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

  • Math -

    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 -

    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 -

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

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