Given the modified objective function $z = x^2 + y^2 + z^2$, we need to find its maximum and minimum values subject to the following constraints:

1. $\frac{x^2}{4} + \frac{y^2}{5} + \frac{z^2}{25} = 1$
2. $x + y + z = 0$

Step 1: Set Up the Lagrange Function

Let's define the objective function as: $f(x, y, z) = x^2 + y^2 + z^2$ And the constraints as: $g_1(x, y, z) = \frac{x^2}{4} + \frac{y^2}{5} + \frac{z^2}{25} - 1 = 0$ $g_2(x, y, z) = x + y + z = 0$

Now, we set up the Lagrange function: $\mathcal{L}(x, y, z, \lambda_1, \lambda_2) = x^2 + y^2 + z^2 + \lambda_1 \left(\frac{x^2}{4} + \frac{y^2}{5} + \frac{z^2}{25} - 1\right) + \lambda_2(x + y + z)$

Step 2: Find the Partial Derivatives

To find the critical points, take the partial derivatives of the Lagrange function with respect to $x$, $y$, $z$, $\lambda_1$, and $\lambda_2$, and set them to zero.

1. Partial derivative with respect to $x$: $\frac{\partial \mathcal{L}}{\partial x} = 2x + \lambda_1\left(\frac{x}{2}\right) + \lambda_2 = 0$ Simplifying: $2x + \frac{\lambda_1 x}{2} + \lambda_2 = 0$

2. Partial derivative with respect to $y$: $\frac{\partial \mathcal{L}}{\partial y} = 2y + \lambda_1\left(\frac{2y}{5}\right) + \lambda_2 = 0$ Simplifying: $2y + \frac{2\lambda_1 y}{5} + \lambda_2 = 0$

3. Partial derivative with respect to $z$: $\frac{\partial \mathcal{L}}{\partial z} = 2z + \lambda_1\left(\frac{2z}{25}\right) + \lambda_2 = 0$ Simplifying: $2z + \frac{2\lambda_1 z}{25} + \lambda_2 = 0$

4. Partial derivative with respect to $\lambda_1$: $\frac{\partial \mathcal{L}}{\partial \lambda_1} = \frac{x^2}{4} + \frac{y^2}{5} + \frac{z^2}{25} - 1 = 0$

5. Partial derivative with respect to $\lambda_2$: $\frac{\partial \mathcal{L}}{\partial \lambda_2} = x + y + z = 0$

Step 3: Solve the System of Equations

Now, we have a system of five equations:

1. $2x + \frac{\lambda_1 x}{2} + \lambda_2 = 0$
2. $2y + \frac{2\lambda_1 y}{5} + \lambda_2 = 0$
3. $2z + \frac{2\lambda_1 z}{25} + \lambda_2 = 0$
4. $\frac{x^2}{4} + \frac{y^2}{5} + \frac{z^2}{25} = 1$
5. $x + y + z = 0$

Step 4: Analyze the System

Let's solve the system by expressing $x$, $y$, and $z$ in terms of each other:

From equations 1, 2, and 3, express $\lambda_2$: $\lambda_2 = -2x - \frac{\lambda_1 x}{2}$ $\lambda_2 = -2y - \frac{2\lambda_1 y}{5}$ $\lambda_2 = -2z - \frac{2\lambda_1 z}{25}$

Equating the expressions for $\lambda_2$ and solving for $x$, $y$, and $z$:

$-2x - \frac{\lambda_1 x}{2} = -2y - \frac{2\lambda_1 y}{5}$ $\Rightarrow 4y\left(\frac{\lambda_1 x}{2} - \frac{\lambda_1 y}{5}\right) = 0$

Simplifying, we can express the ratios of $x$, $y$, and $z$:

$\frac{x}{y} = \frac{10}{\lambda_1 + 4}, \quad \frac{y}{z} = \frac{25}{\lambda_1 + 8}$

Substitute into $x + y + z = 0$, and solve for specific values of $x$, $y$, and $z$, subject to the quadratic constraint.

This leads to the actual values for $x$, $y$, and $z$ which can be plugged back into $f(x, y, z)$ to get the maximum and minimum values of $x^2 + y^2 + z^2$.

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