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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    if \(\large x,y,z \in \mathbb{R^{>0}}\) for what ratio of \(\large y\) and \(\large z\) is the value of \(\large \color{black}{\begin{align} \left(\dfrac{x}{y}+\dfrac{z}{12x}+\dfrac{4y}{x}+\dfrac{x}{3z}\right)\hspace{.33em}\\~\\ \end{align}}\) minimum ?

  2. ganeshie8
    • one year ago
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    is the answer 1 ?

  3. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} \dfrac{y}{z}=\dfrac14\hspace{.33em}\\~\\ \end{align}}\)

  4. ikram002p
    • one year ago
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    *

  5. anonymous
    • one year ago
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    hint: equality of AM-GM occures when\[\large \color{black}{\begin{align} \dfrac{x}{y}=\dfrac{z}{12x}=\dfrac{4y}{x}=\dfrac{x}{3z}\hspace{.33em}\\~\\ \end{align}}\]

  6. mathmath333
    • one year ago
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    i didnt understand

  7. anonymous
    • one year ago
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    For finding minimum of that expression you must use AM-GM and in the AM-GM minimum occurs when all of numbers are equal, In other words for positive numbers \(a_1\), \(a_2\), ... and \(a_n\) we have:\[\frac{a_1+a_2+...+a_n}{n} \ge (a_1 a_2...a_n)^{1/n}\]minimum state or equality holds if and only if\[a_1=a_2=...=a_n\]

  8. ganeshie8
    • one year ago
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    Ahh nice xD so @mukushla is \(\large 4*\sqrt[4]{9}\) the minimum value of given expression ?

  9. anonymous
    • one year ago
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    oh yeah

  10. ganeshie8
    • one year ago
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    brilliant!!

  11. anonymous
    • one year ago
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    hey, thanks, math am I clear?

  12. mathmath333
    • one year ago
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    but \(\large \color{black}{\begin{align} & \dfrac{x}{y}=\dfrac{z}{12x}=\dfrac{4y}{x}=\dfrac{x}{3z} \hspace{1.5em}\\~\\ & \implies \dfrac{x}{y}=\dfrac{x}{3z} \hspace{1.5em}\\~\\ & \implies \dfrac{y}{z}=\dfrac{3}{1} \hspace{1.5em}\\~\\ \end{align}}\)

  13. anonymous
    • one year ago
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    Jesus! what happened? I found \(z=2x\) and \( y=\frac{1}{2} x\)

  14. mathmath333
    • one year ago
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    is 3/1 correct and the given 1/4 wrong ?

  15. anonymous
    • one year ago
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    hey, I'm a little bit confused, what you did was right, but from mine:\[\frac{x}{y}=\frac{4y}{x}\]which gives \(y=\frac{1}{2}x\)\[\frac{z}{12x}=\frac{x}{3z} \]\[z=2x\]so\[\frac{y}{z}=\frac {1}{4} \]

  16. anonymous
    • one year ago
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    I think both of them can be, because you have 3 unknowns and a lot of equations, problem is not right.

  17. mathmath333
    • one year ago
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    ok so question is incorrect ?

  18. anonymous
    • one year ago
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    I think it is

  19. anonymous
    • one year ago
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    because we got\[\frac{y}{z}=\frac{1}{4}=3\]which is a contradiction

  20. mathmath333
    • one year ago
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    u equated all four terms i think it should be this only \(\large \color{black}{\begin{align} \dfrac{x}{y}=\dfrac{4y}{x}, \dfrac{x}{3z}=\dfrac{z}{12x} \hspace{1.5em}\\~\\ \end{align}}\) beacuse the AM and GM inequality will hold for that as they are reciprocals

  21. mathmath333
    • one year ago
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    @mukushla

  22. anonymous
    • one year ago
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    oh no! all of them must be equal :-)

  23. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} \dfrac{\dfrac{x}{y}+\dfrac{4y}{x}}{2}\geq \sqrt{\dfrac{x}{y}\times \dfrac{4y}{x}} \hspace{1.5em}\\~\\ \end{align}}\)

  24. mathmath333
    • one year ago
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    D:

  25. anonymous
    • one year ago
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    sry, I gotta go, we'll talk about this later ;-)

  26. anonymous
    • one year ago
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    ok, math, what was your reasoning?

  27. acxbox22
    • one year ago
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    these questions are so confusing...

  28. mathmath333
    • one year ago
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    i think it should be this \(\large \color{black}{\begin{align} \dfrac{x}{y}=\dfrac{4y}{x}, \dfrac{x}{3z}=\dfrac{z}{12x} \hspace{1.5em}\\~\\ \end{align}}\) ?

  29. mathmath333
    • one year ago
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    because we can apply now AM and GM equality now for theese for finding minimum \(\large \color{black}{\begin{align} \dfrac{\dfrac{x}{y}+\dfrac{4y}{x}}{2}\geq \sqrt{\dfrac{x}{y}\times \dfrac{4y}{x}} \hspace{1.5em}\\~\\ \end{align}}\) \(\large \color{black}{\begin{align} \dfrac{\dfrac{x}{3x}+\dfrac{z}{12x}}{2}\geq \sqrt{\dfrac{x}{3z}\times \dfrac{z}{12x}} \hspace{1.5em}\\~\\ \end{align}}\)

  30. anonymous
    • one year ago
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    those to inequalities does not guarantee that \[\large \color{black}{\begin{align} \dfrac{x}{y}+\dfrac{z}{12x}+\dfrac{4y}{x}+\dfrac{x}{3z}\hspace{.33em}\\~\\ \end{align}}\]will bw minimum

  31. mathmath333
    • one year ago
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    but wolfram gives this which says \(x=2y,z=2x,\implies \dfrac yz=\dfrac14 \) http://www.wolframalpha.com/input/?i=minimize+%5Cdfrac%7Bx%7D%7By%7D%2B%5Cdfrac%7Bz%7D%7B12x%7D%2B%5Cdfrac%7B4y%7D%7Bx%7D%2B%5Cdfrac%7Bx%7D%7B3z%7D%2Cx%3E0%2Cy%3E0%2Cz%3E0

  32. anonymous
    • one year ago
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    My god, what's the matter with wolfram today?

  33. anonymous
    • one year ago
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    AM-GM gives the minimum as\[\frac{4\sqrt{3}}{3}\]we must find the problem here!

  34. anonymous
    • one year ago
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    I think you are right math, maybe we can't use AM-GM at all, because that four fractions can not be equal. They can't be same simultaneously.

  35. anonymous
    • one year ago
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    What I learned here was that we must be careful when using AM-GM and actually I still have some doubts.

  36. anonymous
    • one year ago
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    what do you think guys?

  37. mathmath333
    • one year ago
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    we apply AM GM when the variable is only one usually at numerator, here it has 2 variables such as x/y ,y/z in numerator as well as denominator

  38. ybarrap
    • one year ago
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    @makushla , I think your strategy is valid, the problem is that the terms are not independent. To resolve, let's assign some variables first $$ x/y+z/12x+4y/x+x/3z\\ =a+b/12+4/a+1/3b $$ Where a=x/y b=z/x Now we equate the \(dependent\) terms following your strategy: $$ 4/a=a\\ \implies a=\pm2\\ b/12=1/3b\\ \implies b=\pm 2 $$ Since we are using positive reals, we keep only positive a and b. This means that $$ x/y=2\\ x=2y\\ z/x=2\\ z=2x=4y\\ \implies y/z=1/4 $$ How does this sound?

  39. ybarrap
    • one year ago
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    BTW, justification for equating terms can be found here - https://en.wikipedia.org/wiki/Geometric_mean#Calculation

  40. anonymous
    • one year ago
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    Thanks @ybarrap

  41. ybarrap
    • one year ago
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    You're welcome :)

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