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Finding sums and squares

  1. Mar 25, 2007 #1
    1. The problem statement, all variables and given/known data

    The sum of two numbers is 20. What is the least possible sum of their squares.

    2. The attempt at a solution

    Before I show my work, I'm pretty sure I have the answer. I think it's 200. If you add 10 and 10, you will have 20. If you square 10 you get 100, thus the sum of the squares would be 200. If you used any other numbers to get a sum of 20 (i.e. 1 and 19, 2 and 18, 3 and 17, etc.), and you'd end up with a number over 200. For example, 18 + 2 = 20, 18^2 = 324, 2^2 = 4, 324 + 4 = 328. For all numbers other than 10 and 10, you'll get an answer over 200.

    It's just I'm not exactly sure how to show my work for that. The only thing I can come up with is this:

    The equation to show that the sum of two numbers equalling 20 is x + x = 20.

    To show the squares of those numbers would be x^2 + x^2 = 20.

    Therefore, 2x^2 = 20.

    x^2 = 20/2

    x^2 = 10

    Am I on the right track with this?
     
  2. jcsd
  3. Mar 25, 2007 #2
    I think so...just show that x is the square root of 10 then.
     
  4. Mar 25, 2007 #3
    This problem can be done using derivatives....

    let the numbers be say [tex]x[/tex] and [tex]a[/tex]

    Now [tex]x+a=20[/tex]
    also,

    [tex]a=20-x[/tex]

    now we wish to find the minimum value of [tex](20-x)^2+x^2[/tex]


    if we graph the equation....set any point on the y axis = [tex](20-x)^2+x^2[/tex]

    if we perform [tex]\frac{dy}{dx}[/tex] and set the slope to zero

    we get

    [tex]4x=40[/tex]

    so x=10 and therefore y=10

    i guess.....therefore the minimum sum of their squares is 200
     
    Last edited: Mar 25, 2007
  5. Mar 25, 2007 #4
    your idea is correct.

    In general, you have an inequality:
    [tex]\sqrt{\frac{x^2+y^2}{2}}\ge \frac{x+y}{2}[/tex]

    equality occurs if and only if x=y.
    although this inequality only works for positive x and y. you can put absolute values around them for this problem.
    [tex]\sqrt{\frac{x^2+y^2}{2}}\ge\frac{|x|+|y|}{2}\ge \frac{x+y}{2}[/tex]

    or root mean square >= arithmetic mean

    as for proving it.... just expand and simplify
     
    Last edited: Mar 25, 2007
  6. Mar 25, 2007 #5
    :eek: What a mess! There's a very simple solution.

    If [tex] a + b = 20 [/tex] then

    [tex]a = 10 - c [/tex] and [tex]b = 10 + c[/tex]. The sum of the squares is then

    [tex](10 - c)^2 + (10 + c)^2 = 200 + 2c^{2}[/tex].

    Obviously we have to let c = 0 to get a minimum. Thus a = 10 and b = 10 and the the least possible sum is 200.
     
    Last edited: Mar 25, 2007
  7. Mar 25, 2007 #6

    HallsofIvy

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    Science Advisor

    When you write this, you are assuming x= y which is the same as assuming x=y= 10. "The equation to show that the sum of two numbers equalling 20" is x+ y= 20.

    No, nothing was said about the sum of the squares being 20!! Even you are claiming it is 200!

    [/quote]Therefore, 2x^2 = 20.

    x^2 = 20/2

    x^2 = 10

    Am I on the right track with this?[/QUOTE]
    Well, if x= y and x2= 10, then x2+ y2= 20 which isn't what you originally thought, is it? In fact, it comes from your incorrect equation 2x2= 20!

    You know that x+ y= 20 so y= 20- x. Now, the sum of the squares is x2+ y[/sup]2[/sup]= x2+ (20- x)2= x2+ 400- 40x+ x2= 2x2- 40x+ 400.

    Now, you can "complete" the square to find the maximum possible value of that.
     
    Last edited: Mar 25, 2007
  8. Mar 25, 2007 #7
    Alright I think I got it:

    2x^2 - 40x + 400
    = 2(x^2 - 20x) + 400
    = 2[x^2 - 20x + (20/2)^2 - (20/2)^2] + 400
    = 2[x^2 - 20x + 10^2 - 10^2] + 400
    = 2[(x-10)^2 - 100] + 400
    = 2(x-10)^2 + 400 - 200
    = 2(x-10)^2 + 200

    10 being the two squares
    200 being the least possible sum
     
    Last edited: Mar 25, 2007
  9. Mar 25, 2007 #8
    First, it's x - 10 inside the bracket and second... this is a totally uselessly long way to do it.
     
  10. Mar 25, 2007 #9
    Oh yeah, just saw that.

    And that's the way the teacher taught it to us. What's the shorter way?
     
  11. Mar 25, 2007 #10

    AKG

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    The way Werg22 showed you in post 5 is the best way.
     
  12. Mar 25, 2007 #11
    Or you can find the axis of symmetry once you have y = 2x^2 - 40x + 400
     
    Last edited: Mar 25, 2007
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