Question 684171
<pre>
This is a cross section cut through the center of the sphere:

{{{drawing(200,200,-1,1,-1,1,locate(.3,0,r),locate(.74,.4,h),locate(.3,.5,R),locate(.74,-.24,h),
circle(0,0,1), line(sqrt(2)/2,sqrt(2)/2,sqrt(2)/2,-sqrt(2)/2),
line(-sqrt(2)/2,-sqrt(2)/2,-sqrt(2)/2,sqrt(2)/2),green(line(0,0,sqrt(2)/2,0),
line(0,0,sqrt(2)/2,sqrt(2)/2)))

 )}}}

The lateral area of circular cylinder is

      {{{A}}}{{{""=""}}}{{{2pi*rsdius*height}}}

      {{{A}}}{{{""=""}}}{{{2pi*r*2h}}}

      {{{A}}}{{{""=""}}}{{{4pi*r*h}}}

By the Pythagorean theorem (refer to the drawing): {{{h}}}{{{""=""}}}{{{sqrt(R^2-r^2)}}}, so we substitute:

      {{{A}}}{{{""=""}}}{{{4pi*r*sqrt(R^2-r^2)}}}

Since square roots are difficult to work with, let's square both sides:

      {{{A^2}}}{{{""=""}}}{{{16pi^2*r^2*(R^2-r^2)}}}

The trick here is that if we maximize the SQUARE of the lateral area,
we will also have maximized the lateral area.  So we let S = AČ

      {{{S}}}{{{""=""}}}{{{16pi^2*r^2*(R^2-r^2)}}}

      {{{S}}}{{{""=""}}}{{{16pi^2*R^2*r^2- 16pi^2*r^4}}}

      {{{(dS)/(dr)}}}{{{""=""}}}{{{32pi^2*R^2*r- 64pi^2*r^3}}}

We set that equal to zero:

      {{{32pi^2*R^2*r- 64pi^2*r^3}}}{{{""=""}}}0 

      {{{32pi^2*r(R^2-2r^2)}}}{{{""=""}}}0

Divide through by constant {{{32pi^2}}}

      {{{r(R^2-2r^2)}}}{{{""=""}}}0

      r=0;         RČ - 2rČ = 0
(min, area = 0)        -2rČ = -RČ
                         rČ = {{{R^2/2}}}
                          r = {{{R/sqrt(2)}}} 

So the radius of the cylinder which has maximum
surface area is {{{R/sqrt(2)}}}

       Since h = {{{sqrt(R^2-r^2)}}}
             h = {{{sqrt(R^2-R^2/2)}}}
             h = {{{sqrt(R^2/2)}}}
             h = {{{R/sqrt(2)}}}

   height = 2h = {{{2R/sqrt(2)}}}

Edwin</pre>