We are given the problem of dividing 24 into three parts, say $x$, $y$, and $z$, such that the continued product of the first, square of the second, and cube of the third is maximized. The continued product is defined as:

$P = x \cdot y^2 \cdot z^3$

We need to maximize this product under the constraint:

$x + y + z = 24$

Step 1: Set up the objective function and constraint

The objective function to maximize is:

$P(x, y, z) = x \cdot y^2 \cdot z^3$

The constraint is:

$g(x, y, z) = x + y + z - 24 = 0$

Step 2: Use the method of Lagrange multipliers

We now apply the method of Lagrange multipliers. Define the Lagrangian as:

$\mathcal{L}(x, y, z, \lambda) = x \cdot y^2 \cdot z^3 + \lambda (x + y + z - 24)$

Step 3: Find the partial derivatives

Take the partial derivatives of $\mathcal{L}$ with respect to $x$, $y$, $z$, and $\lambda$:

1. $\frac{\partial \mathcal{L}}{\partial x} = y^2 z^3 + \lambda = 0$
2. $\frac{\partial \mathcal{L}}{\partial y} = 2xy z^3 + \lambda = 0$
3. $\frac{\partial \mathcal{L}}{\partial z} = 3x y^2 z^2 + \lambda = 0$
4. $\frac{\partial \mathcal{L}}{\partial \lambda} = x + y + z - 24 = 0$

Step 4: Solve the system of equations

We now solve the system of equations:

1. From $\frac{\partial \mathcal{L}}{\partial x} = 0$: $\lambda = -y^2 z^3$

2. From $\frac{\partial \mathcal{L}}{\partial y} = 0$: $\lambda = -2xy z^3$

3. From $\frac{\partial \mathcal{L}}{\partial z} = 0$: $\lambda = -3x y^2 z^2$

Equating the expressions for $\lambda$ from the first two equations:

$y^2 z^3 = 2xy z^3$

Canceling $z^3$ (assuming $z \neq 0$):

$y^2 = 2xy$

This simplifies to:

y = 2x \quad \text{(assuming y \neq 0)}

Next, equating the first and third expressions for $\lambda$:

$y^2 z^3 = 3x y^2 z^2$

Canceling $y^2 z^2$ (assuming $y \neq 0$ and $z \neq 0$):

$z = 3x$

Step 5: Substitute into the constraint

We now substitute $y = 2x$ and $z = 3x$ into the constraint $x + y + z = 24$:

$x + 2x + 3x = 24$

$6x = 24$

$x = 4$

Step 6: Find $y$ and $z$

Now that we know $x = 4$:

$y = 2x = 2(4) = 8$ $z = 3x = 3(4) = 12$

Step 7: Verify the maximum

We have found $x = 4$, $y = 8$, and $z = 12$. To verify that this gives the maximum value of the product, we check the second-order conditions or test the critical points within the context of the problem.

Final Answer:

The values of $x$, $y$, and $z$ that maximize the continued product are:

$x = 4, \quad y = 8, \quad z = 12$

Would you like to explore more on this problem or need further clarifications? Here's a list of related questions:

1. How does the Lagrange multiplier method work in optimization problems?
2. What are the second-order conditions for optimization?
3. How can we extend this approach to other similar problems?
4. How would the problem change if the constraint was $x + y + z = 30$?
5. How do we interpret the values of $x$, $y$, and $z$ in terms of the maximum product?

Tip: When using Lagrange multipliers, always remember to check for non-zero values in denominators before canceling terms.