In a problem, I'm asked to find the local maximum of the function: $ \rho_v = (\rho^2 - 10^{-4})z\sin(2\phi) $ over the solid: \begin{align*} 0.005 &\leq \rho \leq 0.02 \\ 0 &\leq \phi \leq \frac{\pi}{2} \\ 0 &\leq z \leq 0.04 \end{align*}
Differentiating, I get the following system: \begin{cases} 2\rho z \sin(2\phi) &= 0 \\ (\rho^2 - 10^{-4}) z \cos(2\phi) &= 0 \\ (\rho^2 - 10^{-4}) \sin(2\phi) &= 0 \end{cases}
While solving, I get from the first two partial derivatives that ($\phi = 0$ or $\phi = \frac{\pi}{2}$) and that $\phi = \frac{\pi}{4}$. Clearly, that's not possible, the original function doesn't have critical points. (Or maybe these two cases are not mutually exclusive, but I learned that between the equations I use an AND.)
However, inspecting the function I can affirm that it have a local maximum with $\rho = 0.02$, $\phi = \frac{\pi}{4}$ and $z = 0.04$. What I don't know is how to find those values explicitly, that is, without guessing.