We are tasked with finding the maximum and minimum values of the function $f(x, y, z) = xyz$ subject to the constraint $x^2 + 2y^2 + 3z^2 = 96$ using the method of Lagrange multipliers.

Step 1: Define the Lagrange function

We introduce the Lagrange multiplier $\lambda$ and form the Lagrange function: $\mathcal{L}(x, y, z, \lambda) = xyz + \lambda (96 - x^2 - 2y^2 - 3z^2)$ where the constraint is $g(x, y, z) = x^2 + 2y^2 + 3z^2 - 96 = 0$.

Step 2: Compute the gradients

We need to find the partial derivatives of $\mathcal{L}$ with respect to $x$, $y$, $z$, and $\lambda$.

1. $\frac{\partial \mathcal{L}}{\partial x} = yz - 2\lambda x = 0$
2. $\frac{\partial \mathcal{L}}{\partial y} = xz - 4\lambda y = 0$
3. $\frac{\partial \mathcal{L}}{\partial z} = xy - 6\lambda z = 0$
4. $\frac{\partial \mathcal{L}}{\partial \lambda} = 96 - x^2 - 2y^2 - 3z^2 = 0$

Step 3: Solve the system of equations

We now solve the system of equations: $yz = 2\lambda x$ $xz = 4\lambda y$ $xy = 6\lambda z$ $x^2 + 2y^2 + 3z^2 = 96$

To approach this, we can look for relationships between $x$, $y$, and $z$. Start by dividing equations to eliminate $\lambda$.

1. Divide equation 1 by equation 2:

$\frac{yz}{xz} = \frac{2\lambda x}{4\lambda y} \implies \frac{y}{x} = \frac{1}{2} \implies y = \frac{x}{2}$

2. Divide equation 2 by equation 3:

$\frac{xz}{xy} = \frac{4\lambda y}{6\lambda z} \implies \frac{z}{y} = \frac{2}{3} \implies z = \frac{2y}{3}$

3. Substitute $y = \frac{x}{2}$ and $z = \frac{2y}{3}$ into the constraint

Substitute these into the constraint equation $x^2 + 2y^2 + 3z^2 = 96$.

Since $z = \frac{2y}{3}$, first substitute $z$ into the equation: $x^2 + 2y^2 + 3\left(\frac{2y}{3}\right)^2 = 96$ $x^2 + 2y^2 + 3\left(\frac{4y^2}{9}\right) = 96$ $x^2 + 2y^2 + \frac{4y^2}{3} = 96$ Multiply through by 3 to eliminate the fraction: $3x^2 + 6y^2 + 4y^2 = 288$ $3x^2 + 10y^2 = 288$ Since $y = \frac{x}{2}$, substitute $y = \frac{x}{2}$ into the equation: $3x^2 + 10\left(\frac{x}{2}\right)^2 = 288$ $3x^2 + 10\left(\frac{x^2}{4}\right) = 288$ $3x^2 + \frac{10x^2}{4} = 288$ Multiply through by 4 to clear the fraction: $12x^2 + 10x^2 = 1152$ $22x^2 = 1152$ $x^2 = \frac{1152}{22} = 52.36$ Thus, $x = \sqrt{52.36} \approx 7.23$

Step 4: Calculate $y$ and $z$

Now, $y = \frac{x}{2} \approx \frac{7.23}{2} \approx 3.62$, and $z = \frac{2y}{3} \approx \frac{2(3.62)}{3} \approx 2.41$.

Step 5: Calculate the function value $f(x, y, z) = xyz$

$f(7.23, 3.62, 2.41) = 7.23 \times 3.62 \times 2.41 \approx 63.02$

Thus, the maximum value of $f(x, y, z)$ is approximately 63.02.

Step 6: Check for minimum

By the nature of the function $f(x, y, z) = xyz$, which has no lower bound (since it can take negative values as well), the minimum value is DNE (Does Not Exist).

Final Answer:

• Maximum value: 63.02
• Minimum value: DNE

Would you like more details or further explanations?

Here are 5 related questions to deepen your understanding:

1. What are Lagrange multipliers, and how are they used in optimization?
2. How would the problem change if the constraint were different?
3. What are the necessary conditions for a function to have both maximum and minimum values?
4. How can you visualize the constraint as a geometric shape?
5. What is the geometric interpretation of the solution found?

Tip: Always check the boundary conditions when using Lagrange multipliers to ensure that all possible extrema are found.