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  1. #1
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    Math help

    Hi. I have a question that reads " Among all pairs of numbers whose sum is 15, find the pair such that the sum of their squares is the smallest possible."

    Could someone please work this problem out for me and explain the steps. Thanks!!!!!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathfailure View Post
    Hi. I have a question that reads " Among all pairs of numbers whose sum is 15, find the pair such that the sum of their squares is the smallest possible."

    Could someone please work this problem out for me and explain the steps. Thanks!!!!!
    since there are not many numbers, let's try and do this by trial and error

    here are the pairs of integers who's sum give 15

    1 + 14 ...........1^2 + 14^2 = 197

    2 + 13 ...........2^2 + 13^2 = 173

    3 + 12 ...........3^2 + 12^2 = 153

    4 + 11 ...........4^2 + 11^2 = 137

    5 + 10 ...........5^2 + 10^2 = 125

    6 + 9 ............6^2 + 9^2 = 117

    7 + 8 ............7^2 + 8^2 = 115 ........this is your guy


    and they repeat with the second number taking the place of the first and the first taking the place of the second

    i'm sure TPH or someone else can come up with a neater way to do this
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  3. #3
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    Quote Originally Posted by Mathfailure View Post
    Hi. I have a question that reads " Among all pairs of numbers whose sum is 15, find the pair such that the sum of their squares is the smallest possible."

    Could someone please work this problem out for me and explain the steps. Thanks!!!!!
    Call one of the numbers x, the other must then be 15 - x

    As x + 15 - x = 15

    Squaring them and adding them together we get

    x^2 + (15 - x)^2 = x^2 + 225 - 30x + x^2 = 2x^2 - 30x + 225

    Call this expression y, ie y = 2x^2 - 30x + 225

    We are looking for the minimum point so differentiate

    dy/dx = 4x - 30

    For minimum dy/dx = 0 so 4x - 30 = 0

    Solving

    4x = 30
    x = 7.5

    So pair is 7.5, 15 - 7.5 = 7.5
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jhevon View Post
    i'm sure TPH or someone else can come up with a neater way to do this
    let me try:

    we want x + y = 15 and x^2 + y^2 a minimum

    Calculus Way:


    since x + y = 15
    => x = 15 - y

    so x^2 + y^2 = (15 - y)^2 + y^2

    = 225 - 30y +y^ + y^2
    = 225 - 30y + 2y^2

    Let's call this S
    that is, S = 225 - 30y + 2y^2
    since this is a parabola, we want S' = 0 for min point

    => S' = -30 + 4y = 0
    => 4y - 30 = 0
    => y = 30/4 = 7 1/2

    but x = 15 - y = 15 - 7 1/2 = 7 1/2

    so x = y = 7.5 is a solution, but these are not integers, so i guess we can shift 0.5 from 1 to the other to get the integers, so the numbers are 8 and 7. if you didn't necessarily want integers, then 7.5 and 7.5 is your answer

    PreCalculus Way:

    since x + y = 15
    => x = 15 - y

    so x^2 + y^2 = (15 - y)^2 + y^2

    = 225 - 30y +y^ + y^2
    = 225 - 30y + 2y^2

    Let's call this S
    that is, S = 225 - 30y + 2y^2

    we want the vertex of S
    that is, we want, y = -b/2a = 30/2(2) = 30/4 = 7.5

    but x = 15 - y = 15 - 7.5 = 7.5

    and the same argument holds
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  5. #5
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    The non-calculus way is to use the theorem about finding max/min points on parabolas.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    The non-calculus way is to use the theorem about finding max/min points on parabolas.
    Nice!
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