Step 1: Understanding the Concept:
This is an optimization problem using derivatives.
We need to define a function for the product described and find its maximum value over a valid domain.
Step 2: Key Formula or Approach:
Let the two parts be $x$ and $20-x$.
The function to maximize is $P(x) = x^3(20 - x)^2$.
Find the critical points by setting $P'(x) = 0$ and verify the maximum using the first or second derivative test.
Step 3: Detailed Explanation:
Let the two parts be $x$ and $y$, so $x + y = 20$, which gives $y = 20 - x$.
We want to maximize the product $P = x^3 y^2 = x^3 (20 - x)^2$.
Differentiating $P$ with respect to $x$:
\[ \frac{dP}{dx} = 3x^2(20 - x)^2 + x^3 \cdot 2(20 - x)(-1) \]
\[ \frac{dP}{dx} = x^2(20 - x) \left[ 3(20 - x) - 2x \right] \]
\[ \frac{dP}{dx} = x^2(20 - x) (60 - 3x - 2x) \]
\[ \frac{dP}{dx} = x^2(20 - x) (60 - 5x) \]
To find critical points, set $\frac{dP}{dx} = 0$:
\[ x^2(20 - x)(60 - 5x) = 0 \]
This gives $x = 0, x = 20, \text{ or } 60 - 5x = 0 \Rightarrow 5x = 60 \Rightarrow x = 12$.
Since the parts must be positive for a meaningful division of 20 in this context, $x = 0$ and $x = 20$ yield a product of 0, which is a minimum.
So, the maximum occurs at $x = 12$.
We can verify it's a maximum using the first derivative test. For $x$ slightly less than 12 (e.g., $x=11$), $\frac{dP}{dx}$ is positive. For $x$ slightly greater than 12 (e.g., $x=13$), $\frac{dP}{dx}$ is negative. Thus, $P(x)$ has a local maximum at $x=12$.
When $x = 12$, the other part is $y = 20 - 12 = 8$.
The two parts are 12 and 8.
Step 4: Final Answer:
The two parts are 12 and 8.