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- Volume and Nets
- Packaging Candy

A net is the two-dimensional representation of a three-dimensional object. For example, you can cut the net of a cube out of paper and then fold it into a cube.

Here are the nets of some “open” boxes — boxes without lids.

- If you were to cut out each net, fold it into a box, and fill the box with cubes, how many cubes would it take to fill the box? Make a quick prediction and then use two different approaches to find the number of cubes. You may want to cut out the actual nets (PDF file), fold them up, and tape them into boxes to help with your predictions.
- What strategies did you use to determine the number of cubes that filled each box?

**Problem A2**

Given a net, generalize an approach for finding the number of cubes that will fill the box created by the net. How is your generalization related to the volume formula for a rectangular prism (length • width • height)?

**Problem A3**

- Imagine another box that holds twice as many cubes as Box A. What are the possible dimensions of this new box with the doubled volume?
- What if the box held four times as many cubes as Box A? What are the possible dimensions of this new box with quadrupled volume?
- What if the box held eight times as many cubes as Box A? What are the possible dimensions of the new box with an eightfold increase in volume?

You may want to start by constructing a solid Box A (2 by 2 by 4) from cubes that can be connected together. Next, double one dimension of the solid and build a new solid. What is its volume? What happens to the volume of the original solid (Box A) if you double two of the dimensions? If you double all three of its dimensions? Try it.

**Problem A4**

If you took Box B and tripled each of the dimensions, how many times greater would the volume of the larger box be than the original box? Explain why.

Divide the new volume by the original volume to see how many times greater it is. Can you figure out why?

**Problem A5**

What is the ratio of the volume of a new box to the volume of the original box when all three dimensions of the original box are multiplied by k? Give an example.

The standard unit of measure for volume is the cubic unit, but we often need to fill boxes with different-sized units or packages. For example, suppose a candy factory has to package its candy in larger shipping boxes. Examine the different-sized packages of candy below. A package is defined for this activity as a solid rectangular prism whose dimensions are anything except 1 • 1 • 1.

**Problem A6**

- Using just one size of package at a time, how many of each package (1-5) will fit into Box B (from Problem A1) so that the box is filled as completely as possible?

**Package Number****Number That Fit**Package 1 ___________ Package 2 ___________ Package 3 ___________ Package 4 ___________ Package 5 ___________ You may want to cut out the net for Box B and fold it into an open box. You can then use the box to help you visualize placing packages inside it.

- Describe your strategy for determining how many of Package 2 fit into Box B.

**Problem A7**

- Notice that not all of the packages fill Box B completely so that there is no leftover space. Design the smallest box (in terms of volume) that could be used to ship all of the candy packages above. The box needs to be of a size and shape that can be completely filled by Packages 1-5 separately.

Think about the dimensions of each package and the possible dimensions of your new box.

- Is there more than one size box that can be used to ship the different candy packages? Explain why or why not.
- How are the dimensions of the packages related to the dimensions of the larger shipping box?
- Generalize the relationship between the dimensions of any package and the volume of any box that can be completely filled by the packages.

**Problem A1
**

- The first net takes 16 cubes. The second net takes 48 cubes. The third net takes 36 cubes.
- Answers will vary. One strategy is to fold up the nets into boxes and fill them with multilink cubes.

**Problem A2**

Count how many cubes it takes to cover the base in a single layer (that is, the area of the base), and then multiply by how many layers it would take to fill up the box (the height of the box). This formula gives V = l • w • h.

**Problem A3**

- There are many possible answers; two possibilities are 2 by 2 by 8 and 2 by 4 by 4. These are found by doubling any one of the three dimensions of the original.
- Of the many possible answers, two possibilities are 2 by 4 by 8 and 4 by 4 by 4. These are found by doubling any two of the three dimensions (or multiplying one dimension by a factor of 4).
- Of the many possible answers, a 4-by-4-by-8 box is similar to the original. This is formed by doubling all three of the original’s dimensions.

**Problem A4**

It would be 27 times greater in volume:

**Volume of Box B:**

(4 • 4 • 3) = 48 cubic units

**Volume of Enlarged Box B:**

(4 • 3) • (4 • 3) • (3 • 3) = (4 • 4 • 3) • (3 • 3 • 3) = 48 • 27 = 1,296 cubic units

There would be three times as many cubic units in each direction, or 3 • 3 • 3 = 27 times as many in the overall volume.

**Problem A5**

The ratio is k^{3}, since there are k times as many cubes in all three dimensions. Problem A3 (c) used k = 2, and Problem A4 used k = 3.

**Problem A6**

Note that Box B is 4 by 4 by 3.

- Only four of Package 1 will fit in Box B.

Sixteen of Package 2 will fit in Box B (vertically), filling the box.

Twelve of Package 3 will fit in Box B, filling the box.

Four of Package 4 will fit in Box B, filling the box.

Package 5 will not fit in Box B at all. Its largest dimension is larger than any of the dimensions of the box! - Answers will vary, but one strategy is to align the packages along matching dimensions. Recognizing that a package is 1 by 1 by 3 helps to fit it into a 4-by-4-by-3 box.

**Problem A7**

- One approach is to think about the dimensions of the new box with respect to the dimensions of Packages 1-5. For example, one dimension of the new box needs to be divisible by 5 (because of Package 5). We need all three dimensions to be divisible by 2 (because of Package 1). We need one dimension divisible by 3 (because of Package 2 and Package 4). Based on these observations, the dimensions of the new box are 2 by (2 • 5) by (2 • 3), which is 2 by 10 by 6. Using this same reasoning, you could also decide on a 2-by-2-by-30 box (which also works for all five packages). These are the smallest (in total volume) that will work.
- We know that a 2-by-10-by-6 box works. If we double any of the sides, for example, then the dimensions are all still divisible by the necessary lengths, so it will work to ship all of the packages. By this reasoning, if the sides of a box are “divisible” by 2, 10, and 6, or “divisible” by 2, 2, and 30 from our second example, then it will work to ship all of the packages.
- The dimensions of the larger box must be divisible by the dimensions of the smaller packages. Given a small package with dimensions l, w, and h, we must have one dimension of the box divisible by l, a different dimension divisible by w, and the third dimension divisible by h. For example, for Package 1 to completely fill the box, we see that each dimension must be divisible by 2 (i.e., it must be even).
- Each dimension of the box has to be a common multiple of unique dimension of each of the packages. So for any package with dimensions l, w, and h, we must have one dimension of the box divisible by l, a different dimension divisible by w, and the third dimension divisible by h.