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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 \(\large \color{black}{\begin{align} (a+1)\times (b+2)\times c\hspace{.33em}\\~\\ \end{align}}\) if \(\large \color{black}{\begin{align} a+b+c=12\hspace{.33em}\\~\\ \end{align}}\) where \(\large \color{black}{\begin{align} \{a,b,c\}\in \mathbb{R},\ \{a,b,c\}>0\hspace{.33em}\\~\\ \end{align}}\)

  2. xapproachesinfinity
    • one year ago
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    what do you think as a start

  3. mathmath333
    • one year ago
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    lol idk i m new to the topic

  4. xapproachesinfinity
    • one year ago
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    hmm still thinking!

  5. anonymous
    • one year ago
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    try AM-GM

  6. mathmath333
    • one year ago
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    this one ? \(\large \color{black}{\begin{align} \dfrac{a+b+c}{3}\geq (abc)^{1/3}\hspace{.33em}\\~\\ \end{align}}\)

  7. anonymous
    • one year ago
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    yes

  8. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} &\dfrac{a+b+c}{3}\geq (abc)^{1/3}\hspace{.33em}\\~\\ &\dfrac{12}{3}\geq (abc)^{1/3}\hspace{.33em}\\~\\ &4\geq (abc)^{1/3}\hspace{.33em}\\~\\ &64\geq abc \end{align}}\)

  9. anonymous
    • one year ago
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    well you want \((a+1)(b+2)c\) in the RHS

  10. anonymous
    • one year ago
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    what you gonna do for that

  11. mathmath333
    • one year ago
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    is am-gm inequality enough ?

  12. anonymous
    • one year ago
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    yes

  13. anonymous
    • one year ago
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    replace \(a\), \(b\) and \(c\) in the AM-GM with desired ones

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

  15. mathmath333
    • one year ago
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    i get this \(64+2ac+bc+2c\)

  16. Michele_Laino
    • one year ago
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    please, trey to apply the lagrange multipliers method to this function: \[G\left( {x,y,z;\lambda } \right) = \left( {x + 1} \right)\left( {y + 2} \right)z - \lambda \left( {x + y + z - 12} \right)\]

  17. Michele_Laino
    • one year ago
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    try*

  18. anonymous
    • one year ago
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    emm, idk, how did you get that? What I meant was this\[\frac{a+1+b+2+c}{3}\ge ((a+1)(b+2)c)^{1/3}\]

  19. xapproachesinfinity
    • one year ago
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    i was thinking of calculus the first sight too but can't use that since math has not taken that yet

  20. Michele_Laino
    • one year ago
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    ok!

  21. anonymous
    • one year ago
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    which gives us:\[(a+1)(b+2)c \le \left(\frac{15}{3} \right)^3=125\]

  22. xapproachesinfinity
    • one year ago
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    wow! very neat

  23. anonymous
    • one year ago
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    equality occurs when \(a+1=b+2=c=5\)

  24. mathmath333
    • one year ago
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    thnks

  25. anonymous
    • one year ago
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    np

  26. anonymous
    • one year ago
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    Lagrange multipliers will work if you familiar with :-)

  27. mathmath333
    • one year ago
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    idk lagrage multiplers

  28. anonymous
    • one year ago
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    ok!

  29. anonymous
    • one year ago
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    1 Attachment
  30. xapproachesinfinity
    • one year ago
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    cool AM-GM turned to be very useful!

  31. xapproachesinfinity
    • one year ago
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    can we do this without that! a different alternative but no calculus

  32. anonymous
    • one year ago
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    there must be some other methods, I can't think of other one now

  33. anonymous
    • one year ago
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    LOl i shared this on facebook

  34. xapproachesinfinity
    • one year ago
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    eh can't think of something else!

  35. anonymous
    • one year ago
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    let's see if this leads us somewhere

  36. anonymous
    • one year ago
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    \[(a+1)(b+2)c=(a+1)(14-a-c)c=-ca^2+(13c-c^2)a+14c-c^2\]

  37. anonymous
    • one year ago
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    which is a downward quadratic (leading coefficient is negative) and maximum occurs at \[a=\frac{13c-c^2}{2c}=\frac{13-c}{2}\]

  38. anonymous
    • one year ago
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    now put this back in the quadratic

  39. anonymous
    • one year ago
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    quadratic simplifies to\[\frac{1}{4} (c-15)^2 c\]

  40. mathmath333
    • one year ago
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    i got new question pls help me there

  41. anonymous
    • one year ago
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    local maximum of this function occurs at \(c=5\) which is \(125\)

  42. anonymous
    • one year ago
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    ok @mathmath333 , we're done

  43. anonymous
    • one year ago
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    second way is much more elementary :-)

  44. xapproachesinfinity
    • one year ago
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    oh this has beat me! i was doing that but didn't not think of quadratic!

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