mathmath333
  • mathmath333
find the maximum value of
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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mathmath333
  • mathmath333
find 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
  • xapproachesinfinity
what do you think as a start
mathmath333
  • mathmath333
lol idk i m new to the topic

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xapproachesinfinity
  • xapproachesinfinity
hmm still thinking!
anonymous
  • anonymous
try AM-GM
mathmath333
  • mathmath333
this one ? \(\large \color{black}{\begin{align} \dfrac{a+b+c}{3}\geq (abc)^{1/3}\hspace{.33em}\\~\\ \end{align}}\)
anonymous
  • anonymous
yes
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} &\dfrac{a+b+c}{3}\geq (abc)^{1/3}\hspace{.33em}\\~\\ &\dfrac{12}{3}\geq (abc)^{1/3}\hspace{.33em}\\~\\ &4\geq (abc)^{1/3}\hspace{.33em}\\~\\ &64\geq abc \end{align}}\)
anonymous
  • anonymous
well you want \((a+1)(b+2)c\) in the RHS
anonymous
  • anonymous
what you gonna do for that
mathmath333
  • mathmath333
is am-gm inequality enough ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
replace \(a\), \(b\) and \(c\) in the AM-GM with desired ones
anonymous
  • anonymous
what did you get?
mathmath333
  • mathmath333
i get this \(64+2ac+bc+2c\)
Michele_Laino
  • Michele_Laino
please, trey to apply the lagrange multipliers method to this function: \[G\left( {x,y,z;\lambda } \right) = \left( {x + 1} \right)\left( {y + 2} \right)z - \lambda \left( {x + y + z - 12} \right)\]
Michele_Laino
  • Michele_Laino
try*
anonymous
  • anonymous
emm, 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
  • xapproachesinfinity
i was thinking of calculus the first sight too but can't use that since math has not taken that yet
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
which gives us:\[(a+1)(b+2)c \le \left(\frac{15}{3} \right)^3=125\]
xapproachesinfinity
  • xapproachesinfinity
wow! very neat
anonymous
  • anonymous
equality occurs when \(a+1=b+2=c=5\)
mathmath333
  • mathmath333
thnks
anonymous
  • anonymous
np
anonymous
  • anonymous
Lagrange multipliers will work if you familiar with :-)
mathmath333
  • mathmath333
idk lagrage multiplers
anonymous
  • anonymous
ok!
anonymous
  • anonymous
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xapproachesinfinity
  • xapproachesinfinity
cool AM-GM turned to be very useful!
xapproachesinfinity
  • xapproachesinfinity
can we do this without that! a different alternative but no calculus
anonymous
  • anonymous
there must be some other methods, I can't think of other one now
anonymous
  • anonymous
LOl i shared this on facebook
xapproachesinfinity
  • xapproachesinfinity
eh can't think of something else!
anonymous
  • anonymous
let's see if this leads us somewhere
anonymous
  • anonymous
\[(a+1)(b+2)c=(a+1)(14-a-c)c=-ca^2+(13c-c^2)a+14c-c^2\]
anonymous
  • anonymous
which is a downward quadratic (leading coefficient is negative) and maximum occurs at \[a=\frac{13c-c^2}{2c}=\frac{13-c}{2}\]
anonymous
  • anonymous
now put this back in the quadratic
anonymous
  • anonymous
quadratic simplifies to\[\frac{1}{4} (c-15)^2 c\]
mathmath333
  • mathmath333
i got new question pls help me there
anonymous
  • anonymous
local maximum of this function occurs at \(c=5\) which is \(125\)
anonymous
  • anonymous
ok @mathmath333 , we're done
anonymous
  • anonymous
second way is much more elementary :-)
xapproachesinfinity
  • xapproachesinfinity
oh this has beat me! i was doing that but didn't not think of quadratic!

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