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2 March, 22:06

Divide the number 20 into two parts (not necessarily integers) in a way that makes the product of one part with the square of the other as large as possible

use calculus methods

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  1. 3 March, 00:59
    0
    13 1/3 and 6 2/3.

    Step-by-step explanation:

    Let the 2 parts be x and (20 - x).

    So x^2 (20 - x) must be a maximum.

    y = x^2 (20 - x)

    y = 20x^2 - x^3

    Finding the derivative):

    y' = 40x - 3x^2 = 0 for maxm/minm value.

    x (40 - 3x) = 0

    40 - 3x = 0

    3x = 40

    x = 40/3.

    This is a maximum because the second derivative y" = 40 - 6x = 40 - 6 (40/3)) is negative.

    So the 2 numbers are 13 1/3 and 6 2/3.
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