Find the maximum value of the multiple variable function Show that the maximum value of$$xy(z-h)\left (\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}\right )$$is$$\left (\frac{2}{5}h\right )^5\frac{ab}{c^4}.$$
 A: Take the partial derivatives with respect to $x$, $y$, and $z$, then set them to $0$. For $x$,
$$y(z-h)\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}\right)+xy(z-h)\frac{2x}{a^2}=0\\y(z-h)\left(\frac{3x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}\right)=0$$
Similarly for $y$:
$$x(z-h)\left(\frac{x^2}{a^2}+\frac{3y^2}{b^2}-\frac{z^2}{c^2}\right)=0$$
For $z$:
$$xy\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}\right)-xy(z-h)\frac{2z}{c^2}=0\\xy\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{3z^2}{c^2}+\frac{2zh}{c^2}\right)=0$$
You know that $x\ne0$, $y\ne 0$, and $(z-h)\ne 0$ (otherwise the expression is zero). Then you have a simpler system:
$$\frac{3x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=0\\
\frac{x^2}{a^2}+\frac{3y^2}{b^2}-\frac{z^2}{c^2}=0\\\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{3z^2}{c^2}+\frac{2zh}{c^2}=0$$
Subtracting the first two equations above will give $$\frac{x^2}{a^2}=\frac{y^2}{b^2}$$
Adding the first two equations yield:
$$4\frac{x^2}{a^2}+4\frac{y^2}{z^2}=2\frac{z^2}{c^2}$$
Therefore $$\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac14\frac{z^2}{c^2}$$
Now plug this expression into the third simplified equation. You will get an equation in $z$. It is quadratic, but one solution $z=0$ is not good (since that yields $x=y=0$). So it's a linear equation that will give you $z$ as a function of $h$ and $c$. Then find $x$ and $y$, and plug them into the original expression.
