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Find two numbers adding to 20 such that the sum of their squares is as small as possible.
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x+y = 20 minimize x^2+y^2
y = 20-x f(x) = x^2 + (20-x)^2 = x^2 + 400 - 40x + x^2
differentiate: f'(x) = 2x -40 + 2x = 0, solve for x. x = 10
test if it's a min: f''(x) = 4 > 0, so x=10 is a min.
y = 20-x=20-10=10
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Thanks a ton I can see where I mess up
np, i thought i was messing up too somewhere.. i kept getting 0 for y' for some reason
me too
bet anything you make them equal right?
yes
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calc will work but you can argue by symmetry, since you cannot tell the difference between x and y
\[x^2+y^2\] \[x+y=20\] switch x and y and get \[y^2+x^2\] \[y+x=20\] meaning you cannot tell the difference between x and y
just thought i would mention it
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