Question

find the maximum value of

Answer 1

find the maximum value of
\(\large \color{black}{\begin{align}

a^3\times b^2\hspace{.33em}\\~\\
\end{align}}\)

if

\(\large \color{black}{\begin{align}

a+b=5\hspace{.33em}\\~\\
\end{align}}\)

where


\(\large \color{black}{\begin{align}
\{a,b\}>0
\end{align}}\)

Answer 2

@mukushla

Answer 3

hmm AM-GM could work again! no?

Answer 4

yes, it works, note that\[a+b=\frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{2}+\frac{b}{2}\]

Answer 5

oh no don't spoil it yet!

Answer 6

ok

Answer 7

I just gave you a hint

Answer 8

i have no clue how you achieved that though haha

Answer 9

oh i see silly me!

Answer 10

I made that partitioning so that we can form that \(a^3 b^2\) on the RHS of AM-GM

Answer 11

i'm trying to come from here first
\[a^3b^2\le (\frac{a^3+b^2}{2})^2\]

Answer 12

what did you get?

Answer 13

i feel dumb!
this what i did \[a^3b^2\leq a^2b\left (\frac{a+b}{2}\right )^2\]
i still have trouble associating your hint with this

Answer 14

apply AM-GM with five numbers

Answer 15

hmm let see

Answer 16

oh i guess you actually give me the answer
\[a^3b^2<27.4=108\]

Answer 17

oh hold on forgot roo 3

Answer 18

oh good still the same though

Answer 19

that's right\[\frac{\frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{2}+\frac{b}{2}}{5} \ge (\frac{a}{3}.\frac{a}{3}.\frac{a}{3}.\frac{a}{3}.\frac{b}{2}.\frac{b}{2})^{1/5}\]

Answer 20

yeah that's what i did!
man i would never thought of that manipulation

Answer 21

\[\frac{\frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{2}+\frac{b}{2}}{5} \ge (\frac{a}{3}.\frac{a}{3}.\frac{a}{3}.\frac{b}{2}.\frac{b}{2})^{1/5}\]

Answer 22

ok, good questions, thanks guys