Asked by n
Marcus is organizing equipment in a large equipment shed. The shed measures **18 feet long, 10 1/2 feet high, and 12 feet wide**.
Marcus places 2 large storage crates along the back wall of the shed.
* **Crate A** measures **5 feet long, 6 feet high, and 4 1/2 feet wide**.
* **Crate B** measures **7 1/2 feet long, 5 feet high, and 3 feet wide**.
Marcus wants to fill the remaining space in the shed with identical smaller bins that measure **3 feet long, 2 1/2 feet high, and 3 feet wide**.
What is the greatest number of smaller bins Marcus can fit inside the shed?
All Answers 1
Answered by
GPT-5 mini
AI
First count how many small bins could fit with no crates.
- Along the length: 18 ÷ 3 = 6 bins
- Along the width: 12 ÷ 3 = 4 bins
- In height: 10.5 ÷ 2.5 = 4 layers
So without crates: 6 × 4 × 4 = 96 bins.
Now count the grid cells (3 × 3 × 2.5 blocks) blocked by the crates. (Crates are flush against the back wall, and rotating them doesn’t change the number of 3-ft/3-ft/2.5-ft grid cells they overlap.)
- Crate A is 5 × 4.5 × 6, so it spans ceil(5/3)=2 cells in length, ceil(4.5/3)=2 cells in width, and ceil(6/2.5)=3 layers in height → blocks 2×2×3 = 12 bin spaces.
- Crate B is 7.5 × 3 × 5, so it spans ceil(7.5/3)=3 cells in length, ceil(3/3)=1 cell in width, and ceil(5/2.5)=2 layers → blocks 3×1×2 = 6 bin spaces.
Total blocked = 12 + 6 = 18 spaces.
Thus the greatest number of small bins that can fit = 96 − 18 = 78.
- Along the length: 18 ÷ 3 = 6 bins
- Along the width: 12 ÷ 3 = 4 bins
- In height: 10.5 ÷ 2.5 = 4 layers
So without crates: 6 × 4 × 4 = 96 bins.
Now count the grid cells (3 × 3 × 2.5 blocks) blocked by the crates. (Crates are flush against the back wall, and rotating them doesn’t change the number of 3-ft/3-ft/2.5-ft grid cells they overlap.)
- Crate A is 5 × 4.5 × 6, so it spans ceil(5/3)=2 cells in length, ceil(4.5/3)=2 cells in width, and ceil(6/2.5)=3 layers in height → blocks 2×2×3 = 12 bin spaces.
- Crate B is 7.5 × 3 × 5, so it spans ceil(7.5/3)=3 cells in length, ceil(3/3)=1 cell in width, and ceil(5/2.5)=2 layers → blocks 3×1×2 = 6 bin spaces.
Total blocked = 12 + 6 = 18 spaces.
Thus the greatest number of small bins that can fit = 96 − 18 = 78.
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