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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align}& \{p,q,r\}\geq 0\hspace{.33em}\\~\\ & p+q+r=10 \hspace{.33em}\\~\\ & \normalsize \text{find the maximum value of} \ (pq+qr+pr+pqr)\hspace{.33em}\\~\\ \end{align}}\)

  2. anonymous
    • one year ago
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    20 i think

  3. anonymous
    • one year ago
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    because it 4 p's, 3 r's and 2 q's and you can only make two sets of (p=q=r)=10

  4. anonymous
    • one year ago
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    if im doin it right lol

  5. freckles
    • one year ago
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    So p,q,r are real numbers?

  6. freckles
    • one year ago
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    We know 70 something can be achieved since \[\text{ if } p=q=r=\frac{10}{3} \\ \text{ then } 3 (\frac{10}{3})^2+(\frac{10}{3})^3=70.37\] but there might be a higher number we can reach then that maybe thinking...

  7. freckles
    • one year ago
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    that should be an approximation symbol there

  8. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & a.)\ \geq 40\ \cap \leq 50 \hspace{.33em}\\~\\ & b.)\ \geq 50\ \cap \leq 60 \hspace{.33em}\\~\\ & c.)\ \geq 60\ \cap \leq 70 \hspace{.33em}\\~\\ & a.)\ \geq 70\ \cap \leq 80 \hspace{.33em}\\~\\ \end{align}}\)

  9. freckles
    • one year ago
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    well if there is a bigger number than 70.37 and you have no other inequalities then process of elimination would mean...

  10. mathmath333
    • one year ago
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    the correct option given by book is option \(c.)\) the book also gave a hint to assume \(p=4,q=3,r=3\)

  11. freckles
    • one year ago
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    that is assuming p,q,r are integers but that was never given

  12. mathmath333
    • one year ago
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    oh sry i forget they are integers given.

  13. freckles
    • one year ago
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    oh I would use the book's hint then :p

  14. mathmath333
    • one year ago
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    but book's solution is kind of trial and error

  15. freckles
    • one year ago
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    I think those numbers came from being close to 10/3

  16. mathmath333
    • one year ago
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    or is thaat the only way in case they are integers

  17. ganeshie8
    • one year ago
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    yeah but this symmetry stuff doesn't work always, remember we saw it failing multiple times before

  18. mathmath333
    • one year ago
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    in case other question comes with restriction given as integers, should i use the same method as described by the book

  19. ganeshie8
    • one year ago
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    that method has no mathematical justification, simply saying "by symmetry" wont do

  20. ganeshie8
    • one year ago
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    you either need to use AM-GM inequality or other standard methods

  21. mathmath333
    • one year ago
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    using AM-GM even in case of integers ?

  22. ganeshie8
    • one year ago
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    it works on reals but you can use it to pick the closest integers

  23. mathmath333
    • one year ago
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    ok thnx.

  24. dan815
    • one year ago
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    69

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