To find the global maxima and minima of the function $f(x, y, z) = x^3 - y^3 + z^2$ subject to the constraint $2x^2 + y^2 + z^2 \leq 2$, we can follow these steps:

1. Define the Problem:

• Objective function: $f(x, y, z) = x^3 - y^3 + z^2$
• Constraint: $g(x, y, z) = 2x^2 + y^2 + z^2 \leq 2$

2. Use Lagrange Multiplier Method:

Since we have a constraint, we can solve this using the method of Lagrange multipliers. The Lagrange multiplier method states that at the maximum or minimum of the function $f(x, y, z)$, the gradient of $f(x, y, z)$ is parallel to the gradient of the constraint $g(x, y, z)$. This gives us the system:

$\nabla f(x, y, z) = \lambda \nabla g(x, y, z)$

3. Compute the Gradients:

• Gradient of $f(x, y, z)$:

$\nabla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (3x^2, -3y^2, 2z)$

• Gradient of $g(x, y, z)$:

$\nabla g(x, y, z) = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z} \right) = (4x, 2y, 2z)$

4. Set Up the System of Equations:

From the method of Lagrange multipliers, we equate the gradients:

$(3x^2, -3y^2, 2z) = \lambda (4x, 2y, 2z)$

This gives us the following system of equations:

1. $3x^2 = 4\lambda x$
2. $-3y^2 = 2\lambda y$
3. $2z = 2\lambda z$

5. Solve the System of Equations:

We now solve each equation:

1. Solve $3x^2 = 4\lambda x$:

If $x \neq 0$, divide both sides by $x$:

$3x = 4\lambda \quad \Rightarrow \quad \lambda = \frac{3x}{4}$

2. Solve $-3y^2 = 2\lambda y$:

If $y \neq 0$, divide both sides by $y$:

$-3y = 2\lambda \quad \Rightarrow \quad \lambda = \frac{-3y}{2}$

3. Solve $2z = 2\lambda z$:

If $z \neq 0$, divide both sides by $2z$:

$1 = \lambda$

6. Analyze the Cases:

We now consider the cases for each variable being zero and non-zero.

Case 1: $z = 0$

From equation (3), $\lambda = 1$.

Now substitute $\lambda = 1$ into the other equations:

• From equation (1): $3x^2 = 4x$, we get $x(3x - 4) = 0$, so $x = 0$ or $x = \frac{4}{3}$.
• From equation (2): $-3y^2 = 2y$, we get $y(3y + 2) = 0$, so $y = 0$ or $y = -\frac{2}{3}$.

So, the possible solutions are $(x, y, z) = (0, 0, 0), \left(\frac{4}{3}, 0, 0\right), (0, -\frac{2}{3}, 0)$.

Case 2: $x = 0$

From equation (1), $3x^2 = 4\lambda x$, this is automatically satisfied since $x = 0$.

• From equation (2): $-3y^2 = 2\lambda y$, we get $y = 0$ or $y = -\frac{2}{3}$.
• From equation (3): $2z = 2\lambda z$, we get $\lambda = 1$, so $z \neq 0$.

Check $g(x, y, z) \leq 2$: