Posted by bobby on Tuesday, December 16, 2008 at 7:52pm.
two numbers : x and 10-x
S = x^2 + (10-x)^2
dS/dx = 2x - 2(10-x)
= 0 for max/min
x = 5
so the numbers are both 5
x and (10-x)
s = x^2 + (10-x)^2
s = x^2 + 100 -20x + x^2
s = 2 x^2 - 20 x + 100
s/2 = x^2 -10 x + 100 we can minimize half the sum easier than the whole sum
That is a parabola and you could find the vertex but since you said this was "calculus" we will take the derivative and set to zero.
0 = 2 x - 10
x = 5
10-x = 5
Interesting the answer is halfway between.
Say a sum of two numbers is s
We want to minimize the sum of squares of x^2 and (s-x)^2
sum = 2 x^2 -2sx
d sum/dx = 0 = 4 x -2s
x = s/2
so it works for any old sum, not just 10
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