Question

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

Answer 1

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

Answer 2

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

Answer 3

Thanks....I figured it out to be (8, -8) also. Thanks again.

Answer 4

The numbers 🤣 are 3 and 5