What are the dimensions of the lightest open-top right circular cylinder can that will hold a volume of 200π cubic cm.?

Answer 1

Let V=πr²h be the volume and M=πr² +2πrh be the surface area. So h=200π/πr² = 200/r² and the surface area becomes M=πr² +2πr(200/r²) =πr² +400π/r and dM/dr = 2πr -400π/r². To find the r let dM/dr =0 and so r=(200)^1/3 or r= 5.847 cm. This is a minimum since d²M/dr² =2π +800π/r² which is >0. using the value for r in the equation for h gives h= 5.85 cm so h≈r.