Let the two numbers= #x# and #y#.
So #x# +#y#= #16# [ sum of the positive numbers] ............#[1]#
Let the sum of the the squares of these two numbers =#S# , say, whereby #S=x^2+y^2#. We can now substitute for #y# from .........#[1]# giving #y=[16-x]#
Therefore the sum of the squares of these numbers #S# =#x^2+[16-x]^2#. Giving terms in only one variable, now we need to differentiate this expression to find max/ min, and we can do this either by expanding the bracket or using the chain rule, and this will result in,
#[dS]/[dx]#=#4x- 32# = #0#, [for max/min].....i.e, #x=8#
So from ........#[1]# #y=8#. The second derivative, #S''#=#4#, which is positive, [ whatever the value of #x#] and thus by the second derivative test confirms that this value of #x# will minimise the sum , #S#.
#y^2+x^2#= #8^2+8^2#= #128#. Hope this was helpful.
