How do you solve this problem?

Question

What are the dimensions of the lightest open-top right circular cylindrical can that will hold a volume of 1000^3?

Answer 1

Let r = radius of the base

     h = height

     V = volume

     S = surface area

 

Given:  V = πr²h = 1000 cubic units   Then, h = 1000/(πr²)

 

Minimize:  S = area of base + area of the side

 

                  = πr² + 2πrh

 

                  = πr² + 2πr(1000)/(πr²)

 

                  = πr² + 2000/r, r>0

 

S' = 2πr - 2000/r² = (2πr³-2000)/r²

 

            S' = 0 when 2πr³ = 2000

 

                                 r³ = 1000/π

 

                                 r = 10/(π)^1/3

 

If 0<r<10/(π)^1/3, S'<0.  So, S is decreasing.

 

If r > 10/(π)^1/3, S'>0.  So, S is increasing.

 

Therefore, S is minimized when r = 10/(π)^1/3

 

                                            h = 1000/(πr²) = 10/(π)^1/3