# Two numbers have a difference of 28. What is the sum of their squares if it is a minimum?thanks!!

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll put the numbers as a and b:

a - b = 28

a = 28 + b

a^2 + b^2 = (28+b)^2 + b^2

We'll expand the square:

784 + 56b + b^2 + b^2

We'll combine the like terms and we'll re-arrange the terms:

2b^2 + 56b + 784

We'll divide by 2:

b^2 + 28b + 392

The minimum value for the given expression is obtained for the value of b that cancels the first derivative

We'll calculate the first derivative:

2b + 28 = 0

2b = -28

b = -14

We'll calculate a:

a = 28 + b

a = 28 - 14

a = 14

The values for a and b that respect the imposed constraints are: {14;14}.

johndoeone | Student, Undergraduate | eNotes Newbie

Posted on

You're question is " What is the sum of their squares if it is a minimum?"

The answer is obviously 392, as shown in the answer given above.

I don't understand why one has mentioned the values -14 and 14 and another has got even that wrong and written 14 and 14.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Since the difference of the  two numbers is 28, we assume that the numbers are x and x+28.

Their sum of sqares   S(x) = x^2+(x+28)^2

To find the minimum of S(x):

To we find x = c for which S'(c) = 0 and S"(c) > 0 so that S(c) would be minumum.

S'(x) = {x^2+(x+28^2)}' = {x^2+x^2+56x+28^2}'

S'(x) = (2x^2+56x+28^2)' = 4x +56. Or  or x = -56/4 = -14.

S''(x) = (4x+56)' = 4 > 0

Therefore x = c= -14

x= -14 and x+28 = 14 are the numbers for which  the sum of the squares is minimum and the difference of the numbers is 28.