We know that the two numbers have a difference of 28.

Let one of the numbers be x. The other is x+ 28.

Now the square of x is x^2 and the square of x+ 28 is (x+28)^2.

The sum of the squares is x^2 + (x+ 28)^2.

x^2 +...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

We know that the two numbers have a difference of 28.

Let one of the numbers be x. The other is x+ 28.

Now the square of x is x^2 and the square of x+ 28 is (x+28)^2.

The sum of the squares is x^2 + (x+ 28)^2.

x^2 + (x+ 28)^2

= x^2 + x^2 + 28^2 + 56x

= 2x^2 + 28^2 + 56x

Now let f(x) = 2x^2 + 28^2 + 56x

f'(x) = 4x + 56

Equating f'(x) to 0.

4x + 56 = 0

=> x= -56/4

=> x = -14

f''(x) = 4 which is positive, therefore at x = -14 we have the minimum value of the function f(x).

f(-14) = 2*(-14)^2 + 28^2 + (-14)*56 = 392.

**Therefore the required value of the sum of their squares if it is a minimum is 392.**