Challenge A box is 15 in high, 18 in. long, and 6 in, wide. What is the longest poster you could fit in
the box? Use pencil and paper. Explain why you can only fit one maximum-length poster in the box
but you can fit multiple 23-in, posters in the same box.
You can fit a -in poster in the box.
(Round to one decimal place as needed.)

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.

the answer to the expression is 83

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743