Asked by larry
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M06_Unit 6_Area & Volume DUA (DUE:2.24.26) — 14 Questions
Stuckey, Larry
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Question 3.
Stephanie is renting a storage unit. The storage unit measures
feet long,
feet high, and
feet wide.
Stephanie places 2 larger boxes in the back of the storage unit.
Box A measures
feet long,
feet high, and
feet wide.
Box B measures
feet long,
feet high, and
feet wide.
Stephanie wants to fill the remaining space of the storage unit with smaller boxes that are the same size:
feet long,
feet high, and
feet wide.
What is the largest number of smaller boxes Stephanie can fit into the storage unit?
90
89
101
102
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All Answers 1
Answered by
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AI
I can solve this, but some numbers are missing from the problem you pasted (the lengths/heights/widths of the storage unit and the boxes). Could you either:
- Type in the missing dimensions for the storage unit, Box A, Box B, and the small box, or
- Upload a clear photo/screenshot of the full problem?
While you do that, here are the two ways I’ll use to get the correct answer once I have the numbers (and which one I’ll try first):
1) Dimension-by-dimension packing (recommended)
- This is the accurate method for box-packing into a rectangular space when all boxes are axis-aligned.
- Compute how many small boxes fit along each dimension: floor(storage_length / small_length) × floor(storage_height / small_height) × floor(storage_width / small_width).
- If the two larger boxes occupy some integral number of those small-box “slots” (often they will if dimensions are multiples), subtract the number of slots occupied by Box A and Box B from the total slots.
- If rotation of the small box is allowed, check the different orientations (permutations of small_length/height/width) and pick the orientation that yields the maximum count.
2) Volume method (only when perfect packing can be assumed)
- Compute remaining volume = storage_volume − volume(Box A) − volume(Box B).
- Compute max small boxes = floor(remaining_volume / volume(small box)).
- Use this only if the problem statement implies ideal/continuous packing (or when all dimensions are multiples so volume division equals slot count). This can overestimate if packing leaves unusable gaps.
If you give me the dimensions now I’ll compute exactly and tell you which multiple-choice answer (90, 89, 101, or 102) is correct.
- Type in the missing dimensions for the storage unit, Box A, Box B, and the small box, or
- Upload a clear photo/screenshot of the full problem?
While you do that, here are the two ways I’ll use to get the correct answer once I have the numbers (and which one I’ll try first):
1) Dimension-by-dimension packing (recommended)
- This is the accurate method for box-packing into a rectangular space when all boxes are axis-aligned.
- Compute how many small boxes fit along each dimension: floor(storage_length / small_length) × floor(storage_height / small_height) × floor(storage_width / small_width).
- If the two larger boxes occupy some integral number of those small-box “slots” (often they will if dimensions are multiples), subtract the number of slots occupied by Box A and Box B from the total slots.
- If rotation of the small box is allowed, check the different orientations (permutations of small_length/height/width) and pick the orientation that yields the maximum count.
2) Volume method (only when perfect packing can be assumed)
- Compute remaining volume = storage_volume − volume(Box A) − volume(Box B).
- Compute max small boxes = floor(remaining_volume / volume(small box)).
- Use this only if the problem statement implies ideal/continuous packing (or when all dimensions are multiples so volume division equals slot count). This can overestimate if packing leaves unusable gaps.
If you give me the dimensions now I’ll compute exactly and tell you which multiple-choice answer (90, 89, 101, or 102) is correct.
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