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

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We assume that the two numbers should be whole numbers.

The sum of squares will be minimum when the two numbers are equal. That is, the two numbers are 8 and 8. For this combination sum of square is equal to:

8^2 + 8^2 = 64 + 64 =128

If we insist that the two numbers should be different than the two numbers will be 9 and 7. For this pair of numbers the sum of square will be equal to:

7^2 +9^2 = 49 + 81 = 130

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.

It's easy to find the numbers, using Minkowski inequality: The square of the sum is bigger that the sum of squares:

(x+y)^2 > (x)^2 + (y)^2

From the statement we have x+y=16. We'll introduce what we know into the Minkowski formula:

(16)^2 > (x)^2 + (y)^2

Also from the statement, we could write x+y=16, so y=16-x, and we'll substitute it into the inequality:

(16)^2 > (x)^2 + (16-x)^2

(16)^2 > (x)^2 + (16)^2- 2*16*x + (x)^2

We'll subtract (16)^2 from the inequality, so that;

0 > 2*(x)^2 - 32*x

We'll have 2x as common factor:

2x(x-16)<0

A product of 2 factors is negative if one factor has opposite sign in relation to the other, so :

2x>0 and x-16<0, x<16

x-16=0 When x=16, but x<16 and x>0 so x could be 8.

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