Answer:
1. 13 rectangular bales
2. 142 cylindrical bales (better value); 1778 rectangular bales
3. 108 cylindrical bales; 1728 rectangular bales (more volume)
Explanation:
1. The volume of a cylindrical bale is given by the formula ...
V = πr²h
V = π(3 ft)²(4 ft) = 36π ft³ ≈ 113.1 ft³
The volume of a rectangular bale is ...
V = LWH
V = (3 ft)(2 ft)(1.5 ft) = 9 ft³
The number of rectangular bales required to match the volume of a cylindrical bale is ...
(113.1 ft³)/(9 ft³) ≈ 12.57
You would need to purchase 13 rectangular bales to have the same amount of hay as one cylindrical bale.
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2. For 16,000 ft³ of hay, you need ...
16000/113.1 = 141.47 ≈ 142 cylindrical bales
16000/9 = 1777.8 ≈ 1778 rectangular bales
The relative cost of a cylindrical bale is ...
20/2.75 = 7.27 . . . times the rectangular bale cost
times the cost of a rectangular bale. Since it has 12.57 times the volume, the cylindrical bale is the better value.
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3. If the cylindrical bales are placed on a 6-foot grid and stacked 3 high, then 6×6×3 = 108 of them can be put in the barn.*
The barn volume is (36 ft)(36 ft)(12 ft) = 15,552 ft³ = 1728 times the 9 ft³ volume of a rectangular bale.
Since there is no wasted space for a rectangular bale in the rectangular space, a greater volume of hay will fit with rectangular bales.
108 cylindrical bales will fit (76% of annual need)
1728 rectangular bales will fit (97% of annual need)
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* The space utilization on a square grid is π/4 ≈ 78.5%. About the best that can be achieved is 90.7% with hexagonal packing. In a rectangular barn, that density is not achievable. Something between those values may be achievable if the cylindrical bales are deformed a little bit.
