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x^2 + y^3 + z^4 = 26. Find The Maximum Value Of xyz. (Important For IIT-JEE)

Sujoy Das

07/02/2018 1 0

Question:

x^2 + y^3 + z^4 = 26. Find the maximum value of xyz.

Answer:

LCM of powers= 12

x should be divided into 12/2 = 6 parts y into 4 parts z into 3 parts

x^2= (1/6*x^2)*6

y^3=(1/4*y^3)*4

z^4=(1/3*z^4)*3

AM= (x^2+y^3+z^4)/13= 2

Similarly find GM and apply AM>=GM

ANS = 2^(9/4) * 3^ (3/4)

x^2/6=y^3/4=z^4/3..use this.no need to use AM>=GM (case2).

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