Many thanks to Andrei Lazanu of School 205 Bucharest for the inspiration for this solution. Well done Andrei. One might have expected the triangular solution to be best. I wonder what happens if you chop off the corners?

Using the following notation:

- for the height of the cup and

- for the diameter of the cup.

Test each of the two possible forms of the box, one with a rectangular base and one with a triangle as its base:

Cuboidal box

As there are six cups they could be ordered in two ways:

1 x 6 cups

and

2 x 3 cups

For any cuboid box, the surface area is: 2 times the top area + 2 times the side area + 2 times the front area:

The total area of a 1 x 6 cups box is:

= 185470

The total area of a 2 x 3 cups box is:

= 157250

For a triangular prism box
This case is illustrated in the figure of the problem.

The base is an equilateral triangle. The surface area is 2 times the area of the equilateral triangle + 3 times the area of each of the faces.

First calculate the side of the equilateral triangle ABC.

This is 2 x BM + 2 x 85~mm.

BM is a tangent to the circumference of a corner cup (centre O, radius d/2 = 85/2~mm).

Also the angle MOB = 60°

Therefore angle MBO =30°.

So the best box is a box of 2 x 3 cups.