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mathmath333
 one year ago
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mathmath333
 one year ago
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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1if \(\large x,y,z \in \mathbb{R^{>0}}\) 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 ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} \dfrac{y}{z}=\dfrac14\hspace{.33em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hint: equality of AMGM occures when\[\large \color{black}{\begin{align} \dfrac{x}{y}=\dfrac{z}{12x}=\dfrac{4y}{x}=\dfrac{x}{3z}\hspace{.33em}\\~\\ \end{align}}\]

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i didnt understand

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For finding minimum of that expression you must use AMGM and in the AMGM 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\[a_1=a_2=...=a_n\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Ahh nice xD so @mukushla is \(\large 4*\sqrt[4]{9}\) the minimum value of given expression ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hey, thanks, math am I clear?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1but \(\large \color{black}{\begin{align} & \dfrac{x}{y}=\dfrac{z}{12x}=\dfrac{4y}{x}=\dfrac{x}{3z} \hspace{1.5em}\\~\\ & \implies \dfrac{x}{y}=\dfrac{x}{3z} \hspace{1.5em}\\~\\ & \implies \dfrac{y}{z}=\dfrac{3}{1} \hspace{1.5em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Jesus! what happened? I found \(z=2x\) and \( y=\frac{1}{2} x\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1is 3/1 correct and the given 1/4 wrong ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hey, 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\[\frac{y}{z}=\frac {1}{4} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think both of them can be, because you have 3 unknowns and a lot of equations, problem is not right.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1ok so question is incorrect ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because we got\[\frac{y}{z}=\frac{1}{4}=3\]which is a contradiction

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1u 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh no! all of them must be equal :)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} \dfrac{\dfrac{x}{y}+\dfrac{4y}{x}}{2}\geq \sqrt{\dfrac{x}{y}\times \dfrac{4y}{x}} \hspace{1.5em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sry, I gotta go, we'll talk about this later ;)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, math, what was your reasoning?

acxbox22
 one year ago
Best ResponseYou've already chosen the best response.0these questions are so confusing...

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i think it should be this \(\large \color{black}{\begin{align} \dfrac{x}{y}=\dfrac{4y}{x}, \dfrac{x}{3z}=\dfrac{z}{12x} \hspace{1.5em}\\~\\ \end{align}}\) ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1because we can apply now AM and GM equality now for theese for finding minimum \(\large \color{black}{\begin{align} \dfrac{\dfrac{x}{y}+\dfrac{4y}{x}}{2}\geq \sqrt{\dfrac{x}{y}\times \dfrac{4y}{x}} \hspace{1.5em}\\~\\ \end{align}}\) \(\large \color{black}{\begin{align} \dfrac{\dfrac{x}{3x}+\dfrac{z}{12x}}{2}\geq \sqrt{\dfrac{x}{3z}\times \dfrac{z}{12x}} \hspace{1.5em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0those 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

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1but 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My god, what's the matter with wolfram today?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0AMGM gives the minimum as\[\frac{4\sqrt{3}}{3}\]we must find the problem here!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think you are right math, maybe we can't use AMGM at all, because that four fractions can not be equal. They can't be same simultaneously.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What I learned here was that we must be careful when using AMGM and actually I still have some doubts.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what do you think guys?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1we 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

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1@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?

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1BTW, justification for equating terms can be found here  https://en.wikipedia.org/wiki/Geometric_mean#Calculation
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