Given $H = \left\{1,3,5,7,8,8,9,10\right\}$ , find $\min$ and $max$ value of $ ab+cd+ef+gh$ which $a,b,c,d,e,f,g,h $ can be any number in $H$ (1 letter pair 1 number only).
This problem I think that is the same way that $H$ will be a box ,put number in box $H$ and draw for $2$ numbers and don't put them back to $H$ box then we have to draw $4$ times ,anyway I cannot start to this problem. Please give me a hint or theorem that relavant about this problem and other problems like this .
Thank you.
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If the numbers in $a,b,c,d,e,f,g,h\in H $ can be repeated we must $ \max_{a,b,c,d,e,f,g,h\in H } (ab+cd+ef+gh ) $ equal to $$ \max{H}\cdot\max{H}+\max{H}\cdot\max{H}+\max{H}\cdot\max{H}+\max{H}\cdot\max{H} $$ If the numbers in $a,b,c,d,e,f,g,h\in H $ are all different we must $ \max_{a,b,c,d,e,f,g,h\in H } (ab+cd+ef+gh ) $ equal to $$ \max{H}\cdot\max{H_1}+\max{H_2}\cdot\max{H_3}+\max{H_4}\cdot\max{H_5}+\max{H_6}\cdot\max{H_7} $$ where the sets $H_1$, $H_2$, $H_3$, $H_4$, $H_5$ $H_6$ and $H_7$ are given by $$ H_1=H-\{\max H\} \\ H_2=H_1-\{\max H_1\} \\ H_3=H_2-\{\max H_2\} \\ H_4=H_3-\{\max H_3\} \\ H_5=H_4-\{\max H_4\} \\ H_6=H_5-\{\max H_5\} \\ H_7=H_6-\{\max H_6\} $$ In an analogous way, we proceed to find $$ \min_{a,b,c,d,e,f,g,h\in H } (ab+cd+ef+gh ) $$
Start with an arbitrary arrangement. As long as you find an instance of $xy+uv$ with $x<u$ and $y> v$, you can increase $H$ by rearranging this to $xv+uy=(xy+uv)+(y-v)(u-x)>xy+uv$. We conclude that in the maximal case, larger factors are paired with larger co-factors, i.e., the maximum is achieved with $$1\cdot 3+5\cdot 7+8\cdot 8+9\cdot 10. $$ By similar reasoning, the minimum is achieved with $$1\cdot 10+3\cdot 9+5\cdot 8+7\cdot 8. $$