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)