What is the maximum area of the shadow cast by a box of sides 1, 2, 3?

Question

Assume that the light rays are parallel and the plane of the shadow is perpendicular to the light rays.

The box can be at an angle to the parallel rays of light.

Answer 1

The areas of faces of the box are 2,3,6.
If e is unit vector in the direction of light,
then area of the shadow of each face
is scalar product of normal to the face and e,
total area of the shadow being

2i e + 3j e + 6k e =
(2i + 3j + 6k) e =
√(2²+3²+6²) cosθ,


Answer:
Maximum area of the shadow is 7,
when e = (2/7, 3/7, 6/7)

Answer 2

Looking at this on and off for the last couple of hours, it seems reasonably clear the maximum shadow is a hexagon that can be inscribed in a circle of radius √14 / 2. You can therefore set an upper limit by assuming a regular hexagon, except this shadow will not be a regular hexagon. I'm guessing the sides will be three pairs in the ratio √5:√10;:√13, but that's just a guess.

That's it for now. Adding a third dimension to this makes it tough. If the three paired sides of the hexagon are x, y, and z, I think you need to maximize √(x² + y²) + √(x² + z²) + √(y² + z²) with the limitation that x² + y² + z² = 14. So how's that done?

Answer 3

Take all the three possibilities of placing the box:

i) With sides 1 &2 facing the bottom.
ii) With sides 2 &3 facing the bottom.
iii) With sides 1 &3 facing the bottom.

Find the area covered by these phases. That covering maximum area is the answer,
i.e., Case ii) has base area = 2 * 3 unit sq.

Hence, 6 unit sq. is the maximum area of the shadow cast by the box of sides 1, 2, 3.

Answer 4

Without going into too much detail (in case I'm wrong),
I think the maximum area is :
sqrt(1^2 + 3^2) * sqrt(1^2 + 2^2)
= 5*sqrt(2)

Answer 5

Insufficient data.