find the maximum value of

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find the maximum value of

Mathematics
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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 \(\large \color{black}{\begin{align} \{a,b,c\}\in \mathbb{R},\ \{a,b,c\}>0\hspace{.33em}\\~\\ \end{align}}\)
what do you think as a start
lol idk i m new to the topic

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

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