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mathmath333

  • one year ago

find the maximum value of

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  1. mathmath333
    • one year ago
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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 \(\large \color{black}{\begin{align} \{a,b\}>0 \end{align}}\)

  2. mathmath333
    • one year ago
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    @mukushla

  3. xapproachesinfinity
    • one year ago
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    hmm AM-GM could work again! no?

  4. anonymous
    • one year ago
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    yes, it works, note that\[a+b=\frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{2}+\frac{b}{2}\]

  5. xapproachesinfinity
    • one year ago
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    oh no don't spoil it yet!

  6. anonymous
    • one year ago
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    ok

  7. anonymous
    • one year ago
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    I just gave you a hint

  8. xapproachesinfinity
    • one year ago
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    i have no clue how you achieved that though haha

  9. xapproachesinfinity
    • one year ago
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    oh i see silly me!

  10. anonymous
    • one year ago
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    I made that partitioning so that we can form that \(a^3 b^2\) on the RHS of AM-GM

  11. xapproachesinfinity
    • one year ago
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    i'm trying to come from here first \[a^3b^2\le (\frac{a^3+b^2}{2})^2\]

  12. anonymous
    • one year ago
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    what did you get?

  13. xapproachesinfinity
    • one year ago
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    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

  14. anonymous
    • one year ago
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    apply AM-GM with five numbers

  15. xapproachesinfinity
    • one year ago
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    hmm let see

  16. xapproachesinfinity
    • one year ago
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    oh i guess you actually give me the answer \[a^3b^2<27.4=108\]

  17. xapproachesinfinity
    • one year ago
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    oh hold on forgot roo 3

  18. xapproachesinfinity
    • one year ago
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    oh good still the same though

  19. anonymous
    • one year ago
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    that's right\[\frac{\frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{2}+\frac{b}{2}}{5} \ge (\frac{a}{3}.\frac{a}{3}.\frac{a}{3}.\frac{a}{3}.\frac{b}{2}.\frac{b}{2})^{1/5}\]

  20. xapproachesinfinity
    • one year ago
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    yeah that's what i did! man i would never thought of that manipulation

  21. anonymous
    • one year ago
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    \[\frac{\frac{a}{3}+\frac{a}{3}+\frac{a}{3}+\frac{b}{2}+\frac{b}{2}}{5} \ge (\frac{a}{3}.\frac{a}{3}.\frac{a}{3}.\frac{b}{2}.\frac{b}{2})^{1/5}\]

  22. anonymous
    • one year ago
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    ok, good questions, thanks guys

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