Question

a company makes japanese-style lunch boxes called bento boxes. each bento box is a cube with \\(\frac{1}{3}\\)-foot-long edges. the company ships the bento boxes in a container in the shape of a rectangular prism that measures \\(\frac{3}{2}\\) feet by \\(\frac{3}{2}\\) feet by \\(\frac{3}{2}\\) feet.
a manager correctly finds the volume of the container using a formula. an employee double-checks the calculation by packing the container full of bento boxes.
use the drop-down menus to explain how both volume calculations compare.

click the arrows to choose an answer from each menu.

the manager can find the volume of the container using the formula choose...
the employee correctly determines that the container can be filled with choose... bento boxes. he can find the volume of the container by multiplying the number of bento boxes that fill the container by the choose... of one bento box.
the volume found by using the formula is choose... the volume of the total number of bento boxes in the container.

Explanation:

Step1: Volume of container (manager's method)

The container is a rectangular prism, so the formula for its volume \( V \) is \( V = l \times w \times h \), where \( l=\frac{3}{2} \), \( w=\frac{3}{2} \), \( h=\frac{3}{2} \). So \( V=\frac{3}{2} \times \frac{3}{2} \times \frac{3}{2}=\frac{27}{8} \) cubic feet.

Step2: Volume of one bento box

Each bento box is a cube with edge length \( s = \frac{1}{3} \) foot. The volume of a cube is \( V_{cube}=s^3 \), so \( V_{cube}=(\frac{1}{3})^3=\frac{1}{27} \) cubic feet.

Step3: Number of bento boxes (employee's method)

To find how many bento boxes fit along each dimension:
- Along length: \( \frac{3/2}{1/3}=\frac{3}{2} \times 3=\frac{9}{2}=4.5 \), but since we can't have half a box, wait, actually, \( \frac{3}{2} \div \frac{1}{3}=\frac{9}{2}=4.5 \), but maybe the problem assumes integer? Wait, no, let's recalculate. Wait, \( \frac{3}{2} \) divided by \( \frac{1}{3} \) is \( \frac{9}{2}=4.5 \), but maybe the container dimensions are such that it's a cube? Wait, the container is \( \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2} \), so it's a cube. The edge length of container is \( \frac{3}{2} \) feet, edge length of bento is \( \frac{1}{3} \) feet. Number of bento boxes along one edge: \( \frac{3/2}{1/3}=\frac{9}{2}=4.5 \)? No, that can't be. Wait, maybe I made a mistake. Wait, \( \frac{3}{2} \) feet is 1.5 feet, \( \frac{1}{3} \) feet is about 0.333 feet. 1.5 divided by 0.333 is about 4.5. But maybe the problem has a typo, or maybe the container is \( \frac{3}{2} \) as in 1.5, but let's check the volume. Wait, the manager's volume is \( (\frac{3}{2})^3=\frac{27}{8}=3.375 \) cubic feet. The volume of one bento is \( (\frac{1}{3})^3=\frac{1}{27}\approx0.037 \) cubic feet. Now, if we calculate how many bento boxes fit: along each edge, \( \frac{3/2}{1/3}=\frac{9}{2}=4.5 \), but since we can't have half, maybe the problem actually has the container as \( \frac{3}{2} \) which is 1.5, and bento edge \( \frac{1}{3} \), but maybe the numbers are such that \( \frac{3}{2} \div \frac{1}{3}= \frac{9}{2} \), but that's not integer. Wait, maybe the container is \( \frac{3}{2} \) feet, which is 1.5, and bento is \( \frac{1}{3} \) feet. Wait, maybe the problem is designed so that the number of bento boxes is \( (\frac{3/2}{1/3})^3=(\frac{9}{2})^3=\frac{729}{8}=91.125 \), but that seems odd. Alternatively, maybe the container is \( \frac{3}{2} \) as 3/2, and bento is 1/3, so the number of bento boxes is \( (\frac{3/2}{1/3}) \times (\frac{3/2}{1/3}) \times (\frac{3/2}{1/3}) = (\frac{9}{2})^3=\frac{729}{8} \), and the volume of all bento boxes is \( \frac{729}{8} \times \frac{1}{27}=\frac{729}{216}=\frac{27}{8} \), which is equal to the manager's volume. Ah! So even though the number of bento boxes along each edge is 4.5, when we multiply the number of bento boxes (including fractions, but in reality, since volume is continuous, the total volume of bento boxes (treating them as filling the space) should equal the container's volume. So:

- The manager uses \( V = lwh \) (volume of rectangular prism).
- The employee finds the number of bento boxes: along each edge, \( \frac{3/2}{1/3}=\frac{9}{2} \), so total number is \( (\frac{9}{2})^3=\frac{729}{8} \). Then, volume of container is number of bento boxes times volume of one bento box (since each bento box has volume \( (\frac{1}{3})^3=\frac{1}{27} \), so \( \frac{729}{8} \times \frac{1}{27}=\frac{27}{8} \), same as manager's.

So the blanks:

1. The manager can find the volume of the container using the formula \( \boldsymbol{V = l \times w \times h} \) (volume of rectangular prism).

2. The employee correctly determines that the container can be filled with \( \boldsymbol{\frac{729}{8}} \) (or 91.125, but as a fraction \( \frac{729}{8} \)) bento boxes. Wait, no, let's recalculate: \( \frac{3}{2} \div \frac{1}{3} = \frac{9}{2} \), so along each edge, \( \frac{9}{2} \) boxes, so total number is \( (\frac{9}{2})^3 = \frac{729}{8} \).

3. He can find the volume of the container by multiplying the number of bento boxes that fill the container by the \( \boldsymbol{volume} \) of one bento box.

4. The volume found by using the formula is \( \boldsymbol{equal\ to} \) the volume of the total number of bento boxes in the container.

Answer:

1. The manager's formula: \( V = l \times w \times h \)
2. Number of bento boxes: \( \frac{729}{8} \) (or 91.125)
3. Multiply by: volume
4. Comparison: equal to

(Note: For the drop - down menus, the first should be "Volume of a rectangular prism (\( V = lwh \))", the second is the number of bento boxes (calculated as \( (\frac{3/2}{1/3})^3=\frac{729}{8} \)), the third is "volume", and the fourth is "equal to".)