Two questions actually.
If you are trying to minimize the surface area of a cylinder with a given volume (500), is there better way to solve this than the following:
 Solve for height in terms of radius using the equation for volume
V=pir^2h
500=pir^2h
500/(pi*r^2)=h
 Substitute that value for height in the surface area equation.
SA= 2pir^2 + 2pirh
SA= 2pir^2 + 2pir(500/(pir^2)
SA= 2pi*r^2 + (1000/r)

Then you graph the last formula, and trace the line to find the yminimum.

Then plug in the r value at the ymin to the volume equation to find the height.
Ok, now I know that will eventually give you the correct answer if you are discerning enough to find the minimum correctly, but is there another way to solve this without calculus. I know you can set the derivative to 0, but my students do not know calculus yet.
Also, if a question specifies that the top and bottom portions of the can (the parts represented in the SA equation by 2pir^2) cost twice as much as the sides, would you just double that portion of the equation when graphing it to figure out the proper height/radius? Thanks in advance.