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# Prove that maximum value of a^2*b^3*c^4 subject to a+b+c=18 is 4^2*6^3*8^4. 1).....without using  AM,GM equality thing...2).......using  AM,GM equality thing...PLS EXPLAIN!!!!!!

9 years ago

Dear Shivam,

Arithmetic mean is always greater than or equal to geometric mean

A.M. >= G.M.

Lets say a,b,c are 3 number such that a+b+c=m

Now you have to find maximum value of (a^2) x (b^3) x(c^4)

Since it is given that a+b+c=m

=> a/2 +a/2 +b/3 +b/3 +b/3 +c/4 +c/4 +c/4 +c/4 = m

=> (a/2 +a/2 +b/3 +b/3 +b/3 +c/4 +c/4 +c/4 +c/4)/9 = m/9

We know that AM >= GM

Max value of GM= AM, which is attained when all the numbers are equal

Max of [ (a/2)^2 x (b/3)^3 x(c/4)^4] ^1/9= (a/2 +a/2 +b/3 +b/3 +b/3 +c/4 +c/4 +c/4 +c/4)/9 = m/9

And for maxima, a/2=b/3=c/4

And max (a/2)^2 x (b/3)^3 x(c/4)^4 = (m/9)^9

max of (a)^2 x (b)^3 x(c)^4= (2^2) (3^3)(4^4) (m/9)^9

Best Of luck

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