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mathmath333
 one year ago
find the maximum value of
mathmath333
 one year ago
find the maximum value of

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0find 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 \(\large \color{black}{\begin{align} \{a,b\}>0 \end{align}}\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2hmm AMGM could work again! no?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, it works, note that\[a+b=\frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{2}+\frac{b}{2}\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2oh no don't spoil it yet!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just gave you a hint

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2i have no clue how you achieved that though haha

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2oh i see silly me!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I made that partitioning so that we can form that \(a^3 b^2\) on the RHS of AMGM

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2i'm trying to come from here first \[a^3b^2\le (\frac{a^3+b^2}{2})^2\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2i 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0apply AMGM with five numbers

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2hmm let see

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2oh i guess you actually give me the answer \[a^3b^2<27.4=108\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2oh hold on forgot roo 3

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2oh good still the same though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's right\[\frac{\frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{2}+\frac{b}{2}}{5} \ge (\frac{a}{3}.\frac{a}{3}.\frac{a}{3}.\frac{a}{3}.\frac{b}{2}.\frac{b}{2})^{1/5}\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.2yeah that's what i did! man i would never thought of that manipulation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{\frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{2}+\frac{b}{2}}{5} \ge (\frac{a}{3}.\frac{a}{3}.\frac{a}{3}.\frac{b}{2}.\frac{b}{2})^{1/5}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, good questions, thanks guys
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