Hatakekakashi
DavidTutorexamPAL
Bunuel, just for the sake of nitpicking, why can't the cans be arranged in a triangle with another one on top? They would still all be 'on their bases', as required.
even though the question is not directed towards me I think if it's a triangle all 4 sides of the box won't bulge
So I did the math and actually you would lose very little height by arranging in a triangle, the answer would still be (C).
Basically if you put two cans on the base of the box = the base of the triangle they still need 8 inches of space for width.
And if you put the third right above them in the middle then you lose about 0.54 inches in total length giving a length of about 7.46.
So a 30-inch-perimeter box with a side of 7.5 inches would still likely bulge on all sides and at any rate is the only relevant answer.
Since we're on the topic, consider an arrangement where two cans are in opposite corners and are tangent to each other.
The other two cans are on top of them.
We have no information on the total height of the box vs. height of the cans so this could work.
Then the side of the minimal square is 4 + 2*sqrt(2) and the perimieter is about 27.3 inches
This could also be made to bulge on 4 sides quite easily by decreasing the perimeter slightly (but there is no relevant answer).
Anyways, for anyone reading please note that the above is not really relevant for your exam, feel free to skip over it