Question 962508
The volume is described by
{{{pi*r^2*h=2pi}}}
so
{{{h=(2pi)/(pi*r^2)=2/r^2}}}
The surface area is described by
{{{SA=2pi*h+2*pi*r}}}
{{{SA=2pi*(2/r^2)+2*pi*r}}}
{{{SA=(4pi)/r+2*pi*r^2}}}
Taking the derivative, we get
{{{d/(dr)*SA=-4pi/r^2+4pi*r}}}
{{{d/(dr)*SA=(-4pi+4pi*r^3)/r^2}}}
which is equal to zero when
{{{-4pi+4pi*r^3=0}}}
{{{-1+r^3=0}}}
{{{r^3=1}}}
{{{r=1}}}
If you do the first derivative test, you'll find that this is indeed a minimum value. Thus
{{{h=2/1^2=2}}}
so the optimal dimensions are 1in radius by 2in height.