if x+y+z=6 , and x>0 , y>0 , z>0 then find the maximum value of xy^3z^2

Question

Answer 1

x=1, z=2,y=3 ans is 108

Answer 2

x=1, z=2,y=3 ans is 108

Answer 3

x + y/3 + y/3 + y/3 + z/2 + z/2 = 6

AM >= GM

(x + y/3 + y/3 + y/3 + z/2 + z/2)/6 >= $$(x*y/3*y/3*y/3*z/2*z/2)^{1/6}$$

=> 1 >= $$(x*y/3*y/3*y/3*z/2*z/2)^{1/6}$$
=> 1 >= $$x*y^3*z^2 / (3^3*2^2)$$
=> $$x*y^3*z^2$$ <= 27*4 = 108

Max value = 108

Answer 4

x + y/3 + y/3 + y/3 + z/2 + z/2 = 6

AM >= GM

(x + y/3 + y/3 + y/3 + z/2 + z/2)/6 >= $$(x*y/3*y/3*y/3*z/2*z/2)^{1/6}$$

=> 1 >= $$(x*y/3*y/3*y/3*z/2*z/2)^{1/6}$$
=> 1 >= $$x*y^3*z^2 / (3^3*2^2)$$
=> $$x*y^3*z^2$$ <= 27*4 = 108

Max value = 108

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if x+y+z=6 , and x>0 , y>0 , z>0 then find the maximum value of x*y^3*z^2

if x+y+z=6 , and x>0 , y>0 , z>0 then find the maximum value of xy^3z^2