Posted by **Anonymous** on Monday, August 20, 2012 at 12:36am.

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

- Math -
**Steve**, Monday, August 20, 2012 at 4:31am
minimize x^2+y^2 where x-y=16

x=8 y=-8

That is, f(x) = x^2 + (x-16)^2

f' = 4(x-8)

f'=0 when x=8

- Math -
**Bosnian**, Monday, August 20, 2012 at 4:55am
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

- Math -
**Anonymous**, Monday, August 20, 2012 at 10:31pm
Thanks....I figured it out to be (8, -8) also. Thanks again.

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