if a+b+c+d+e+f=1 then the maximum value of ab+bc+cd+de+ef . Given that a,b,c,d,e,f are positive...???

Question

if a+b+c+d+e+f=1 then the maximum value of ab+bc+cd+de+ef . Given that a,b,c,d,e,f  are positive...???

anonymous · Answer

1

anonymous · Answer

$$(a+b+c+d+e+f)^2=1\a^2+b^2+c^2+d^2+e^2+f^2+2(ab+bc+cd+de+ef)=1$$To minimize $a^2+b^2+c^2+d^2+e^2+f^2$ we want all to be equal (so the 'weight' is distributed evenly) so presume $a=b=c=d=e=f=\frac16$ and thus we have $a^2+b^2+c^2+d^2+e^2+f^2=\frac16$:$$2(ab+bc+cd+de+ef)=\frac56\ab+bc+cd+de+ef=\frac5{12}$$

anonymous · Answer

I believe this should work...

zarkon · Answer

are you sure that is correct

zarkon · Answer

what is $(a+b+c+d+e+f)^2$ again...

zarkon · Answer

are you sure that problem states "Given that a,b,c,d,e,f  are positive"

if that is the case then you are trying to maximize a function over a non-compact set.  (you are not guaranteed to find an answer)