find max and min value

This one is kind of old.. (I think you have missed out the thing that all are +Ve ..
otherwise the value would be unbounded.. )

x+y+z = x/2+x/2+y/3+y/3+y/3+z/4+z/4+z/4+z/4

now

x/2+x/2+y/3+y/3+y/3+z/4+z/4+z/4+z/49≥(x/2.x/2.y/3./3.y/3.z/4.z/4.z/4.z/4)^1/9

So 189≥(x/2.x/2.y/3./3.y/3.z/4.z/4.z/4.z/4)^1/9

Hence x²y³z⁴≤ 2 . 2² . 3³ . 4⁴

if the numbers are +ve then the minimum will be zero ;)

no it is not in syllabus..

it is used in multivariate calculus and is generally taught at the 1st/ 2nd yr at engineering.