Conditional Maximum Value

Firstly, it has to be recognised that finding the maximum value of $x^2 + y^2 + z^2$ is the same as finding the maximum value of $k$ for $x^2 + y^2 + z^2 = k$ to have at least one solution consistent with the remaining equations.

The equations can then be labelled as follows:

\left\{ \begin{array}{1 1} x^2 + y^2 + z^2 = k \quad \quad \textbf{Equation 1} \\ \frac {x^2}{4} + \frac {y^2}{5} + \frac {z^2}{25} = 1 \quad \; \; \, \textbf{Equation 2} \\ z = x + y \quad \quad \quad \quad \; \; \; \; \textbf{Equation 3} \end{array} \right.

Substituting Equation 3 into Equation 1 and Equation 2 :

\begin{array}{1 1} \quad \,\, x^2 + y^2 + (x + y)^2 = k \\ \Rightarrow x^2 + y^2 + x^2 + 2xy + y^2 = k \\ \Rightarrow 2x^2 + 2xy + 2y^2 = k \qquad \qquad \qquad \qquad \quad \textbf{Equation 4} \end{array}

\begin{array}{1 1} \quad \,\, \frac {x^2}{4} + \frac {y^2}{5} + \frac {(x + y)^2}{25} = 1 \\ \Rightarrow 25x^2 + 20y^2 + 4(x^2 + 2xy + y^2) = 100 \\ \Rightarrow 29x^2 + 8xy + 24y^2 = 100 \qquad \qquad \qquad \;\;\;\: \textbf{Equation 5} \end{array}

Now, since Equation 1 represents a hollow sphere, Equation 2 represents an ellipsoid and Equation 3 represents a plane in a three-dimensional Cartesian space, therefore when $k$ is at the maximum value, the three surfaces share the same tangent.

Therefore, it is also true that the curves represented by Equation 4 and Equation 5 have the same gradient when $k$ is at the maximum value.

Differentiating Equation 4 and Equation 5 with respect to $x$ (using the product rule) respectively:

\begin{array}{1 1} \quad \,\, \frac {\mathrm d}{\mathrm d x} (2x^2 + 2xy + 2y^2) = \frac {\mathrm d}{\mathrm d x} k \\ \Rightarrow 4x + (2y + 2x \frac {\mathrm d y}{\mathrm d x}) + 4y \frac {\mathrm d y}{\mathrm d x} = 0 \\ \Rightarrow \frac {\mathrm d y}{\mathrm d x} = - \frac {4x + 2y}{2x + 4y} = - \frac {2x + y}{x + 2y} \qquad \qquad \qquad \quad \;\; \textbf{Equation 6} \end{array}

\begin{array}{1 1} \quad \,\, \frac {\mathrm d}{\mathrm d x} (29x^2 + 8xy + 24y^2) = \frac {\mathrm d}{\mathrm d x} 100 \\ \Rightarrow 58x + (8y + 8x \frac {\mathrm d y}{\mathrm d x}) + 48y \frac {\mathrm d y}{\mathrm d x} = 0 \\ \Rightarrow \frac {\mathrm d y}{\mathrm d x} = - \frac {58x + 8y}{8x + 48y} = - \frac {29x + 4y}{4x + 24y} \qquad \qquad \qquad \;\, \textbf{Equation 7} \end{array}

Therefore at the maximum value of $k$ :

\begin{array}{1 1} \quad \,\, - \frac {2x + y}{x + 2y} = - \frac {29x + 4y}{4x + 24y} \\ \Rightarrow (2x + y)(4x+ 24y) = (29x + 4y)(x + 2y) \\ \Rightarrow 8x^2 + 52xy + 24y^2 = 29x^2 + 62xy + 8y^2 \\ \Rightarrow 16y^2 - 10xy - 21x^2 = 0 \\ \Rightarrow 16^2 y^2 - 2 \cdot 5 \cdot 16xy - 336x^2 = 0 \\ \Rightarrow 16^2 y^2 - 2 \cdot 16y \cdot 5x + 5^2 x^2 = 361x^2 \\ \Rightarrow (16y - 5x)^2 = 19^2 x^2 \\ \Rightarrow 16y - 5x = \pm 19x \\ \Rightarrow y = \frac {5x \pm 19x}{16} \qquad \qquad \qquad \qquad \qquad \qquad \;\;\,\, \textbf{Equation 8} \end{array}

This means that $\boxed{y = - \frac {7}{8} x}$ or $\boxed{y = \frac {3}{2} x}$ .

Substituting $y = - \frac {7}{8} x$ into Equation 5 :

\begin{array}{1 1} \quad \,\, 29x^2 + 8x(- \frac {7}{8} x) + 24(- \frac {7}{8} x)^2 = 100 \\ \Rightarrow \frac {323}{8} x^2 = 100 \\ \Rightarrow x^2 = \frac {800}{323} \\ \Rightarrow \boxed{x = \pm 20 \sqrt{\frac {2}{323}}} \qquad \qquad \qquad \qquad \qquad \;\;\, \textbf{Equation 9} \end{array}

Substituting Equation 9 into $y = - \frac {7}{8} x$ , $\boxed{y = \mp \frac {35}{\sqrt{646}}}$ . Therefore, from Equation 4 :

\begin{array}{1 1} \quad \,\, k = 2(\pm 20 \sqrt{\frac {2}{323}})^2 + 2(\pm 20 \sqrt{\frac {2}{323}})(\mp \frac {35}{\sqrt{646}}) + 2(\mp \frac {35}{\sqrt{646}})^2 \\ \Rightarrow k = \frac {1600}{323} - \frac {1400}{323} + \frac {1225}{323} \end{array} \\ \Rightarrow \boxed{k = \frac {75}{17}}

Similarly, substituting $y = \frac {3}{2} x$ into Equation 5 :

\begin{array}{1 1} \quad \,\, 29x^2 + 8x(\frac {3}{2} x) + 24(\frac {3}{2} x)^2 = 100 \\ \Rightarrow 95 x^2 = 100 \\ \Rightarrow x^2 = \frac {20}{19} \\ \Rightarrow \boxed{x = \pm 2 \sqrt{\frac {5}{19}}} \qquad \qquad \qquad \qquad \qquad \quad \;\: \textbf{Equation 10} \end{array}

Substituting Equation 10 into $y = \frac {3}{2} x$ , $\boxed{y = \pm 3 \sqrt{\frac {5}{19}}}$ . Therefore, from Equation 4 :

\begin{array}{1 1} \quad \,\, k = 2(\pm 2 \sqrt{\frac {5}{19}})^2 + 2(\pm 2 \sqrt{\frac {5}{19}})(\pm 3 \sqrt{\frac {5}{19}}) + 2(\pm 3 \sqrt{\frac {5}{19}})^2 \\ \Rightarrow k = \frac {40}{19} + \frac {60}{323} + \frac {90}{19} \end{array} \\ \Rightarrow \boxed{k = 10}

Since $10 > \frac {75}{17}$ , therefore the maximum value of $x^2 + y^2 + z^2$ is 10 .

First get rid of $z$ . The condition is now $\frac{x^2}{4} + \frac{y^2}{5} + \frac{(x+y)^2}{25} = 1$ and you want to find $2(x^2+xy+y^2)$ .

Let $n = x^2 + xy + y^2$ so you want to find $2n$ .

Rewriting the condition gives $29x^2 + 8xy + 24y^2 = 100$

$21x^2 + 16y^2 + 8n = 100$

$9x^2 + 4y^2 + 12(x^2+y^2) + 8n = 100$

Applying AM-GM gives $9x^2+4y^2 \ge 12xy$ .

Substituting that back in the equation gives $12xy + 12(x^2+y^2) + 8n \le 100$ or $20n \le 100$ .

The maximum value of $n$ is $5$ , so the answer is $2 \times 5 = \boxed{10}$

This is an example of a large class of problems that can be tackled with Lagrange multipliers. I calculate the necessary partial derivatives, and solve the resulting system of equations, using the sympy library in the Python language but this is unnecessary (if your algebra is better than mine!).