<s>[INST] Maximization problem with partial derivative Problem: Find the volume of the largest rectangular box (with faces parallel to the coordinate planes) that can be inscribed inside the ellipsoid \begin{align*} \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \end{align*}
Attempt at a solution: Since the box is inside the ellipsoid, the volume of the box is given by $V= 2x \cdot 2y \cdot 2z = 8xyz$. The box and the ellipsoid have several coordinates in common (at the respective vertices), so choose $(x,y,z)$ in the first octant. We can solve the equation for the ellipsoid for $z$ so that we are left with only two variables instead of three. \begin{align*} 1 &= \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \\ z^2 &= c^2 (1- \frac{y^2}{b^2} - \frac{x^2}{a^2}) \\ z &= c \sqrt{1 - \frac{y^2}{b^2} - \frac{x^2}{a^2}} \end{align*} We put this in our equation for the volume and get: \begin{align*} V &= 8xyc \sqrt{1 - \frac{y^2}{b^2} - \frac{x^2}{a^2}} \\ V^2 &= 64x^2 y^2 c^2 \Big(1- \frac{y^2}{b^2}- \frac{x^2}{a^2} \Big) \\ &= 64c^2 \Big(x^2 y^2 - \frac{x^2 y^4}{b^2} - \frac{x^4 y^2}{a^2} \Big) \end{align*} I guess now we need to take partials with respect to $x$ and $y$. \begin{align*} \frac{\partial V^2}{\partial x} &= 64c^2 \Big(x^2 y^2 - \frac{x^2 y^4}{b^2} - \frac{x^4 y^2}{a^2} \Big)' \\ &= 64c^2 \Big(2xy^2 - \frac{2xy^4}{b^2} - \frac{4x^3 y^2}{a^2}\Big) \\ &= 128c^2 xy^2 \Big(1- \frac{y^2}{b^2} - \frac{2x^2}{a^2} \Big) \\ \frac{\partial V^2}{\partial y} &= 128c^2x^2 y \Big(1- \frac{2y^2}{b^2} - \frac{x^2}{a^2} \Big) \end{align*} How can I proceed now? I know I have to equate the partials with zero and solve, but there's just too many variables?
 [/INST] Since $x,y>0$, you can conclude that $\displaystyle\frac{2x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\displaystyle\frac{x^2}{a^2}+\frac{2y^2}{b^2}=1$.
Subtracting these two equations gives $\displaystyle\frac{x^2}{a^2}=\frac{y^2}{b^2}$, and then substituting back into either equation gives
$\;\;\;\displaystyle\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{1}{3}$, so $\displaystyle\frac{z^2}{c^2}=\frac{1}{3}$ and $\displaystyle x=\frac{a}{\sqrt{3}}, y=\frac{b}{\sqrt{3}}, z=\frac{c}{\sqrt{3}}.$
Therefore the largest volume is given by $\displaystyle V=8xyz=\frac{8abc}{3\sqrt{3}}$.

Here is an alternate approach:
We want to maximize $V=8xyz$ subject to $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$, so using Lagrange multipliers gives
$\displaystyle\;\;\;8yz=\lambda\cdot\frac{2x}{a^2}, \;\;8xz=\lambda\cdot\frac{2y}{b^2}, \;\;8xy=\lambda\cdot\frac{2z}{c^2}$.
Multiplying by x in the first equation, y in the second, and z in the third gives
$\;\;\displaystyle\lambda\left(\frac{2x^2}{a^2}\right)=\lambda\left(\frac{2y^2}{b^2}\right)=\lambda\left(\frac{2z^2}{c^2}\right)$, so $\lambda\ne0\implies\displaystyle\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}$.
Since $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1,\;\;\; \displaystyle\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{1}{3}\implies x=\frac{a}{\sqrt{3}}, y=\frac{b}{\sqrt{3}}, z=\frac{c}{\sqrt{3}}$.
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