To find the global maximum of the function $f(x, y, z) = xyz$ subject to the constraint $2x + y + 2z = 243$, we can use the method of Lagrange multipliers.

Step 1: Set up the Lagrange multiplier equations

We define the Lagrange multiplier function: $\mathcal{L}(x, y, z, \lambda) = f(x, y, z) - \lambda \cdot g(x, y, z)$ where $f(x, y, z) = xyz$ and $g(x, y, z) = 2x + y + 2z - 243$ is the constraint.

Thus, $\mathcal{L}(x, y, z, \lambda) = xyz - \lambda (2x + y + 2z - 243)$

Step 2: Compute partial derivatives

We need to find the partial derivatives of $\mathcal{L}(x, y, z, \lambda)$ and set them equal to zero.

For $x$:

$\frac{\partial \mathcal{L}}{\partial x} = yz - \lambda \cdot 2 = 0$ $yz = 2\lambda$

For $y$:

$\frac{\partial \mathcal{L}}{\partial y} = xz - \lambda \cdot 1 = 0$ $xz = \lambda$

For $z$:

$\frac{\partial \mathcal{L}}{\partial z} = xy - \lambda \cdot 2 = 0$ $xy = 2\lambda$

For $\lambda$ (the constraint equation):

$\frac{\partial \mathcal{L}}{\partial \lambda} = -(2x + y + 2z - 243) = 0$ $2x + y + 2z = 243$

Step 3: Solve the system of equations

Now we solve the system of equations:

1. $yz = 2\lambda$
2. $xz = \lambda$
3. $xy = 2\lambda$
4. $2x + y + 2z = 243$

From equations (1) and (2), we can express $\lambda$ in terms of $x, y, z$: $\lambda = xz$ Substitute into equation (1): $yz = 2xz$ This simplifies to: $y = 2x$

Now, substitute $y = 2x$ into equation (3): $x(2x) = 2xz$ $2x^2 = 2xz$ Cancel the common factor of 2: $x^2 = xz$ If $x \neq 0$, we can divide by $x$: $x = z$

Thus, we have: $y = 2x \quad \text{and} \quad z = x$

Step 4: Substitute into the constraint

Substitute $y = 2x$ and $z = x$ into the constraint $2x + y + 2z = 243$: $2x + 2x + 2x = 243$ $6x = 243$ $x = \frac{243}{6} = 40.5$

Step 5: Find the values of $y$ and $z$

Since $y = 2x$ and $z = x$, we have: $y = 2(40.5) = 81$ $z = 40.5$

Step 6: Find the value of the objective function

Now substitute $x = 40.5$, $y = 81$, and $z = 40.5$ into the objective function $f(x, y, z) = xyz$: $f(40.5, 81, 40.5) = 40.5 \times 81 \times 40.5$ $f(40.5, 81, 40.5) = 40.5^2 \times 81$ $40.5^2 = 1640.25$ $1640.25 \times 81 = 132562.25$

Thus, the global maximum value of $f(x, y, z) = xyz$ subject to the constraint is $\boxed{132562.25}$.

Additional Questions:

1. How do the Lagrange multipliers help in solving constrained optimization problems?
2. What is the role of the constraint $2x + y + 2z = 243$ in this problem?
3. What would happen if we had different constraints?
4. How do we confirm that this is indeed the global maximum?
5. How can the method of Lagrange multipliers be extended to more variables?

Tip:

Always check if the values you get make sense within the context of the problem—sometimes the mathematical solution might need interpretation or verification with boundary conditions.