Using time period for a physical pendilum
`T = 2 pi sqrt((I)/(mgx)) = 2 pi sqrt((m[l^(2)//12] + x^(2)]/(mgx)`
`= Ksqrt((l^(2)/(12x + x)))`
For T to be maximum or minimum
`dT//dx = 0`
`x = l//sqrt12`
Further for `x = l//sqrt12`
`(d^(2)T)/(dx^(2)) = + ve`
Hence T is minimum.