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mathmath333
 one year ago
find the maximum value of
mathmath333
 one year ago
find the maximum value of

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0find the maximum value of \(\large \color{black}{\begin{align} (a+1)\times (b+2)\times c\hspace{.33em}\\~\\ \end{align}}\) if \(\large \color{black}{\begin{align} a+b+c=12\hspace{.33em}\\~\\ \end{align}}\) where \(\large \color{black}{\begin{align} \{a,b,c\}\in \mathbb{R},\ \{a,b,c\}>0\hspace{.33em}\\~\\ \end{align}}\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0what do you think as a start

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0lol idk i m new to the topic

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0hmm still thinking!

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0this one ? \(\large \color{black}{\begin{align} \dfrac{a+b+c}{3}\geq (abc)^{1/3}\hspace{.33em}\\~\\ \end{align}}\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} &\dfrac{a+b+c}{3}\geq (abc)^{1/3}\hspace{.33em}\\~\\ &\dfrac{12}{3}\geq (abc)^{1/3}\hspace{.33em}\\~\\ &4\geq (abc)^{1/3}\hspace{.33em}\\~\\ &64\geq abc \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well you want \((a+1)(b+2)c\) in the RHS

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what you gonna do for that

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0is amgm inequality enough ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0replace \(a\), \(b\) and \(c\) in the AMGM with desired ones

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i get this \(64+2ac+bc+2c\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0please, trey to apply the lagrange multipliers method to this function: \[G\left( {x,y,z;\lambda } \right) = \left( {x + 1} \right)\left( {y + 2} \right)z  \lambda \left( {x + y + z  12} \right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0emm, idk, how did you get that? What I meant was this\[\frac{a+1+b+2+c}{3}\ge ((a+1)(b+2)c)^{1/3}\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i was thinking of calculus the first sight too but can't use that since math has not taken that yet

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which gives us:\[(a+1)(b+2)c \le \left(\frac{15}{3} \right)^3=125\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0wow! very neat

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0equality occurs when \(a+1=b+2=c=5\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lagrange multipliers will work if you familiar with :)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0idk lagrage multiplers

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0cool AMGM turned to be very useful!

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0can we do this without that! a different alternative but no calculus

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there must be some other methods, I can't think of other one now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0LOl i shared this on facebook

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0eh can't think of something else!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let's see if this leads us somewhere

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(a+1)(b+2)c=(a+1)(14ac)c=ca^2+(13cc^2)a+14cc^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which is a downward quadratic (leading coefficient is negative) and maximum occurs at \[a=\frac{13cc^2}{2c}=\frac{13c}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now put this back in the quadratic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0quadratic simplifies to\[\frac{1}{4} (c15)^2 c\]

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i got new question pls help me there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0local maximum of this function occurs at \(c=5\) which is \(125\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok @mathmath333 , we're done

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0second way is much more elementary :)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0oh this has beat me! i was doing that but didn't not think of quadratic!
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