IIT JEE , AIEEE Forum - Mathematics , Algebra

neeraj_agarwal_1990

The greatest value of P=a²b³c⁴,when a+b+c=18 is?
[ans:4².6^3.8⁴]

amaron

You have to know that when sum of terms is constant,product is maximum when all terms are equal.This is a direct consequence of AM>=GM.

just see how you have to split a+ b+ c so that on Am >Gm you get a²b³c⁴ in the numerator.

this can obviously be done of a + b+ c is written as a/2 + a/2 + b/3 + b/3 +b/3 +c/4 + c/4 + c/4 +c/4 =18

now apply Am>gm you will get a/2 =b/3 =c/4 implies a=4,b=6,c=8

therefore on substituting,max value of a²b³c⁴=4².6³.8⁴

That solves the problem!

abhijeet_0201

your method looks gr8 but i am not able to understand it.plz be elaborate esp the A.M>GM part
thanks

narengoiit

Hi amaron, wonderful attempt, but i would like to have further explanation of the problem for better understanding.

iitkgp_bipin

Split a+b+c in a/2+a/2+b/3+b/3+b/3+c/4+c/4+c/4+c/4

Now using AM>=GM

(a+b+c)/9 >= (a/2xa/2xb/3xb/3xb/3xc/4xc/4xc/4xc/4)^1/9

a²b³c⁴ <= (18/9)⁹2²3³4⁴

a²b³c⁴ <= 4²6³8⁴

Hence maximum value of a²b³c⁴ is 4²6³8⁴

In such type of problems greatest value can be found by taking ratios of a,b,c powers.

a:b:c: = 2:3:4
a+b+c=18
a=4 b=6 c=8

a²b³c⁴ = 4²6³8⁴ (maximum value)

vish0001

excellent , great thinking amaron and sir !! wow !!! that was simply superb !

rebel

~~~~~~~hey that was a splendid way to kill this question ! ~
~~~~~~~hey amaron, u need to be my friend ~~
~~~~~~~take a salute for killing it ~~

ruhi

ya amaron nice approach

umang

Thanks amaron and bipin sir for this wonderful technique !!!!

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