Question

find the maximum value of

Answer 1

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


\(\large \color{black}{\begin{align}
\{a,b,c\}\in \mathbb{R},\ \{a,b,c\}>0\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

what do you think as a start

Answer 3

lol idk i m new to the topic

Answer 4

hmm still thinking!

Answer 5

try AM-GM

Answer 6

this one ?

\(\large \color{black}{\begin{align}
\dfrac{a+b+c}{3}\geq (abc)^{1/3}\hspace{.33em}\\~\\
\end{align}}\)

Answer 7

yes

Answer 8

\(\large \color{black}{\begin{align}
&\dfrac{a+b+c}{3}\geq (abc)^{1/3}\hspace{.33em}\\~\\
&\dfrac{12}{3}\geq (abc)^{1/3}\hspace{.33em}\\~\\
&4\geq (abc)^{1/3}\hspace{.33em}\\~\\
&64\geq abc
\end{align}}\)

Answer 9

well you want \((a+1)(b+2)c\) in the RHS

Answer 10

what you gonna do for that

Answer 11

is am-gm inequality enough ?

Answer 12

yes

Answer 13

replace \(a\), \(b\) and \(c\) in the AM-GM with desired ones

Answer 14

what did you get?

Answer 15

i get this \(64+2ac+bc+2c\)

Answer 16

please, trey to apply the lagrange multipliers method to this function:

\[G\left( {x,y,z;\lambda } \right) = \left( {x + 1} \right)\left( {y + 2} \right)z - \lambda \left( {x + y + z - 12} \right)\]

Answer 17

try*

Answer 18

emm, idk, how did you get that? What I meant was this\[\frac{a+1+b+2+c}{3}\ge ((a+1)(b+2)c)^{1/3}\]

Answer 19

i was thinking of calculus the first sight too
but can't use that since math has not taken that yet

Answer 20

ok!

Answer 21

which gives us:\[(a+1)(b+2)c \le \left(\frac{15}{3} \right)^3=125\]

Answer 22

wow! very neat

Answer 23

equality occurs when \(a+1=b+2=c=5\)

Answer 24

thnks

Answer 25

np

Answer 26

Lagrange multipliers will work if you familiar with :-)

Answer 27

idk lagrage multiplers

Answer 28

ok!

Answer 29

cool AM-GM turned to be very useful!

Answer 30

can we do this without that! a different alternative but no calculus

Answer 31

there must be some other methods, I can't think of other one now

Answer 32

LOl i shared this on facebook

Answer 33

eh can't think of something else!

Answer 34

let's see if this leads us somewhere

Answer 35

\[(a+1)(b+2)c=(a+1)(14-a-c)c=-ca^2+(13c-c^2)a+14c-c^2\]

Answer 36

which is a downward quadratic (leading coefficient is negative) and maximum occurs at \[a=\frac{13c-c^2}{2c}=\frac{13-c}{2}\]

Answer 37

now put this back in the quadratic

Answer 38

quadratic simplifies to\[\frac{1}{4} (c-15)^2 c\]

Answer 39

i got new question pls help me there

Answer 40

local maximum of this function occurs at \(c=5\) which is \(125\)

Answer 41

ok @mathmath333 , we're done

Answer 42

second way is much more elementary :-)

Answer 43

oh this has beat me! i was doing that but didn't not think of quadratic!