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Given a cone with height $H$ and radius $R$ we want to find the optimal radius $r$ and height $h$ of a cylinder placed inside the cone to maximize the volume of the cylinder.

This can be written as: $$f(\lambda, r, h)=\pi r^2h-\lambda \left(h-(1-r/R)H\right)$$

where $h=(1-r/R)H$ is out constraint.

Using the optimization via Lagrange multipliers we find:

$$0=\partial_hf=\pi r^2-\lambda\Rightarrow\lambda=\pi r^2$$ this we can insert into $\partial_rf=0$: $$0=\partial_rf=2\pi rh-\pi r^2\frac{H}{R}\Rightarrow r=2h\frac{R}{H}$$ inserted into the constraint gives us $h$: $$h=(1-r/R)H=\frac{1}{3}H$$ which in turn gives us finally $r$: $$r=\frac{2}{3}R$$

So we should have a local maximum for $(r^*,h^*)=(\frac{2}{3}R,\frac{1}{3}H)$ and $V^*=\frac{4}{27}\pi R^2H$ (which I also get using mathematica).

If I now want to confirm this result using the second derivative test I calculate the bounded hessian matrix:

$$H_f=\begin{pmatrix} \frac{\partial^2f}{\partial\lambda^2} & \frac{\partial^2f}{\partial\lambda\partial h} & \frac{\partial^2f}{\partial\lambda\partial r} \\ \frac{\partial^2f}{\partial h\partial\lambda} & \frac{\partial^2f}{\partial r^2} & \frac{\partial^2f}{\partial r\partial h} \\ \frac{\partial^2f}{\partial r\partial\lambda} & \frac{\partial^2f}{\partial h\partial r} & \frac{\partial^2f}{\partial h^2} \end{pmatrix} =\begin{pmatrix} 0 & -1 & -\frac{H}{R} \\ -1 & 0 & 2\pi r \\ -\frac{H}{R} & 2\pi r & 2\pi h\\ \end{pmatrix}$$

which has $det H_f(r^*,h^*)=2\pi H >0$ but which is not positive definite (as $det\begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}=-1<0$).

Where is the problem?

Update: I just tested another geometrical optimization problem (cuboid inside ellipsoid) where I face the same problem so I conclude that my bounded hessian matrix should be wrong.

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I the first place you don't need second derivatives here. The radius $r$ of the cylinder satisfies $0\leq r\leq R$, and the volume is obviously zero when $r=0$ or $r=R$. It follows that the maximum is taken at an interior point and has to show up using Lagrange's method. Since you have found just one conditionally stationary point this has to be the maximum.

The Hessian can be used to analyze "fully" stationary points, where all involved $x$-variables are independent. There is also a "Hessian test" for conditionally stationary points, but it is more involved than you think.

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    1. Isn't in this case the "hessian theorem" stated incomplete (in most sources e.g. wikipedia, lectures) if it does not apply to "all" optimization problems? 2. What do I need to search for to find the "hessian test for conditionally stationary points"?2017-02-07
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    In my professional life I never had reason to go into this. In the article about the "bordered Hessian" in Wikipedia there is a reference [5] where the answer to your question might be found.2017-02-07