Answer: the numbers are 21 and -21
Explanation:
The difference of two numbers is equal to 42; so we have that:
IX - YI = 42
we want to know the sum of their squares if it is a minimum.
to find this, we have that:
X^2 + Y^2 must be a minimum.
now, let's do this to remove the module:
(X - Y)^2 = 42^2
now, expand the left side:
X^2 - 2*X*Y + Y^2 = 42^2
now, we can write this as:
X^2 + Y^2 = 42^2 + 2*X*Y
Here you can see that the minimum of the addition is when 42^2 + 2*X*Y is also a minimum. Then is easy to see that X and Y must be of a different sign, such that the product 2*X*Y is minimum.
Now, knowing that one number must be positive, we can took X positive, and Y negative, in this way we have:
X - Y = 42.
X = 42 + Y
now we can replace it in:
2*X*Y = 2*(42 + Y)*Y = 84*Y + 2*Y^2
now we want to find the minimum of this equation, and we will only work with one variable.
for this, we derivate the function and find the zero.
f' = 84 + 4*Y = 0
Y = -84/4 = -21
and now we have that X must be equal to 21
knowing that the square relation grows faster than any linear relation, this must be the value that we are looking for.
X^2 + Y^2 = (-21)^2 + 21^2 = 2*(21)^2