To maximize or minimize the function $f(x, y, z) = x + z$ under the constraint $g(x, y, z) = x^2 + y^2 + z^2 = 1$, we can apply the method of Lagrange multipliers.

Step 1: Set up the Lagrange multiplier equations

The method of Lagrange multipliers states that to find the extremum of $f(x, y, z)$ subject to the constraint $g(x, y, z) = 1$, we introduce a Lagrange multiplier $\lambda$ and solve the following system of equations:

$\nabla f = \lambda \nabla g$ along with the constraint $g(x, y, z) = 1$.

Step 2: Compute the gradients of $f$ and $g$

The gradient of $f(x, y, z)$ is:

$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (1, 0, 1)$

The gradient of $g(x, y, z)$ is:

$\nabla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z} \right) = (2x, 2y, 2z)$

Step 3: Set up the Lagrange multiplier equation

Now we apply the condition $\nabla f = \lambda \nabla g$, which gives:

$(1, 0, 1) = \lambda (2x, 2y, 2z)$

This results in the following system of equations:

1. $1 = 2\lambda x$
2. $0 = 2\lambda y$
3. $1 = 2\lambda z$

Step 4: Solve the system of equations

From equation (2), we get $\lambda y = 0$. Since $\lambda \neq 0$ (otherwise, the equations become trivial), we conclude that $y = 0$.

Substituting $y = 0$ into the constraint $g(x, y, z) = 1$, we get:

$x^2 + z^2 = 1$

From equations (1) and (3), we have:

• $\lambda = \frac{1}{2x}$
• $\lambda = \frac{1}{2z}$

Equating these two expressions for $\lambda$, we get:

$\frac{1}{2x} = \frac{1}{2z}$

Thus, $x = z$.

Step 5: Substitute into the constraint

Now that we know $x = z$, substitute this into the constraint $x^2 + z^2 = 1$:

$2x^2 = 1 \quad \Rightarrow \quad x^2 = \frac{1}{2} \quad \Rightarrow \quad x = \pm \frac{1}{\sqrt{2}}$

Thus, $z = \pm \frac{1}{\sqrt{2}}$ as well.

Step 6: Find the values of $f(x, y, z)$

Since $f(x, y, z) = x + z$, we now compute the values of $f$ for the two possible cases for $x$ and $z$:

1. If $x = \frac{1}{\sqrt{2}}$ and $z = \frac{1}{\sqrt{2}}$, then $f = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.
2. If $x = -\frac{1}{\sqrt{2}}$ and $z = -\frac{1}{\sqrt{2}}$, then $f = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -\sqrt{2}$.

Step 7: Conclusion

The maximum value of $f(x, y, z)$ is $\sqrt{2}$, and the minimum value is $-\sqrt{2}$.

Would you like any further details or have additional questions on this method? Here are some follow-up questions to deepen your understanding:

1. How do you determine the Lagrange multiplier $\lambda$ in other optimization problems?
2. What happens if the constraint $g(x, y, z)$ is not a circle but a different shape, like an ellipse?
3. How can you apply Lagrange multipliers in higher dimensions, say for a function of four or more variables?
4. Can you explain how the solution changes if the function $f(x, y, z)$ is modified, such as using $f(x, y, z) = x^2 + z^2$?
5. What happens when there are multiple constraints in the optimization problem?

Tip: Always check if the constraint leads to a solution that satisfies all necessary conditions.