Minimal lateral surface of a cylinder 
Inscribe in a given sphere a cylinder such that its lateral surface
  (without the bases) shall be maximal.

So lateral surface is = $2 \pi rh  $ $ \Rightarrow 4 \pi x \sqrt{r^2-x^2}$
Now take derivative ? or what should i do?
 A: *

*There is a set of cylinders that fit into a sphere of radius $r$. Consider those that just fit.






*Try to parametrize them by some parameter $t$, e.g. radius of cylinder or height of cylinder or half height of cylinder.

*Calculate the lateral surface for these cylinders in terms of the parameter $S_L(t)$.

So lateral surface is = $2 \pi rh  $ $ \Rightarrow 4 \pi x
 \sqrt{r^2-x^2}$

That looks like a parametrization in terms of $x$:
$$
S_L(x) = 2 \pi R h = 2 \pi R (2|x|) = 4\pi \sqrt{r^2-x^2} x
$$
where $R$ is the radius of the cylinder, and we consider $x \ge 0$ only.
Note: The above images have the cylinder axis aligned with the $z$ axis, this would be the parametrization of cylinders aligned with the $x$-axis.

(Large version)


*Now use simple calculus to find the maximum of $S_L(x)$ for some $x^*$.


Because of symmetry consider $x \in [0, r]$. Local maxima can occur at critical points where $S_L'(x) = 0$. Otherwise extrema for continious functions can occur at the interval boundaries.
Determining the critical points:
$$
0 = S_L'(x) 
= 4 \pi \left( \frac{1}{2\sqrt{r^2 - x^2}} (-2x) x + \sqrt{r^2 - x^2} \right) \iff \\
0 
= \frac{r^2 - x^2 - x^2}{\sqrt{r^2 - x^2}} 
= \frac{r^2 - 2x^2}{\sqrt{r^2 - x^2}} \Rightarrow \\
r^2 = 2 x^2 \Rightarrow \\
x = r / \sqrt{2}
$$
Further examination would give that this is a local maximum. I updated the graph to show it.
A: No calculus required:
Let's call the radius of the cylinder $x$, and the radius of the sphere $r$. 
To simplify the problem, let $x=r\cos\theta$. Then the height of the cylinder, $h$, is $2r\sin\theta$. 
So the lateral surface area of the sphere is $2\pi xh = 4r^2\pi cos\theta\sin\theta=2r^2\pi \sin2\theta$. This expression is maximized when $\theta=\frac{\pi}{4}$, which is when $x=r\frac{\sqrt{2}}{2}$.
