Two numbers have a difference of 42. what is the sum of their squares if it is a minimum?

The numbers are -21 and 21

Step-by-step explanation:

Let

x ----> one number

y ----> another number

S----> the sum of their squares

we know that

----> equation A

----> equation B

substitute equation A in equation B

This is the equation of a vertical parabola open upward

The vertex is a minimum

Find the coordinates of the vertex

Using a graphing tool

The vertex is the point (21,882) ----> see the attached figure

we have

x=21

Find the value of y

therefore

The numbers are -21 and 21

the numbers are 21 and -21

Step-by-step 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

cosine

step-by-step explanation: