Two numbers have a sum of 13. Find the minimum of the sum of their squares.
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The square root of 9 is 3
The square root of 4 is 2
I hope that helps.
Let x and y be the numbers.
Adding the numbers yields 13: x+y = 13
The sum of the squares is `x^2 + y^2 = S`
Write the sum of squares in terms of x.
`S(x) = x^2 + (13-x)^2`
Expanding the binomial yields:
`S(x) = x^2 + 169 - 26x + x^2 =gt` `S(x) = 2x^2 - 26x + 169`
The sum of squares is minimum if `S'(x) = 0` .
Calculating `S'(x)` yields:`S'(x) = 4x - 26` .
If `S'(x) = 0 =gt 4x - 26 = 0 =gt 4x = 26 =gt x = 26/4`
`y = 13 - 26/4 =gt y = (52-26)/4 =gt y = 26/4`
If `x=y=26/4` , the sum of the squares of the numbers is minimum.
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