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Algebra

himanshu2006

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10 Dec 2007 16:09:04 IST

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484

solve it

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if x,y,z >0 such that x+2y+3z=15,the maximum value of 6(1+x)yz + x(2y+3z) is:-

a)115

b)120

c)200

d)240

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himanshu2006

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28 Dec 2007 22:03:13 IST

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ABE KOI TO REPLY KARO YAAR!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!.

rishabh29288

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29 Dec 2007 00:00:23 IST

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IS the answer is 200.........????????

siddhartha satpathi

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30 Dec 2007 15:21:43 IST

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let a=x, b=2y , c=3z so a+b+c=15

6(1+x)yz + x(2y+3z) = 6yz + 6xyz + 2xy + 3xz = ab+ bc+ca + abc

by AM-GM ab<=(a²+b²)/2 similarly bc<= ..........,ac<=.....

on adding ab+bc+ca<= a²+b²+c²

ab+bc+ca<=(a+b+c)²/3 ( adding 2ab+2bc+2ca both sides)

= 75................1

by AM-GM abc <= ((a+b+c)/3)³= 125........................2

adding 1 and 2 max value is 125+75=200

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