1. **State the problem:** We need to divide 24 into three parts $x$, $y$, and $z$ such that $x + y + z = 24$ and the product $P = x \cdot y^2 \cdot z^3$ is maximized. 2. **Express the problem mathematically:** We want to maximize $$P = x y^2 z^3$$ subject to $$x + y + z = 24.$$ 3. **Use the constraint to express one variable in terms of the others:** $$x = 24 - y - z.$$ 4. **Rewrite the product in terms of $y$ and $z$ only:** $$P = (24 - y - z) y^2 z^3.$$ 5. **Set up the problem for optimization:** We want to find $y$ and $z$ that maximize $$P(y,z) = (24 - y - z) y^2 z^3,$$ with $y > 0$, $z > 0$, and $x = 24 - y - z > 0$. 6. **Use logarithms to simplify differentiation:** Let $$\ln P = \ln(24 - y - z) + 2 \ln y + 3 \ln z.$$ 7. **Find partial derivatives and set them to zero:** $$\frac{\partial \ln P}{\partial y} = -\frac{1}{24 - y - z} + \frac{2}{y} = 0,$$ $$\frac{\partial \ln P}{\partial z} = -\frac{1}{24 - y - z} + \frac{3}{z} = 0.$$ 8. **From the first equation:** $$-\frac{1}{24 - y - z} + \frac{2}{y} = 0 \implies \frac{1}{24 - y - z} = \frac{2}{y} \implies y = 2(24 - y - z).$$ 9. **From the second equation:** $$-\frac{1}{24 - y - z} + \frac{3}{z} = 0 \implies \frac{1}{24 - y - z} = \frac{3}{z} \implies z = 3(24 - y - z).$$ 10. **Rewrite the two equations:** $$y = 48 - 2y - 2z \implies 3y + 2z = 48,$$ $$z = 72 - 3y - 3z \implies 3y + 4z = 72.$$ 11. **Solve the system:** Subtract the first from the second: $$(3y + 4z) - (3y + 2z) = 72 - 48 \implies 2z = 24 \implies z = 12.$$ 12. **Find $y$:** From $3y + 2z = 48$, $$3y + 2(12) = 48 \implies 3y + 24 = 48 \implies 3y = 24 \implies y = 8.$$ 13. **Find $x$:** $$x = 24 - y - z = 24 - 8 - 12 = 4.$$ 14. **Check positivity:** All parts $x=4$, $y=8$, $z=12$ are positive. 15. **Calculate maximum product:** $$P = 4 \times 8^2 \times 12^3 = 4 \times 64 \times 1728 = 442,368.$$ **Final answer:** The three parts are $4$, $8$, and $12$ and the maximum product is $442,368$.