Optimization Problem?

Find two positive real numbers such that they sum to 102 and the product of the first times the square of the second is a maximum.

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  • Philip
    Lv 6
    2 months ago

    Put numbers = x & (102-x).

    Put f(x) = x(102-x)^2 = x(x-102)^2 = x[x^2 -204x + a^2], where a = 102.

    Then f(x) = x^3 -204x^2 + a^2x.

    f'(x) = 3x^2 - 408x + a^2.

    f''(x) = 6x - 408  = 6(x-68).

    Extremums occur where f'(x) = 0, ie., where x^2 -136x + (1/3)[3*34]^2 = 0, ie., where

    g(x) = x^2 - 136x + 3(34)^2 = 0...(1). for g(x) = 0, 2x = 136(+/-)D, where D^2 = (136)^2 - 4*3*(34)^2 = 16(34)^2 - 12(34)^2 = 4(34)^2. Then D =2*34 =68 and

    2x = 136(+/-)68 = 68[2(+/-)1]. Then x = 34(1 or 3) = 34...(2) or x = 102...(3). Clearly,

    root x = 102 is invalid. Therefore x = 34.

    At x = 34, f''(x) = 6*34 - 408 = -(408-204) = -204, < 0 ---> extremum at x = 34 is a

    maximum.

    f(34) = 34(102-34)^2 = 34[3*34 - 34] = 34[2*34]  = 2*17[4*17] = 8*(17)^2 = 8*(289) =

    8*(290-1) = 2320-8 = 2312.

    The 2 numbers are 34 and 68 such that f(34) = maximum = 2312 = 34*68. 

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  • david
    Lv 7
    2 months ago

    n = 1st number

    102 - n = 2nd number

    f(n) = n(102-n)^2  <<< derivative of this

     f '(n)  = 3(n^2 - 136 n + 3468)

              n^2 - 136 n + 3468 = 0

             (n-102)(n-34) = 0

            n = 34  and  n = 102<< not possible, 2ns number would be 0

       n = 34 ...  102 - n = 68  <<<  answers

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  • sepia
    Lv 7
    2 months ago

    Find two positive real numbers such that they sum to 102 

    and the product of the first times the square of the second is a maximum.

    x + y = 102

    x(y^2) = 101^2

    (102 - y)y^2 = 101^2

    y^3 - 102y^2 + 101^2 = 0

    y = 101 and x = 1

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  • 2 months ago

    Find two positive real numbers such that they sum to 102 

    and the product of the first times the square of the second is a maximum.

    x + y = 102

    x(y^2) = 51^3

    (102 - y)y^2 = 51^3

    y^3 - 102y^2 + 51^3 = 0

    y = 51

    The two positive real numbers are 51 and 51

    • Philip
      Lv 6
      2 months agoReport

      Krisnamurthy, would you please be so kind as to check my solution for errors?

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