Minimize the sum of the squares

  • Thread starter star321
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  • #1
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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...
 

Answers and Replies

  • #2
HallsofIvy
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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.
 
  • #3
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leave answer with variables?

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 :smile:
 
  • #4
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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 :smile:
 
  • #5
VietDao29
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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? :)
 

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