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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2008 22:30:07 IST
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f(x,y) = cos(x) +cos(y) - cos(x+y)
What are the maximum and minimum values? Differentiation and graphical method not allowed. Only by analytical transformations.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2008 22:32:09 IST
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min value is obviously -3..when x=y=pi
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Nitwit Blubber Odment Tweak
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2008 22:36:26 IST
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How can that be proved?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2008 22:39:34 IST
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edited
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2008 22:41:37 IST
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by jensen's inequality
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2008 22:45:09 IST
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How's that? And what's the max value?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2008 22:48:00 IST
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Note that
By jensen's we get

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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2008 23:05:37 IST
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Thanks a lot sandeep... however I am a bit confused about jensen's... what exactly does it state?
How can the max value be found?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2008 23:06:21 IST
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clearly min. of sum of three cosines can be -3. Now question is can v get it if x= and y = for max value i will think
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 13:16:00 IST
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u can solve this question by quardadic equation or by using graph for quardadic p =cosx+cosy - cos(x+y) = (cosx+cosy) - cos(x+y) =2cos(x+y/2)cos(x-y/2) - 2cos2(x+y/2)+1 cos(x+y/2) = a than 2acos(x-y/2)-2a2+1= p 2a2_2acos(x-y/2)+(p-1) = 0 if the root of this equation are real than D>=0 4cos(x-y/2)2 - 8(p-1)>=0 cos(x-y/2)2 >= 2(p-1) 2(p-1)<=cos(x-y/2)2<= 1 (p-1)<=1/2 p<=3/2
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 13:22:14 IST
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yeah the solution above is right....although the max and min values are mostly evident by inspection only....min can obviously not be lesser than -3, and we see tha -3 is possible for max, each term in the sum "should be" (1/2), which we see is also possible, for eg, at x=y=pi/3........basically the solution is evident by symmetry....at any extremum, x must equal y since the expression is symmetric with respect to x and y!!! many other such problems can be solved this way by simple inspection
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 13:26:29 IST
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i think max value is when x=y=45 here it will be 2/root 2 which is >1 otherwise if x=y=90, ans. will be 1 but still 2/root 2 will be greater
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 13:29:37 IST
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at least care to read the solution posted just above urs  max is at x=y=pi/3 and is equal to 1.5 and 1.5 is greater than root2 = 1.414
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 13:30:53 IST
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oh i'm sorry..ur right abt that
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Mar 2008 13:38:19 IST
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