Optimization problem: what are the dimensions of the lightest open-top right circularcylindrical can that will hold a volume of 1000cm3? compare the result with the result in example 2.

So Volume of cylinder is pi*r^2*h = 1,000 

Then lightest one means you have the smallest surface area. Which is one base and then the area of the surface. SA = pi*r^2 + 2pi*r*h 

So now you have 2 equations, so: 

h = 1,000/(pi*r^2) 
So then SA = pi*r^2 + 2pi*r*(1,000/(pi*r^2) = pi*r^2 + 2,000/r 

Derivative of SA is then 2pi*r -2,000/r^2. Set to 0 

2pi*r-2,000/r^2 =0 --> 2pi*r^3 = 2,000 --> r^3 = 1,000/pi --> r = 10/pi^(1/3) 

Now go back to the volume function: pi*r^2*h =1,000 --> 1,000/(pi*100/pi^(2/3)) = h 
h = 10 / pi^(1/3)