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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.

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    @osmano807 Of course $\rho = 0$ is not in your allowed values, but the method is correct. For the second step, let $z=0$. This satisfies the first two equations. For the third you need either $\rho = \pm 10^{-2}$ or $\phi = 0$ or $\phi = \pi/2.$ Thus $(\rho,z,\phi) = (-0.01, 0,0)$, $(\rho,z,\phi) = (0.01, 0,0)$, $(\rho,z,\phi) = (-0.01, 0,\pi/2)$ and $(\rho,z,\phi) = (0.01, 0,\pi/2)$ are more solutions. Not all of them lie in your prescribed intervals.2012-12-20

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We are told to find the local maxima of the function $f(r,z,\phi):=(r^2-10^{-4})z\sin(2\phi)$ in a certain box region in $(r,z,\phi)$-space. Looking at the data it becomes obvious that such maxima are taken in points where $z=0.04$, $\ \phi={\pi\over4}$, because for every allowed point $(r,z,\phi)$ with $z<0.04$ or $\phi\ne{\pi\over4}$ there is a nearby allowed point $(r',z',\phi')$ with $f(r',z',\phi')>f(r,z,\phi)$.

It follows that it suffices to consider the univariate function $g(r):=r^2-10^{-4}\qquad(0.005\leq r\leq0.02)\ .$ Solving $g'(r)=2r=0$ shows that $g$ has no critical point in the interval $[0.005, 0.02]$. Furthermore $g(0.005)=-0.000075<0.0003=g(0.02)\ .$ It follows that $g$ takes its maximum at the right endpoint of its domain. Therefore the point $\bigl(0.02,0.04,{\pi\over4}\bigr)$ is the unique point in the allowed region where $f$ is locally maximal, and $f\bigl(0.02,0.04,{\pi\over4}\bigr)=0.000012$ is the global maximum of $f$ in this region.