## mathmath333 one year ago find the maximum value of

1. mathmath333

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

@mukushla

3. xapproachesinfinity

hmm AM-GM could work again! no?

4. anonymous

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

5. xapproachesinfinity

oh no don't spoil it yet!

6. anonymous

ok

7. anonymous

I just gave you a hint

8. xapproachesinfinity

i have no clue how you achieved that though haha

9. xapproachesinfinity

oh i see silly me!

10. anonymous

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

11. xapproachesinfinity

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

12. anonymous

what did you get?

13. xapproachesinfinity

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

apply AM-GM with five numbers

15. xapproachesinfinity

hmm let see

16. xapproachesinfinity

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

17. xapproachesinfinity

oh hold on forgot roo 3

18. xapproachesinfinity

oh good still the same though

19. anonymous

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

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

21. anonymous

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

22. anonymous

ok, good questions, thanks guys