Answer
$\pi{r^{2}}(1+\sqrt 5)$
Work Step by Step
The cylinder has surface area
$S=2$(base area)+(lateral surface area) = $2\pi{(radius)^{2}}+2\pi{(radius)(height)}$ = $2\pi{y^{2}}+2\pi{y(2x)}$
$x^{2}+y^{2}$ = $r^{2}$
$y^{2}$ = $r^{2}-x^{2}$
$y$ = $\sqrt {r^{2}-x^{2}}$
so the surface area is
$S(x)$ = $2\pi{(r^{2}-x^{2})}+4\pi{x(\sqrt {r^{2}-x^{2}})}$, $0$ $\leq$ $x$ $\leq$ $r$
$S'(x)$ = $2\pi{r^{2}}-2\pi{x^{2}}+4\pi{(x\sqrt {r^{2}-x^{2}})}$
$S'(x)$ = $-4\pi{x}+4\pi{\left[x\left(\frac{1}{2}\right)(r^{2}-x^{2})^{1\frac{1}{2}}(-2x)+(r^{2}-x^{2})^{\frac{1}{2}}(1)\right]}$
$S'(x)$ = $4\pi{\left[\frac{-x\sqrt {r^{2}-x^{2}}-x^{2}+r^{2}-x^{2}}{\sqrt {r^{2}-x^{2}}}\right]}$
$S'(x)$ =$0$
$x\sqrt {r^{2}-x^{2}}$ = $r^{2}-2x^{2}$
$5x^{4}-5r^{2}x^{2}+r^{4}$ = $0$
This is quadratic equation in $x^{2}$.
By the quadratic formula
$x^{2}$ = $\frac{(5±\sqrt 5)r^{2}}{10}$
We reject the root with the $+$ since it does not satisfy $r^{2}-2x^{2}>0$, so
$x$ = $\sqrt {\frac{(5-\sqrt 5)}{10}}r$
Since $S(0)$ = $S(r)$ = $0$
the maximum surface area occurs at the critical number and
$x^{2}$ = $\frac{(5-\sqrt 5)r^{2}}{10}$
$y^{2}$ = $r^{2}-\frac{(5-\sqrt 5)r^{2}}{10}$ = $\frac{(5+\sqrt 5)r^{2}}{10}$
The surface area is
$2\pi{[\frac{(5+\sqrt 5)r^{2}}{10}]}+4\pi{\sqrt {\frac{(5-\sqrt 5)}{10}}\sqrt {\frac{(5+\sqrt 5)}{10}}r^{2}}$ = $\pi{r^{2}}(1+\sqrt 5)$
