Answer:
a. 24.2 in
b. Paths of the internal diameter, where the longest poster is placed cross each other
Step-by-step explanation:
a. The longest poster that can be fitted into the box is the size of the internal diagonal of the box, that is
√((8² + 6²) +15²) = 3·√65 = 24.2 in
b. Since the above give the diagonal of the box, there can only be one poster of the dimension 24.2 in as due to path interference of the longest poster that is placed at the internal diagonal of the volume of the box
However, for the surface diagonal we have the dimension given by;
√(8² + 15²) = 23.4 in. Therefore since the side diagonal is parallel to the side of the box, one can fit several 23 in. posters which are lesser than the length of the diagonal of the side of the box side by side in the box.
