Math
posted by Anonymous .
Find two numbers whose difference is 16, and the sum of whose squares a minimum.

minimize x^2+y^2 where xy=16
x=8 y=8
That is, f(x) = x^2 + (x16)^2
f' = 4(x8)
f'=0 when x=8 
x = first number
y = second number
x  y = 16
that is equivalent of
y = x  16
x ^ 2 + y ^ 2 = x ^ 2 + ( x  16 ) ^ 2 =
x ^ 2 + x ^ 2  2 * x * 16 + 16 ^ 2 =
2 x ^ 2  32 x + 256
Quadratic equation a x ^ 2 + b x + c
vhen a is positive in point :
x =  b / 2 a
have MINIMUM
x =  b / 2 a =  (  32 ) / ( 2 * 2 ) = 32 / 4 = 8
When x = 8 minimum value of 2 x ^ 2  32 x + 256 =
2 * 8 ^ 2  32 * 8 + 256 = 2 * 64 + 256 + 256 = 128
x = 8
y = x  16 = 8  16 =  8
x ^ 2 + y ^ 2 = 8 ^ 2 + (  8 ) ^ 2 = 64 + 64 = 128 
Thanks....I figured it out to be (8, 8) also. Thanks again.
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