In a given sphere of radius $R$, it is required to find the cylinder with maximum surface area that we can inscribe in this sphere.
Using that the radius of the cylinder is $r$, with Pythagoras Theorem we can find the height of the cylinder $H =2 \sqrt{R^2 – r^2} $ and we can find the critic points of the function $S(r) = 4 \pi r \sqrt{R^2 – r^2} + 2 \pi (R^2 – r^2)$, which are $ r = \sqrt{ \frac{5 ^+_- \sqrt{5}}{10}}R$, since $r>0$.The answer gives only with the positive sign, since the negative one do not satisfies the equation for $R > 0$, I think. But how can I show that it is indeed a maximum point (or that the function, at least, assume a maximum) without calculating the second derivative? Because if it, indeed, assumes maximum and one of the critic points do not satisfies what I want, then the other one must be what I am looking for.
Maybe it is not the better way to do this exercise too, but with lagrange multipliers it just looks much harder.
Thanks in advance!
Best Answer
The surface area is $0$ at $r=0$ and $\pi R^2$ at $r=R$. If your calculated area is greater than $\pi R^2$ it must be a maximum.
Alpha gets $r \approx 0.812815 R$, with a very messy exact expression if you change $1$ to $R$ here