mathmath333
  • mathmath333
Question
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

mathmath333
  • mathmath333
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 ?
ganeshie8
  • ganeshie8
is the answer 1 ?
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} \dfrac{y}{z}=\dfrac14\hspace{.33em}\\~\\ \end{align}}\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ikram002p
  • ikram002p
*
anonymous
  • anonymous
hint: equality of AM-GM occures when\[\large \color{black}{\begin{align} \dfrac{x}{y}=\dfrac{z}{12x}=\dfrac{4y}{x}=\dfrac{x}{3z}\hspace{.33em}\\~\\ \end{align}}\]
mathmath333
  • mathmath333
i didnt understand
anonymous
  • anonymous
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\[a_1=a_2=...=a_n\]
ganeshie8
  • ganeshie8
Ahh nice xD so @mukushla is \(\large 4*\sqrt[4]{9}\) the minimum value of given expression ?
anonymous
  • anonymous
oh yeah
ganeshie8
  • ganeshie8
brilliant!!
anonymous
  • anonymous
hey, thanks, math am I clear?
mathmath333
  • mathmath333
but \(\large \color{black}{\begin{align} & \dfrac{x}{y}=\dfrac{z}{12x}=\dfrac{4y}{x}=\dfrac{x}{3z} \hspace{1.5em}\\~\\ & \implies \dfrac{x}{y}=\dfrac{x}{3z} \hspace{1.5em}\\~\\ & \implies \dfrac{y}{z}=\dfrac{3}{1} \hspace{1.5em}\\~\\ \end{align}}\)
anonymous
  • anonymous
Jesus! what happened? I found \(z=2x\) and \( y=\frac{1}{2} x\)
mathmath333
  • mathmath333
is 3/1 correct and the given 1/4 wrong ?
anonymous
  • anonymous
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\[\frac{y}{z}=\frac {1}{4} \]
anonymous
  • anonymous
I think both of them can be, because you have 3 unknowns and a lot of equations, problem is not right.
mathmath333
  • mathmath333
ok so question is incorrect ?
anonymous
  • anonymous
I think it is
anonymous
  • anonymous
because we got\[\frac{y}{z}=\frac{1}{4}=3\]which is a contradiction
mathmath333
  • mathmath333
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
mathmath333
  • mathmath333
anonymous
  • anonymous
oh no! all of them must be equal :-)
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} \dfrac{\dfrac{x}{y}+\dfrac{4y}{x}}{2}\geq \sqrt{\dfrac{x}{y}\times \dfrac{4y}{x}} \hspace{1.5em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
D:
anonymous
  • anonymous
sry, I gotta go, we'll talk about this later ;-)
anonymous
  • anonymous
ok, math, what was your reasoning?
acxbox22
  • acxbox22
these questions are so confusing...
mathmath333
  • mathmath333
i think it should be this \(\large \color{black}{\begin{align} \dfrac{x}{y}=\dfrac{4y}{x}, \dfrac{x}{3z}=\dfrac{z}{12x} \hspace{1.5em}\\~\\ \end{align}}\) ?
mathmath333
  • mathmath333
because we can apply now AM and GM equality now for theese for finding minimum \(\large \color{black}{\begin{align} \dfrac{\dfrac{x}{y}+\dfrac{4y}{x}}{2}\geq \sqrt{\dfrac{x}{y}\times \dfrac{4y}{x}} \hspace{1.5em}\\~\\ \end{align}}\) \(\large \color{black}{\begin{align} \dfrac{\dfrac{x}{3x}+\dfrac{z}{12x}}{2}\geq \sqrt{\dfrac{x}{3z}\times \dfrac{z}{12x}} \hspace{1.5em}\\~\\ \end{align}}\)
anonymous
  • anonymous
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
mathmath333
  • mathmath333
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
anonymous
  • anonymous
My god, what's the matter with wolfram today?
anonymous
  • anonymous
AM-GM gives the minimum as\[\frac{4\sqrt{3}}{3}\]we must find the problem here!
anonymous
  • anonymous
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.
anonymous
  • anonymous
What I learned here was that we must be careful when using AM-GM and actually I still have some doubts.
anonymous
  • anonymous
what do you think guys?
mathmath333
  • mathmath333
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
ybarrap
  • ybarrap
@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
  • ybarrap
BTW, justification for equating terms can be found here - https://en.wikipedia.org/wiki/Geometric_mean#Calculation
anonymous
  • anonymous
Thanks @ybarrap
ybarrap
  • ybarrap
You're welcome :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.