find the maximum value of

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find the maximum value of

Linear Algebra
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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 \(\large \color{black}{\begin{align} \{a,b\}>0 \end{align}}\)
hmm AM-GM could work again! no?

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yes, it works, note that\[a+b=\frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{2}+\frac{b}{2}\]
oh no don't spoil it yet!
ok
I just gave you a hint
i have no clue how you achieved that though haha
oh i see silly me!
I made that partitioning so that we can form that \(a^3 b^2\) on the RHS of AM-GM
i'm trying to come from here first \[a^3b^2\le (\frac{a^3+b^2}{2})^2\]
what did you get?
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
apply AM-GM with five numbers
hmm let see
oh i guess you actually give me the answer \[a^3b^2<27.4=108\]
oh hold on forgot roo 3
oh good still the same though
that's right\[\frac{\frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{2}+\frac{b}{2}}{5} \ge (\frac{a}{3}.\frac{a}{3}.\frac{a}{3}.\frac{a}{3}.\frac{b}{2}.\frac{b}{2})^{1/5}\]
yeah that's what i did! man i would never thought of that manipulation
\[\frac{\frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{2}+\frac{b}{2}}{5} \ge (\frac{a}{3}.\frac{a}{3}.\frac{a}{3}.\frac{b}{2}.\frac{b}{2})^{1/5}\]
ok, good questions, thanks guys

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