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.
