If DE≅EF, CD = v + 45, and CF = 4v, what is the value of v?
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Answer:
564 ft²
Step-by-step explanation:
To account for the extra space between units, we can add 2" to every unit dimension and every box dimension to figure the number of units per box.
Doing that, we find the storage box dimensions (for calculating contents) to be ...
3 ft 2 in × 4 ft 2 in × 2 ft 2 in = 38 in × 50 in × 26 in
and the unit dimensions to be ...
(4+2)" = 6" × (6+2)" = 8" × (2+2)" = 4"
A spreadsheet can help with the arithmetic to figure how many units will fit in the box in the different ways they can be arranged. (See attached)
When we say the "packing" is "462", we mean the 4" (first) dimension of the unit is aligned with the 3' (first) dimension of the storage box; the 6" (second) dimension of the unit is aligned with the 4' (second) dimension of the storage box; and the 2" (third) dimension of the unit is aligned with the 2' (third) dimension of the storage box. The "packing" numbers identify the unit dimensions, and their order identifies the corresponding dimension of the storage box.
We can see that three of the four allowed packings result in 216 units being stored in a storage box.
If storage boxes are stacked 4 deep in a 9' space, the 2' dimension must be the vertical dimension, and the floor area of each stack of 4 boxes is 3' × 4' = 12 ft². There are 216×4 = 864 units stored in each 12 ft² area.
If we assume that 2 weeks of production are 80 hours of production, then we need to store 80×500 = 40,000 units. At 864 units per 12 ft² of floor space, we need ceiling(40,000/864) = 47 spaces on the floor for storage boxes. That is ...
47 × 12 ft² = 564 ft²
of warehouse floor space required for storage.
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The second attachment shows the top view and side view of units packed in a storage box.
