I need help with additional mathematics project about aluminium cans.?

Question

The Muhibbah Company is a manufacturer of cylindrical aluminium tins.The manager plans to reduce the cost of production. The production cost is proportional to the area of aluminium sheet used. The volume that each tin can hold is 1000 cm³( 1 litre )
1.Determine the value of h, r and hence calculate the ratio of h/r when the total surface area of each tin is minimum. Here, h cm denotes the height and r cm the radius of the tin.
2. The top and bottom pieces of the tin of height h cm are cut from square-shaped aluminium sheets.
Determine the value for r,h and hence calculate the ratio h/r so that the total area of the aluminium sheets used for making the tin is minimum. ( refer to the diagram below )
3. Investigate cases where the top and bottom surfaces are cut from
i) Equilateral triangle
ii) Regular hexagon
Find the ratio of h/r for each case.

Answer 1

1.) The cylinder with the lowest surface area per unit volume will be cubelike.

Surface area for a cylinder is 2πr^2 + 2πrh, where the 2πr^2 is for the base, and 2πrh is for the tube itself.

On a similar cube, the base would be 2 sides (top and bottom), and the tube would be remaining 4 sides. Thus, for a cubelike cylinder, 2πrh will equal 2 * 2πr^2 (since the tube accounts for twice the area as the base).

2πrh = 2 * 2πr^2
h = 2r (h/r = 2 for your ratio)

V = πr^2h = 2πr^3
1000 = 2πr^3
500 / π = r^3 = 159.15
r = 5.419 cm, so h = 10.838 cm

SA = 2πr^2 + 2πrh = 2πr^2 + 4πr^2 = 6πr^2
6πr^2 = 553.52 cm^2

Check:
r = 5.418
V = πr^2h = 1000
h = 1000 / π * (5.418 ^2) = 10.844
SA = 2πr^2 + 2πrh = 2π (5.418 ^2) + 2π (5.418 * 10.844) = 553.60

r = 5.42
V = πr^2h = 1000
h = 1000 / π * (5.42 ^2) = 10.836
SA = 2πr^2 + 2πrh = 2π (5.42 ^2) + 2π (5.42 * 10.836) = 553.56

Without the diagram, it'll be hard to answer 2 and 3.