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

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### 3 Answers

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.

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.

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