Maximum surface area of cylinder (1-variable)

Question

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!

Answer 1

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

Answer 2

We have the constraint $\dfrac{H^2}4 + r^2 = R^2$ and $S(r ) = 2 \pi r H + 2 \pi r^2$ to maximise.

So we can maximise $r(H+r)$, subject to $H^2+4r^2 = 4R^2$. In fact to simplify, scale all variables down by $R$, to maximise $x(h+x)$ s.t. $h^2+4x^2=4$.

Now, we can set the Lagrangian to $L =x^2+xh - \lambda(h^2+4x^2-4)$.

Trust you can continue from there to solve $\dfrac{\partial L}{\partial x}=0, \dfrac{\partial L}{\partial h}=0,$...