Find the largest value of the function 𝑧3 + 𝑦5 − 𝑥2 = )𝑧 ,𝑦 ,𝑥(𝑓 over the surface 4 = 𝑧^2 + 𝑦^2 + 𝑥^2.

We are asked to find the largest value of the function \( f(z, y, x) = z^3 + y^5 - x^2 \) over the surface defined by \( z^2 + y^2 + x^2 = 4 \).

To solve this, we'll use the method of **Lagrange multipliers**. This method is applicable when we need to find the extrema of a function subject to a constraint.

### Step 1: Define the functions
- The objective function to maximize is:
  \[
  f(z, y, x) = z^3 + y^5 - x^2
  \]
- The constraint function is:
  \[
  g(z, y, x) = z^2 + y^2 + x^2 - 4 = 0
  \]

### Step 2: Set up the Lagrange equations
We introduce a Lagrange multiplier \( \lambda \) and set up the equations:
\[
\nabla f = \lambda \nabla g
\]
where \( \nabla \) denotes the gradient. This gives us the system of equations:
1. \( \frac{\partial f}{\partial z} = 3z^2 = \lambda \cdot 2z \)
2. \( \frac{\partial f}{\partial y} = 5y^4 = \lambda \cdot 2y \)
3. \( \frac{\partial f}{\partial x} = -2x = \lambda \cdot 2x \)

### Step 3: Solve the system of equations

1. **From the first equation** \( 3z^2 = 2\lambda z \):
   - If \( z \neq 0 \), we can divide both sides by \( z \), giving:
     \[
     3z = 2\lambda \quad \text{or} \quad \lambda = \frac{3z}{2}
     \]
   - If \( z = 0 \), substitute into the constraint equation later.

2. **From the second equation** \( 5y^4 = 2\lambda y \):
   - If \( y \neq 0 \), we can divide both sides by \( y \), giving:
     \[
     5y^3 = 2\lambda \quad \text{or} \quad \lambda = \frac{5y^3}{2}
     \]
   - If \( y = 0 \), substitute into the constraint equation later.

3. **From the third equation** \( -2x = 2\lambda x \):
   - If \( x \neq 0 \), we can divide both sides by \( x \), giving:
     \[
     -2 = 2\lambda \quad \text{or} \quad \lambda = -1
     \]
   - If \( x = 0 \), substitute into the constraint equation later.

### Step 4: Solve for critical points

From the above, we have the following cases:

1. **Case 1: \( z = 0 \), \( y = 0 \)**:
   - The constraint becomes \( x^2 = 4 \), so \( x = \pm 2 \).
   - The function value is \( f(0, 0, \pm 2) = -(\pm 2)^2 = -4 \).

2. **Case 2: \( x = 0 \), \( y = 0 \)**:
   - The constraint becomes \( z^2 = 4 \), so \( z = \pm 2 \).
   - The function value is \( f(\pm 2, 0, 0) = (\pm 2)^3 = 8 \).

3. **Case 3: \( z = 0 \), \( x = 0 \)**:
   - The constraint becomes \( y^2 = 4 \), so \( y = \pm 2 \).
   - The function value is \( f(0, \pm 2, 0) = (\pm 2)^5 = 32 \).

### Step 5: Conclusion

Among the function values computed:
- \( -4 \) for \( f(0, 0, \pm 2) \),
- \( 8 \) for \( f(\pm 2, 0, 0) \),
- \( 32 \) for \( f(0, \pm 2, 0) \),

The largest value is \( \boxed{32} \), which occurs at \( (z, y, x) = (0, \pm 2, 0) \).

Would you like more details or have any questions about this solution?

### Related Questions:
1. How can the method of Lagrange multipliers be used in optimizing functions with multiple constraints?
2. What is the geometric interpretation of the surface \( z^2 + y^2 + x^2 = 4 \)?
3. How do you determine the nature (maxima, minima, saddle point) of critical points in constrained optimization?
4. Can the gradient method be used when the function has a non-smooth constraint surface?
5. How does the symmetry of the function and constraint influence the optimization process?

### Tip:
When solving for critical points using Lagrange multipliers, always consider the cases where variables might be zero separately, as they can lead to different solutions.

Discover how to maximize the function z^3 + y^5 - x^2 over the surface defined by z^2 + y^2 + x^2 = 4 using Lagrange multipliers. Explore detailed steps and solutions to find the largest value. Suitable for advanced mathematics enthusiasts.

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