The sum of two numbers is 16. The sum of their squares is a minimum. Determine the numbers.

Since the sum of the two numbers is 16, we can always the write the two numbers, like, x and 16-x  or 8+h and 8-h, etc.

If you choose x and 16-x, then the sum of the squares , s = x^2+(16-x)^2 and  the solution of ds/dx = 0 which makes d2S/dx positive is the minimum.

But ds/dx =[ (x^2+(16-x)^2]' = 0 gives:2x+2(16-x)(-2) = 0 gives: 4x=32 or x=8.  The two numbers are 8 and 8 .If you want the two numbers to be different, choose the numbers such that 8+h and 8-h, where h is any arbitrarily small depending upon your accuracy.

For integral solution, choose h =1, then 8+1 =9 and 8-1 =7 are the solutions.

For the accuracy to the level of 1st decimal, 8+*1=8.1 and 8-0.1 =7.9 are the solution, For the level of 2nd decimal accuracy, 8+0.01=8.01 and 8-0.01 = 7.99 are the solutions.