Use Lagrange multipliers to find the maximum and minimum values of the function
f (x, y, z) = x4 + y4 + z4 subject to the constraint x2 + y2 + z2 = 1.

To find the maximum and minimum values of the function \( f(x, y, z) = x^4 + y^4 + z^4 \) subject to the constraint \( g(x, y, z) = x^2 + y^2 + z^2 - 1 = 0 \), we use the method of Lagrange multipliers. 

1. **Set up the Lagrange function**:
   \[
   \mathcal{L}(x, y, z, \lambda) = x^4 + y^4 + z^4 + \lambda (1 - x^2 - y^2 - z^2)
   \]

2. **Compute the gradients**:
   \[
   \nabla f = \left( 4x^3, 4y^3, 4z^3 \right)
   \]
   \[
   \nabla g = \left( -2x, -2y, -2z \right)
   \]

3. **Set up the equations**:
   The method of Lagrange multipliers gives us the equations:
   \[
   \nabla f = \lambda \nabla g
   \]
   This results in the system of equations:
   \[
   4x^3 = -2\lambda x, \quad 4y^3 = -2\lambda y, \quad 4z^3 = -2\lambda z
   \]
   Along with the constraint:
   \[
   x^2 + y^2 + z^2 = 1
   \]

4. **Analyze the equations**:
   From \( 4x^3 + 2\lambda x = 0 \), we can factor:
   \[
   x(4x^2 + 2\lambda) = 0 \implies x = 0 \quad \text{or} \quad 4x^2 + 2\lambda = 0 \implies \lambda = -2x^2
   \]
   Similar analysis applies to \( y \) and \( z \).

5. **Case analysis**:
   - **Case 1**: \( x = 0 \)
     - Then \( y^2 + z^2 = 1 \).
     - Let \( y = \cos \theta \) and \( z = \sin \theta \):
       \[
       f(0, \cos \theta, \sin \theta) = \cos^4 \theta + \sin^4 \theta
       \]
     - This can be simplified using \( \cos^4 \theta + \sin^4 \theta = \left( \cos^2 \theta + \sin^2 \theta \right)^2 - 2 \cos^2 \theta \sin^2 \theta = 1 - \frac{1}{2} \sin^2(2\theta) \).

   - **Case 2**: \( y = 0 \) or \( z = 0 \)
     - The analysis is similar, leading to the same results for symmetry.

   - **Case 3**: None of \( x, y, z \) are zero.
     - Set \( x = y = z \):
       \[
       3x^2 = 1 \implies x = \pm \frac{1}{\sqrt{3}} \implies (x, y, z) = \left(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}\right)
       \]
       Calculate \( f \):
       \[
       f\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) = 3 \left(\frac{1}{\sqrt{3}}\right)^4 = 3 \cdot \frac{1}{9} = \frac{1}{3}
       \]

6. **Evaluate extrema**:
   - \( f(1, 0, 0) = 1^4 + 0 + 0 = 1 \)
   - \( f(0, 1, 0) = 1 \)
   - \( f(0, 0, 1) = 1 \)
   - \( f\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) = \frac{1}{3} \)

7. **Conclusion**:
   The maximum value is \( 1 \) and occurs at the points \( (1, 0, 0) \), \( (0, 1, 0) \), and \( (0, 0, 1) \). The minimum value is \( \frac{1}{3} \) and occurs at the points \( \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \).

### Summary
- **Maximum**: \( 1 \) at \( (1, 0, 0), (0, 1, 0), (0, 0, 1) \)
- **Minimum**: \( \frac{1}{3} \) at \( \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \)

Do you want details on any specific part or have any questions? Here are some related questions:

1. How can Lagrange multipliers be used in different optimization problems?
2. What are some geometric interpretations of the method of Lagrange multipliers?
3. Can you explain the role of the constraint in optimization problems?
4. How does the method differ for inequality constraints?
5. What happens if the function or constraint is modified? 

**Tip**: When applying Lagrange multipliers, always check the endpoints and any critical points found through the method for global extrema.

Use Lagrange multipliers to find the maximum and minimum values of the function f(x, y, z) = x^4 + y^4 + z^4 subject to the constraint x^2 + y^2 + z^2 = 1.

This solution demonstrates the use of Lagrange multipliers to find maximum and minimum values of the function f(x, y, z) = x^4 + y^4 + z^4 under the constraint x^2 + y^2 + z^2 = 1. Key steps include setting up the Lagrange function, computing gradients, and analyzing equations. Suitable for undergraduate students studying multivariable calculus and optimization.

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