Question

find two positive numbers whose sum is 8 and sum of whose squares is minimum​

Answer 1

Step-by-step explanation:

Let one number be x.

So, other number will be 8 - x.

Now, taking sum of their squares

x² + ( 8 - x )² = 2x² - 16x + 64

For minimum positive value of the equation both numbers must be equal.

So,

x = 8 - x

2x = 8

x = 4

So, the numbers are 4 and 4.

Method 2

Differentiate the equation and equal it to zero for minimum value of sum of squares.

Differentiating you will get,

4x = 16

x = 4

Once again we get answer as 4 and 4.

Thank you.