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

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.