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Homework Help: Minimize the sum of the squares

  1. Jul 10, 2006 #1
    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...
     
  2. jcsd
  3. Jul 10, 2006 #2

    HallsofIvy

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    Science Advisor

    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.
     
  4. Jul 10, 2006 #3
    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:
     
  5. Jul 11, 2006 #4
    n is the only variable.The answer should be in terms of n :smile:
     
  6. Jul 11, 2006 #5

    VietDao29

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    Homework Helper

    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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