Real-World Mathematics: Packaging, Cutting and Material Waste

Application of Mathematics in Real-World Usage · Worked application 7 · BTT Mathematics Hub

A large carton has enough volume for thirty small boxes. Does that mean thirty boxes will fit? Not necessarily. The empty space may be in the wrong shape, individual dimensions may block a placement, or the arrangement may require rotations that the product cannot tolerate. Volume supplies a useful bound, not a packing diagram.

Packaging is a practical meeting point for arithmetic, geometry, inequalities and optimisation. It asks how much space an object needs, how much material surrounds it, how repeated objects can be arranged, what each cut removes and which constraints a proposed saving must preserve. A correct quotient can be only the beginning of the solution.

This guide develops original worked examples from box dimensions through packing certificates and cutting plans. All prices, dimensions, material weights and yields are hypothetical teaching data. The calculations are not carrier quotations, tested packaging specifications or claims that a particular package protects an item. Classroom exploration can use drawn rectangles, paper nets or supplied measurements without operating cutting machinery.

Begin by naming what the package must do

“Use less material” is not a complete problem. A package also has a job: contain an item, keep it in an allowed orientation, preserve required clearance and satisfy any relevant handling or protection conditions. The mathematical objective should be stated alongside those constraints.

In a classroom model, we might minimise the number of cartons while requiring every small box to be packed without overlap. Another model might minimise total sheet area while producing a specified number of labels. A third could minimise mass while holding an idealised internal volume constant. These are different optimisation problems, even when all involve a box.

The NEA announcement of sustainable e-commerce packaging guidelines provides a real-world context for examining packaging choices. The numerical work here is narrower: it tests what can be deduced from stated geometry and constraints. It does not turn a smaller calculated area into a complete environmental or product-protection assessment.

Internal dimensions and external dimensions answer different questions

A box with internal dimensions 20 cm × 15 cm × 10 cm has an idealised usable rectangular volume of 3,000 cm³, or 3 L. That calculation describes its interior before adding any padding, dividers or other obstructions. It does not describe the external space occupied on a shelf.

Suppose, purely for a geometry model, each wall adds 0.5 cm outward on its side. Each complete dimension increases by 1 cm, giving external dimensions 21 cm × 16 cm × 11 cm. External bounding volume is 3,696 cm³. Compared with the 3,000 cm³ interior, that is 696 cm³ more, or 23.2% of the internal volume.

The difference is not an error. Internal volume and external bounding volume measure different spaces. For putting a product inside a box, interior dimensions matter. For fitting boxes into a larger carton, the small boxes’ external dimensions matter. A model that divides a large carton’s external volume by the small boxes’ internal volume compares the wrong boundaries.

Actual wall geometry, folds and lids may require a more detailed model. The purpose of this simplified example is to make the boundary distinction visible. Label each set of dimensions before multiplying them.

Volume, surface area and material mass must remain separate

For an ideal thin-walled closed cuboid with lengths l, w and h, volume is lwh, while surface area is 2(lw + lh + wh). Volume measures enclosed space. Surface area measures six ideal faces. Neither formula alone gives the complete manufacturing blank, because seams, flaps, overlap and trimming may add material.

Compare two ideal boxes: 20 × 15 × 10 cm and 30 × 10 × 10 cm. Both have volume 3,000 cm³. Their surface areas are 1,300 cm² and 1,400 cm² respectively. The same volume can therefore require different ideal surface areas. Shape matters even before manufacturing details enter.

Suppose the first ideal surface is represented by a sheet material with a hypothetical mass of 400 g/m². Convert 1,300 cm² to 0.13 m² before multiplying: 0.13 × 400 = 52 g. Adding a modelled 12% material allowance gives 58.24 g. The allowance is a budgeting assumption, not a proof that an actual net can be made with precisely that area.

Keeping the layers separate prevents three common overclaims: volume is not material quantity, ideal surface area is not a production blank, and a percentage allowance is not a tested design. Each extra claim needs an extra model or measurement.

A net must exist, not merely an area total

A cuboid has six faces, but knowing the sum of their areas does not specify how they are connected on a sheet. A valid net must join the faces in a way that folds into the intended solid without overlap where overlap is not allowed. A production layout may need additional tabs and a permitted cutting direction.

In a teaching task, ask for both the area calculation and a labelled net. This adds a constructive check. If a student reports an economical total but cannot show an arrangement, the result may be a lower bound rather than a usable design.

The same reasoning applies to cutting many shapes from a larger sheet. Total piece area cannot exceed sheet area, but that necessary condition is not sufficient. Shape, orientation, gaps and access for cuts can prevent a layout that the area ratio alone appears to permit.

A three-dimensional packing case

Suppose a large carton has internal dimensions 60 cm × 40 cm × 30 cm. Each small rigid box has external dimensions 19 cm × 14 cm × 9 cm. In this first model, small boxes are axis-aligned, may be rotated into permitted right-angle orientations, need no gaps and do not deform.

With 19 cm along the large carton’s 60 cm direction, 14 cm along its 40 cm direction and 9 cm along its 30 cm direction, a regular grid holds floor(60/19) × floor(40/14) × floor(30/9) = 3 × 2 × 3 = 18 boxes. The floor operation means taking the greatest whole number not exceeding the quotient.

Rotate every small box so that 14 cm runs along 60 cm, 19 cm along 40 cm and 9 cm along 30 cm. The grid now holds 4 × 2 × 3 = 24 boxes. A changed orientation improves the demonstrated capacity by six boxes without changing either volume.

The useful conclusion is that at least 24 boxes fit under the stated assumptions. This does not prove that 24 is the global maximum. Mixed orientations or more complex arrangements may outperform a uniform grid. A searched family must not be confused with all possible arrangements.

What a volume bound actually proves

The large carton has volume 72,000 cm³. Each small box occupies 19 × 14 × 9 = 2,394 cm³. The volume ratio is a little over 30, so no non-overlapping arrangement can contain more than 30 complete boxes. Thirty boxes would occupy 71,820 cm³.

We now have two different kinds of evidence. A constructed arrangement proves a lower bound of 24 on the maximum packable count. The volume argument proves an upper bound of 30. The true maximum lies somewhere from 24 to 30 unless further constraints or a better argument narrow the interval.

The demonstrated 24-box arrangement uses 24 × 2,394 = 57,456 cm³, or 79.8% of the large interior volume. That utilisation describes this arrangement, not the best physically possible one. Reporting 99.75% utilisation for thirty boxes would require first demonstrating that thirty can actually be arranged.

This is an important habit in applied mathematics: label a number by what it establishes. A bound, a feasible example and a proven optimum are not interchangeable. A calculator can give all three numbers the same neat appearance.

A counterexample to “the volume is enough”

Take a 10 cm × 10 cm × 10 cm container and identical 6 cm cubes. Four cubes have total volume 4 × 216 = 864 cm³, below the container’s 1,000 cm³. Yet under axis-aligned packing, even two such cubes cannot fit without overlap.

For two axis-aligned boxes not to overlap internally, their occupied intervals must be separated along at least one coordinate direction. But two intervals each 6 cm long cannot be disjoint inside a 10 cm span: together they require 12 cm. That is true along every axis. Thus at most one cube fits in the stated axis-aligned model.

This counterexample does not need a complicated search. It identifies a missing constraint that the volume quotient ignores. In a lesson, it is often more powerful than several routine packing calculations because it shows exactly why a tempting rule fails.

Clearance and tolerance can erase a theoretical fit

Return to the large carton and add 1 cm of lining on each interior side. Available dimensions become 58 cm × 38 cm × 28 cm. The 14-by-19-by-9 orientation still gives a uniform-grid count of 4 × 2 × 3 = 24. But the 38 cm width is now exactly filled by two nominal 19 cm boxes.

An exact nominal fit leaves no room for positive dimensional variation or a required insertion gap. If each box could be 19.1 cm wide, two require 38.2 cm and no longer fit across a 38 cm available width. The multiplication was correct; the inputs were not adequate for the stronger claim of a reliable fit.

A conservative fit calculation should compare the smallest allowed usable container dimension with the largest allowed packed-item dimensions and required gaps. This is an application of interval reasoning, not a reason to abandon geometry. The assumptions determine how strong the conclusion can be.

For the underlying treatment of units and bounds, return to Measurement, Units, Scale and Estimation. Here the bounds have an immediate consequence: a design that fits only at central values may fail at permitted extremes.

Cutting consumes material even when the pieces are correct

Suppose a stock length is 1,000 mm and each finished piece must be 300 mm. Model the width removed by each separating cut as 3 mm. Assume each piece is cut from the remaining stock, so three finished pieces require three cuts, including the cut separating the last piece from the final offcut. No additional end trimming is included.

Three pieces consume 3 × 300 + 3 × 3 = 909 mm, leaving a 91 mm offcut. Four pieces would consume 1,212 mm and cannot be obtained from the stock. The conservation check is 900 mm of finished pieces + 9 mm removed by cuts + 91 mm offcut = 1,000 mm.

The number of cuts is not automatically the number of pieces minus one. That count applies only to a different geometry in which all pieces are separated between already suitable outer boundaries and no final offcut must be detached. State the actual cut model. A three-millimetre omission can decide whether a borderline plan is feasible.

This is a mathematical planning example, not instruction to operate tools. The important educational task is to track finished material, material removed and retained offcuts as separate quantities.

An offcut is not automatically waste

The 91 mm remainder may be unusable for the current 300 mm demand but useful for a later smaller requirement. Whether it is waste depends on the definition and the future demand being considered. If the model counts all offcuts as discarded, say so. If it retains them as inventory, record their dimensions and do not count them as both saved and discarded.

Two cutting plans can have the same total leftover length but different usefulness. One may leave a single 90 mm piece; another may leave nine 10 mm fragments. Equal total remainder does not imply equal value. This is another example of why arrangement matters beyond a scalar total.

For an honest material balance, distinguish useful output, cutting loss, reusable offcuts and discarded remainder. A claimed reduction should specify which category changed. Renaming a leftover item does not, by itself, reduce material consumption.

One-dimensional bin packing: a small problem with a complete proof

For a simpler model, ignore geometry and cutting loss and consider bins of capacity 10 units. Four items weigh 6 units each and four weigh 4 units each. Total weight is 40, so at least four bins are necessary. A feasible arrangement places one 6-unit item with one 4-unit item in each bin. Four bins are therefore sufficient as well as necessary.

That is a complete optimality certificate: a lower bound of four and a construction using four. The bound and construction meet. The Google OR-Tools bin-packing guide defines the corresponding optimisation job as placing all items into the fewest capacity-limited bins. Our small instance is solved transparently without relying on a software result.

Now consider six items of weight 6. Total weight is 36, so the total-weight lower bound is only four bins. But each item exceeds half the bin capacity, so no bin can hold two. Six bins are necessary and sufficient. The stronger structural bound is more informative than total weight alone.

A reasonable greedy method can still miss the optimum

Use bins of capacity 10 and item weights 6, 5, 3, 2, 2, 2. A simple method sorts items from largest to smallest and places each in the first existing bin with enough space, opening a new bin only when required.

That method first places 6 in bin 1 and 5 in bin 2. The 3 joins the 6, making 9. The next two 2s join the 5, making 9. The final 2 fits neither bin and opens bin 3. The method found a feasible three-bin arrangement.

But two bins suffice: 6 + 2 + 2 = 10 and 5 + 3 + 2 = 10. The total weight is 20, so fewer than two bins are impossible. Two is therefore optimal. The greedy construction was sensible but not optimal for this instance.

The lesson is not that heuristics are useless. A quick method may be valuable when the problem is large. The lesson is that “the method stopped” is not the same as “no better answer exists.” Distinguish a feasible solution from an optimal solution in any report.

Sheet cutting can also produce a matching bound and construction

A sheet measures 120 cm by 80 cm. Required rectangles measure 30 cm by 20 cm, with zero gaps and no cutting loss in this idealised example. Sheet area is 9,600 cm² and each rectangle uses 600 cm², so at most sixteen rectangles fit by area.

A regular four-by-four layout achieves sixteen: four widths of 30 cm exactly span 120 cm, and four heights of 20 cm exactly span 80 cm. Because an arrangement achieves the area upper bound, sixteen is optimal for the stated model.

Add a required gap, a damaged edge or a nonzero cut width and the proof no longer binds to the changed problem. The old answer may still happen to be correct, but it needs to be checked again. Assumptions are part of the mathematical certificate, not optional decoration around it.

An open box joins geometry to calculus

Take an ideal rectangular sheet 30 cm by 20 cm. Remove equal squares of side x from its corners and imagine folding the sides upward, with no material thickness or joining allowance. The open box has height x, length 30 − 2x and width 20 − 2x. Its volume is V(x) = x(30 − 2x)(20 − 2x), with 0 < x < 10.

At x = 3 cm, volume is 3 × 24 × 14 = 1,008 cm³. At x = 4 cm, volume is 4 × 22 × 12 = 1,056 cm³. Increasing the cut height initially helps, but it also shrinks the base. The competition creates an interior maximum.

Expand V = 600x − 100x² + 4x³. Differentiating gives V′ = 600 − 200x + 12x². Setting this to zero yields x = (25 ± 5√7)/3. Only the smaller value, approximately 3.92375 cm, lies inside the physical domain. It gives approximately 1,056.31 cm³.

The derivative changes from positive to negative at that interior point, and the limiting volumes at the domain endpoints are zero, so this is the global maximum for the idealised family. For cuts restricted to whole centimetres, comparing the allowed values gives x = 4 cm. The continuous optimum and the permitted manufacturing choices are related but different problems.

Neither result is a ready-to-manufacture package. Thickness, stiffness, fastening and protective performance were intentionally excluded. The mathematical achievement is precise: we optimised one stated geometric family under its stated assumptions.

Yield converts demand into a planning quantity, not a guarantee

Suppose a fictional process has a planning yield of 90% acceptable finished units. To target an expected 500 acceptable units, divide by 0.90: 500/0.90 is approximately 555.56, so plan 556 starts if starts must be whole numbers. At the assumed yield, expected acceptable output is 500.4.

This is not a guarantee of at least 500 acceptable units when 90% is an estimated average. Actual output may vary. A service-level or reliability claim needs a probability model or a justified minimum-yield bound. The ceiling operation fixes the integer arithmetic; it does not eliminate process uncertainty.

If each starting blank costs a hypothetical $0.08, 556 blanks cost $44.48. That cost calculation depends on the planning assumption. Comparing alternatives should also account for whether rejected units can be reworked, whether setup losses change with batch size and which costs are fixed. Do not load all of those effects into a single unexplained “wastage percentage.”

Material reduction is one measure, not the whole decision

A redesign might use less board but more filler, reduce external volume but increase damaged products, or save sheet area while increasing labour. Whether it is an improvement depends on the complete objective and the required function. The EPA’s waste-reduction guidance gives broad reduce-and-reuse context; our calculations do not substitute for a full comparison of impacts.

In a classroom model, require a decision table with separate quantities: material area or mass, carton count, usable capacity, assumed rejection rate and any untested performance requirement. A smaller number in one column should not automatically erase a worse result in another.

This develops a useful form of restraint. Mathematics should make trade-offs clearer, not disguise them behind a single score. State the result at the strength supported by the evidence: “This arrangement uses four bins,” “This count is optimal in the one-dimensional model,” or “Protection performance remains untested.”

Practice: twenty questions about fit, material and proof

Use idealised rigid, axis-aligned rectangles and cuboids unless specified otherwise. Ignore thickness, gaps and cutting loss only when the question explicitly allows it.

1. Find the volume of a 12 cm × 8 cm × 5 cm cuboid.
2. Find the surface area of the closed ideal cuboid in question 1.
3. Convert 720 cm² to m² and find mass at a hypothetical 250 g/m².
4. A package has internal dimensions 10 × 8 × 6 cm. Add 0.2 cm wall thickness on each side. Find external dimensions.
5. A carton has inner dimensions 50 × 40 × 30 cm. Small boxes are 10 × 8 × 6 cm in that orientation. Find the regular-grid count.
6. For question 5, use the volume bound to decide whether the achieved count is optimal under the stated no-gap model.
7. Two 6 cm cubes have total volume below 1,000 cm³. Can both fit axis-aligned in a 10 cm cube?
8. A 1,000 mm bar provides 300 mm pieces, with one 3 mm separating cut per piece. Find the maximum count and remaining offcut.
9. In question 8, separate useful length, cut loss and offcut, and check their sum.
10. With zero cut loss, pack weights 7, 7, 3, 3 into bins of capacity 10. Prove the minimum bin count.
11. Five items each weigh 6 units and bins hold 10. Compare the total-weight lower bound with the actual minimum count.
12. A 100 × 60 cm sheet supplies 25 × 20 cm rectangles without gaps. Find and prove the maximum count.
13. Useful cut area is 7,200 cm² from a 9,000 cm² sheet. Find area utilisation.
14. An ideal surface area of 800 cm² receives a 15% allowance. Find the budgeted area.
15. At an assumed average yield of 0.8, how many starts target an expected 240 acceptable units? Is it a guarantee?
16. Two successive conditional stage yields are 0.95 and 0.96. Find the combined yield under the stated model.
17. For the 30 × 20 cm open-box model, calculate volume at x = 2 cm.
18. Why is x = 12 cm invalid in that model even though the formula can be evaluated?
19. A gap of at least 2 mm is required between two nominally 190 mm boxes. The usable width is 381 mm. Does the stated arrangement fit?
20. A search finds a 24-box arrangement and a volume bound of 30. What is justified, and what remains unproved?

Worked answers and proof checks

1. 480 cm³. Multiply all three independent dimensions. This measures space, not board area.
2. 392 cm². Calculate 2(12 × 8 + 12 × 5 + 8 × 5) = 2(96 + 60 + 40). The six faces are counted.
3. 0.072 m² and 18 g. Divide square centimetres by 10,000 before multiplying by 250 g/m².
4. 10.4 × 8.4 × 6.4 cm. Each dimension gains thickness on two opposite sides. Adding only 0.2 cm would omit one side.
5. 125 boxes. The grid count is 5 × 5 × 5. It is a construction, not merely an area or volume estimate.
6. Yes, within the stated model. Large volume is 60,000 cm³ and each box is 480 cm³; the upper bound is 125 and the grid achieves it.
7. No. Their intervals cannot be separated along any axis within a 10 cm span. Total volume is a necessary but insufficient condition.
8. Three pieces and 91 mm offcut. Three times 303 mm is 909 mm; four times 303 mm exceeds the stock length.
9. 900 + 9 + 91 = 1,000 mm. The balance distinguishes useful pieces, material removed and the remaining stock.
10. Two bins. Total weight 20 requires at least two. Pair 7 with 3 twice to achieve the bound.
11. The weight bound is three, but five bins are required. No two 6-unit items fit in one 10-unit bin, so each needs its own.
12. Twelve. A four-by-three grid achieves twelve, and the area ratio 6,000/500 = 12 proves that more are impossible.
13. 80%. Divide useful area by supplied area. The unused 20% may include several different categories of remainder.
14. 920 cm². Multiply by 1.15. This is an allowance estimate, not a demonstrated production net.
15. 300 starts; not a guarantee from an average yield. The expected output is 300 × 0.8 = 240. A reliability claim requires additional evidence.
16. 0.912, or 91.2%. The second stage acts on the survivors of the first. The conditional interpretation matters; percentages are not simply added.
17. 832 cm³. Use 2 × 26 × 16. Check that every resulting dimension is positive.
18. It makes the width 20 − 24 negative. The physical domain is 0 < x < 10. A symbolic value outside it does not describe the intended box.
19. No. Required width is 190 + 2 + 190 = 382 mm, exceeding 381 mm even before any dimensional tolerance is added.
20. At least 24 fit and no more than 30 can fit under the model. The exact optimum is not established. Do not label the found arrangement globally optimal.

Teach the certificate as well as the answer

A strong lesson asks the learner to supply two things: a proposed arrangement and a reason it should be trusted. At an introductory level, that means labelled dimensions and a no-overlap check. At a higher level, it means a bound and a construction that meet, or an honest interval when the optimum remains unresolved.

Use the greedy counterexample to separate method execution from mathematical judgement. A student can carry out the chosen rule perfectly and still obtain a non-optimal result. Ask what additional evidence would justify the stronger claim. Then change a gap, a rotation permission or a cut width and ask which earlier proof no longer applies.

This builds transferable habits: preserve units, identify boundaries, distinguish a constraint from an objective, test counterexamples and state exactly what has been established. They are useful far beyond cartons because many real decisions involve arranging finite objects in limited space.

Sources and the next application

For the real-world context and formal optimisation definitions, see NEA: New Guidelines to Reduce E-Commerce Packaging, Google OR-Tools: The Bin Packing Problem, and EPA: What You Can Do to Reduce Plastic Waste. The worked geometries, counterexamples, prices and exercises on this page are original educational constructions.

Continue to Scheduling, Critical Paths and Resources to arrange tasks in limited time rather than objects in limited space. Use Water, Flow, Tanks and Rainfall for capacity through time, or Electricity, Energy, Power and Bills for rates and accumulated resource use. The functions and optimisation guide supplies the mathematical background. Return to the BTT Mathematics Hub for the full learning map.