The Storage Bench Volume Decision
The STEM lab at Riverside Elementary needs a storage bench for its foam blocks, and the fifth-grade design team has two candidate boxes. The bench must slide into a wall space that is 4.5 feet long, and a bulletin board above it means nothing taller than 2.75 feet will fit. Design A has a base of 4 feet by 3 feet and a height of 2.5 feet. Design B has a base of 5 feet by 2 feet and a height of 3 feet. Both are rectangular prisms with no lid. The lab owns 250 foam cubes, each with an edge length of 1/2 foot, and the team wants to fill the chosen bench completely - no gaps, no cubes sticking out. A teammate argues that the two designs must hold different amounts because Design B "is longer." The team needs a recommendation with numbers behind it.
1. Find the volume of each design and the number of 1/2-foot cubes each one holds. Does the teammate's claim hold up?
Read the worked answer
Volumes
- Design A: 4 × 3 × 2.5 = 30 cubic feet
- Design B: 5 × 2 × 3 = 30 cubic feet
The two designs have the same volume, so the teammate's claim fails - a longer box can be narrower or shorter and still hold the same amount.
Cube counts (counting along each edge)
- Design A: 4 ÷ 1/2 = 8 cubes long, 3 ÷ 1/2 = 6 wide, 2.5 ÷ 1/2 = 5 tall → 8 × 6 × 5 = 240 cubes
- Design B: 5 ÷ 1/2 = 10 long, 2 ÷ 1/2 = 4 wide, 3 ÷ 1/2 = 6 tall → 10 × 4 × 6 = 240 cubes
Cross-check with volume of one cube: each cube is 1/2 × 1/2 × 1/2 = 1/8 cubic foot, and 30 ÷ 1/8 = 240. Both methods agree, which is the important confirmation - the dimension method and the volume method must give the same count, or a dimension was divided incorrectly.
2. Which design should the team build, given the wall space and height limit? Support the choice with the measurements.
Read the worked answer
Test each constraint separately; failing either one disqualifies a design.
- Design A: length 4 ft ≤ 4.5 ft ✔ and height 2.5 ft ≤ 2.75 ft ✔ → fits.
- Design B: length 5 ft > 4.5 ft ✘ (and height 3 ft > 2.75 ft ✘) → does not fit.
Design B fails on both constraints, so no amount of rearranging saves it. The team should build Design A.
This is where volume reasoning alone would mislead: the two boxes hold exactly the same 240 cubes, so if the team picked by capacity it would have no reason to prefer either one. The deciding information is the space constraint, not the volume. Note also that Design A leaves 0.5 ft of wall space and 0.25 ft of clearance under the bulletin board, so the bench can actually be slid into place rather than lifted.
3. To save plywood, a student proposes lowering Design A's height from 2.5 feet to 2 feet while keeping the 4-by-3 base. How does that change the cube count, and how many of the 250 cubes would then be left over?
Read the worked answer
New dimensions: 4 ft × 3 ft × 2 ft.
Cubes along each edge: 8 long × 6 wide × 4 tall = 192 cubes.
Volume check: 4 × 3 × 2 = 24 cubic feet, and 24 ÷ 1/8 = 192. ✔
Change from the original: 240 − 192 = 48 fewer cubes. Cutting half a foot of height removes a full layer of 8 × 6 = 48 cubes, which is exactly what the layer picture predicts.
Leftover cubes: 250 − 192 = 58 cubes left over (versus 10 left over with the 2.5-foot design).
The trade-off is real: the shorter bench still fits the wall space, but it stores 20% less and leaves more than five times as many cubes without a home. If all 250 cubes must be stored, the 2-foot version is not acceptable even though it satisfies the size constraints.