-20)^2} = \frac{2u(u-20) - u^2}{(u-20)^2}=\frac{u^2 - 40u}{(u-20)^2}

    Now this is where you need to simplify, set it equal to 0, notice that you can eliminate the denominator, and see the easy solution.

    Spoiler:
    s' = \frac{u(u-40)}{(u-20)^2} \stackrel{set}{=} 0 \iff u(u-40) = 0 \iff u = 0 \mathrm{\ or\ } u = 40


    But as emptyset pointed out, we have to be careful of our solutions. If u = 0 we have a problem with our original constraint that u > 0. Therefore, we have found our single extrema. We can also find the value of v under this value of u because we have a statement of v in terms of u. However, we have s in terms of u, so we can find s by directly substituting u.

    Spoiler:
    s = \frac{u^2}{u-20} = \frac{40^2}{40 - 20} = \frac{1600}{20} = 80,\ \ v = \frac{20u}{40-20} = u = 40
    Last edited by bryangoodrich; June 4th 2011 at 12:08 AM. Reason: Included spoilers and answer and cleaned up latex
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