First, let's recognize that we must find the dimensions of a cylinder that yields a volume of #500# #cm^3# and has the least surface area possible. This means that we must minimize the quantity #2pir^2# + #2pirh#.
Since the volume of a cylinder is #pir^2h#, we have
#pir^2h# = #500#.
#h# = #500/(pir^2)#.
Plugging back into our equation for surface area yields
#A(r)# = #2pir^2# + #1000/r#.
Since we want to find the minimum of this function, we must take the derivative with respect to #r# and set the derivative equal to 0.
#0# = #4pir# + #-1000/r^2#.
Now, all we have to do to solve for #r#. Rearranging yields
#1000/r^2# = #4pir#
#1000# = #4pir^3#
#r^3# = #250/pi#
#r = root(3)(250/pi)#.
Solving for #h# with this value of #r# yields
#h = 500/(piroot(3)(250/pi)^2#
#h = 500/(pi(250^(2/3)/pi^(2/3)))#
#h = (2times250^(3/3))/(250^(2/3)times(pi^(3/3)/pi^(2/3)))#
#h = 2times250^(1/3)/pi^(1/3)#
#h = 2root(3)(250/pi)#.
So our final answer is a cylinder with a radius of #4.3# #cm# and height of #8.6# #cm#.
