If x,y,z>0 & xyz=32.Find Max. & Min. vaues of

x+2y+z.

if x,y,z>0

& (x)(y)(z)=32

 then if we assume y=32

we can have x=1/(big value) & z=(big value)

it satisfies x,y,z>0 & xyz=32

because in xyz=32

LHS= =32

RHS=32

therefore LHS=RHS

so max. val. of x+2y+z=1/(big value)+2(32)+(big value)=1/infinity+64+infinity=infinity

therefore the max value of x+2y+z=infinity

the min value can b determined using AM>=gm which gives us the min value as12 and the max value tends to infinity