Finding the best diameter and height for a tin that will contain 1 litre of paint.

Ok, I have a maths assignment due tomorrow, and I do not know how to do it. The question is mainly about finding the best diameter and hight for a tin that will contain 1 litre of paint.
What is the "efficient" proportion of diameter to height.?

Does this proportion hold for a 5 litre or 10 letre tim of paint?

The dimensions of one brand of paint in 2 litre tins are diameter 14 cm by height 15 cm. How does this compare with your most "efficient" proportions?

These 2 litre tins of paint are packed 5 to a box. The box is square. Determine the dimensions of the box.

Please help me

I assume that the "efficient" dimensions are those which require the minimum amount of tin to hold that volume of paint. Look at what you need to make the tin: Two pieces of radius r to make the top and bottom, and a rectangular piece 2 * pi * r * h to make the sides.

Surface area of a tin (A) = surface area of two ends + area of cylindrical sides
A = 2 * pi * r² + 2 * pi * r * h ... [Equation 1]

Volume of a cylinder (V) = pi * r² * h = 1 litre [(dm^3)]
Rearrange:

h = 1 / (pi * r²) ... [Equation 2]
Substitute for h in [Eq 1]
A = 2*pi*r² + (2*pi*r) / (pi*r²)
A = 2*pi*r² + 2 / r

dA/dr = 4*pi*r - 2/r²

For max or min A, dA/dr = 0 [differentiate again to confirm minimum]

4*pi*r - 2/r² = 0
Multiply through by r²
4 * pi * r^3 - 2 = 0
2 * pi * r^3 = 1
r^3 = 1/(2 * pi)
r = 0.542 dm

Substitute for r in [Eq. 2]

h = 1 /(pi * 0.542²)

h = 1.084 dm

d/h = 2 * 0.542 / 1.084 = 1.00