Build the model before differentiating
An open-topped rectangular box is made by removing a square of side x from each corner of a 30 cm by 20 cm sheet, then folding up the sides. With negligible cardboard thickness, the base measures (30 − 2x) by (20 − 2x) and the height is x. Its theoretical volume is V(x) = x(30 − 2x)(20 − 2x) = 600x − 100x² + 4x³.
Because both base dimensions must be positive, the feasible domain is 0 < x < 10 cm. Without the domain, a derivative equation can produce a mathematically valid but physically meaningless stationary point.
Check the stationary point and reject the impossible root
- Differentiate: V'(x) = 600 − 200x + 12x².
- Set the derivative equal to zero. Solving 12x² − 200x + 600 = 0 gives approximately 3.92375 cm and 12.74292 cm.
- The second root is outside the feasible domain, so reject it before interpreting its mathematical meaning.
- The second derivative is V''(x) = 24x − 200. It is negative at x ≈ 3.92375 cm, so the feasible stationary point is a maximum.
- Substitution gives the idealized maximum volume about 1,056.31 cm³. The volume approaches zero toward either boundary, supporting a global maximum in this domain.
How much precision matters physically?
A cutting tool may make 4.0 cm easier than 3.92375 cm. Use the model to find the actual loss from rounding instead of claiming that a rounded result is automatically poor. This table is calculated, not measured.
| Cut | Predicted volume | Consequence |
|---|---|---|
| 3.50 cm | 1,046.50 cm³ | 9.81 cm³ less than theoretical maximum |
| 3.924 cm | ≈1,056.31 cm³ | Theoretical optimum in the ideal model |
| 4.00 cm | 1,056.00 cm³ | 0.31 cm³ below optimum |
| 4.50 cm | 1,039.50 cm³ | 16.81 cm³ less than optimum |
The 4.0 cm cut loses roughly 0.3 cm³ of the predicted maximum, a tiny amount relative to 1,056 cm³. This suggests the modelled outcome is relatively insensitive to a small rounding decision around its peak. It does not demonstrate that the real box will perform identically once fold thickness, tears and seams are considered.
Turn calculation into a mathematical exploration
Record the assumptions used to create V(x). Decide which are realistic for your own cardboard, then measure actual dimensions and capacity if you perform the construction. Compare the measured values with the mathematical predictions. Do not present theoretical numbers as empirical results. An original IA needs personally selected mathematics, clear notation, an explicit connection to the problem and meaningful reflection rather than merely reproducing a textbook optimization.
A worthwhile extension is to require a minimum base width, allow a fold seam, or optimize packaging cost rather than volume. Each modification changes the constraints and may change the preferred solution. For other mathematics IA tasks, consult research-question planning.
Sources & verification
Primary-source references used to verify key assessment facts on this page. Last source check: October 14, 2026.
Original hypothetical mathematics example. The volumes are calculated idealized outcomes, not measured results or an IB-scored exemplar.
- IB Mathematics: Analysis and Approaches examiner guidanceOfficial description of exploration, mathematical communication and reflection; assessment guidance for the 2019 syllabus.
- IB Mathematics: Analysis and Approaches updatesOfficial first-assessment transition: May 2029 for the subsequent course.