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Solutions for Session 9, Part B
See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6 | B7 | B8 | B9 | B10 | B11 | B12
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Problem B1 | |
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Here is the completed table:
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Length |
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Width |
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Height |
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Volume |
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Surface Area |
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1 |
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1 |
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24 |
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24 |
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98 |
1 |
2 |
12 |
24 |
76 |
1 |
3 |
8 |
24 |
70 |
1 |
4 |
6 |
24 |
68 |
2 |
2 |
6 |
24 |
56 |
2 |
3 |
4 |
24 |
52 |
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<< back to Problem B1
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Problem B2 | |
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The least amount of material is required by the 2-by-3-by-4 box. The greatest amount of material is required by the 1-by-1-by-24 box. The amount of packaging material needed is important, since charges for such material are figured in square inches or square feet, and reducing the amount of needed material will reduce the cost of shipping. Also, the cost of making the box itself will be lower if there is a smaller surface area.
In practical terms, however, you may want to use more packaging material so that the truffles don't get damaged. Also, customers may feel that they're getting more for their money if there is more packaging and the box is larger.
<< back to Problem B2
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Problem B3 | |
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The rectangular prisms with the greatest surface area have dimensions that are far apart (1 by 1 by 24), whereas the prisms with smaller surface area have dimensions close to each other (e.g., 2 by 3 by 4).
<< back to Problem B3
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Problem B4 | |
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Here is the completed table:
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Size of Cube |
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Surface Area |
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Volume |
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Ratio SA:V |
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1 by 1 by 1 |
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6 • 12 = 6 |
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13 = 1 |
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6:1 |
2 by 2 by 2 |
6 • 22 = 24 |
23 = 8 |
6:2 (or 3:1) |
3 by 3 by 3 |
6 • 32 = 54 |
33 = 27 |
6:3 (or 2:1) |
4 by 4 by 4 |
6 • 42 = 96 |
43 = 64 |
6:4 (or 3:2) |
5 by 5 by 5 |
6 • 52 = 150 |
53 = 125 |
6:5 |
6 by 6 by 6 |
6 • 62 = 216 |
63 = 216 |
6:6 (or 1:1) |
7 by 7 by 7 |
6 • 72 = 294 |
73 = 343 |
6:7 |
8 by 8 by 8 |
6 • 82 = 384 |
83 = 512 |
6:8 (or 3:4) |
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In general, the ratio is 6:s, since the surface area formula is 6s2 and the volume formula is s3.
Here is the graph that shows what happens to the ratio of surface area to volume (expressed as a decimal) as the volume increases:
<< back to Problem B4
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Problem B5 | |
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The superstore's surface area-to-volume ratio is as small as possible, compared to smaller stores. This gives a company the ability to put more products inside the store (since volume is relatively large), in comparison to how much it may cost to build the store (since surface area is relatively small).
<< back to Problem B5
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Problem B6 | |
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Answers will vary, but the baby has a higher surface area-to-volume ratio than the adult.
<< back to Problem B6
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Problem B7 | |
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Since babies have a higher surface area-to-volume ratio, and water is a volume-based measurement, babies lose proportionately more water per minute than adults do. As a result, babies will dehydrate more quickly.
<< back to Problem B7
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Problem B8 | |
a. | Answers will vary. The measurements of your height and arm span should be pretty close to each other. |
b. | Answers will vary, but the measurements should be pretty close. |
<< back to Problem B8
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Problem B9 | |
a. | Answers will vary. |
b. | The formula might be based on empirical measurements only, but it treats the entire body as though it had the same circumference as the thighs. It is somewhat similar to the surface-area formula for a cylinder. |
c. | It might, but if babies' thighs are not in the same proportions as adults' thighs (and generally they aren't), then this isn't a good measure. |
<< back to Problem B9
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Problem B10 | |
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Answers will vary. One example is someone with the height of 170 cm and weight of 65 kg, who has the approximate surface area of 17,600 cm2.
<< back to Problem B10
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Problem B11 | |
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Answers will vary. For example, a person with the weight of 65 kg will have the volume of 58.5 L.
<< back to Problem B11
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Problem B12 | |
a. | Answers will vary. Using our previous example, the surface area-to-volume ratio is 17,600:58.5. |
b. | Converting the measures to the metric system, the child's height is about 101.6 cm, and the weight is about 25 kg. Using the nomograph, the surface area is about 8,000 cm2. The volume is 22.5 L. The surface area-to-volume ratio is 8,000:22.5. |
c. | Converting the two ratios into decimals, we get the ratio for an adult to be about 300.86 and for the child 355.56. As you might expect, the child's surface area-to-volume ratio is higher than an adult's. |
<< back to Problem B12
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