# Minimize the sum of the squares

Hello
i need help with a question, other people tried to help me, i just cannot get it! its driving me crazy:grumpy:

Two positive numbers have sum n. What is the smallest value possible for the sum of their squares?

so i have n=x+y
x>0 y>0

y=n-x

we want to minimize S S=x^2+y^2
S=x^2+(n-x)^2
S'=2x-2(n-x)
S'=2x-2n+2x
S'=4x-2n

now ???????? The above is what I have been shown to do... the red part doesnt make much sense to me... Can someone please help me to continue on...

HallsofIvy
Homework Helper
What class is this problem in? If it is a calculus problem, then you should know how to take the derivative of x^2 and (n-x)^2 (using the chain rule for the latter). Since the derivative measrures how fast a function is increasing, if S' is positive, for a specific x, the function is increasing- choosing x smaller will make S smaller so that value is not the minimum. If S' is negative, the function is decreasing- choosing x larger will make S smaller so that value is not the minimum. The minimum must occur where the derivative S' is 0. S'= 4x- 2n= 0 give 4x= 2n or x= n/2. Then y= n- n/2= n/2. The minimum occurs when x= y= n/2. The smallest possible value of x^2+ y^2 is n^2/4+ n^2/4= n^2/2.

If this is NOT for a calculus class and you do not know how to take the derivative, you can still do the problem by completing the square:
S= x^2+ (n-x)^2= x^2+ n^2- 2nx+ x^2= 2x^2- 2nx+ n^2= 2(x^2- nx)+ n^2. We complete the square by adding (and subtracting) (-n/2)^2= n^2/4. 2(x^2- nx+ n^2/4- n^2/4)+ n^2= 2(x^2- nx+ n^2/4)- n^2/2+ n^2= 2(x- n/2)^2+ n^2/2. Since a square is never negative, S is n^2/2 plus something unless x= n/2 in which case it is n^2/2+ 0= n^2/2.

Hi

So this problem is for first year university calculus.

I got confused about the derivative... i do know how to do derivates... too much calculus in one day:yuck: i guess.

So from the steps shown, I followed them and they seem pretty clear.
So do I not calculate an actual number then... just leave answers with variables?
Thanks again for the help So do I not calculate an actual number then... just leave answers with variables?
n is the only variable.The answer should be in terms of n VietDao29
Homework Helper
star321 said:
we want to minimize S S=x^2+y^2
S=x^2+(n-x)^2
S'=2x-2(n-x)
S'=2x-2n+2x
S'=4x-2n
Yes, this is good. :)
Ok, you should notice that 0 <= x <= n, right?
If the derivative of a function at some point, say x0, is positive, then the function is increasing at that point. Otherwise, if it's negative, then the function is decreasing.
So for 0 <= x < n / 2, S' < 0, right? So the function S is decreasing in that interval.
For n / 2 < x < n, S' > 0, the function S is increasing.
So for x = n / 2, S takes the smallest value. Can you find that value?
Is everything clear now? :)