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\(\large \color{black}{\begin{align}& \{p,q,r\}\geq 0\hspace{.33em}\\~\\ & p+q+r=10 \hspace{.33em}\\~\\ & \normalsize \text{find the maximum value of} \ (pq+qr+pr+pqr)\hspace{.33em}\\~\\ \end{align}}\)
20 i think
because it 4 p's, 3 r's and 2 q's and you can only make two sets of (p=q=r)=10

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if im doin it right lol
So p,q,r are real numbers?
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...
that should be an approximation symbol there
\(\large \color{black}{\begin{align} & a.)\ \geq 40\ \cap \leq 50 \hspace{.33em}\\~\\ & b.)\ \geq 50\ \cap \leq 60 \hspace{.33em}\\~\\ & c.)\ \geq 60\ \cap \leq 70 \hspace{.33em}\\~\\ & a.)\ \geq 70\ \cap \leq 80 \hspace{.33em}\\~\\ \end{align}}\)
well if there is a bigger number than 70.37 and you have no other inequalities then process of elimination would mean...
the correct option given by book is option \(c.)\) the book also gave a hint to assume \(p=4,q=3,r=3\)
that is assuming p,q,r are integers but that was never given
oh sry i forget they are integers given.
oh I would use the book's hint then :p
but book's solution is kind of trial and error
I think those numbers came from being close to 10/3
or is thaat the only way in case they are integers
yeah but this symmetry stuff doesn't work always, remember we saw it failing multiple times before
http://www.wolframalpha.com/input/?i=maximize+x%5E4%2By%5E4%2Bz%5E4%2C+x%5E2%2By%5E2%2Bz%5E2%3D1%2Cx%3E0%2Cy%3E0%2Cz%3E0
in case other question comes with restriction given as integers, should i use the same method as described by the book
that method has no mathematical justification, simply saying "by symmetry" wont do
you either need to use AM-GM inequality or other standard methods
using AM-GM even in case of integers ?
it works on reals but you can use it to pick the closest integers
ok thnx.
69

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