An open-top box can be made by cutting an equal square from every corner of a rectangular sheet and folding the remaining side flaps upward. The corner size controls three dimensions simultaneously: increasing the cut makes the box taller but shortens both the base length and width. Additional Mathematics differentiation identifies which cut creates the largest volume.
For a sheet 30 cm long and 20 cm wide, cut squares of side x cm. The resulting box has height x, base length 30−2x and base width 20−2x, so V(x)=x(30−2x)(20−2x). Its physically permitted interval is 0 Did you know that the optimal cut for a rectangle is not generally one-sixth of its shorter side? That familiar proportion applies to a square sheet. This guide derives the rectangular solution, checks both stationary roots, handles whole-number cuts and explains why choosing the tallest possible box gives almost no volume. The starting material is a rectangular sheet of length 30 cm and width 20 cm. Four identical squares are removed from the corners. After cutting, the four remaining flaps are folded up at right angles to the centre rectangle. The cut-out square side length x becomes the height of each side wall. This construction is commonly used to model a simple open-top tray or packaging box. It assumes negligible material thickness and exact right-angle folds, so the mathematical dimensions are straightforward. The key is that x changes both the wall height and the dimensions of the base. Assigning volume as 30×20×x would ignore the material removed from the corners and describe a different object. A square of side x is removed from each of the four corners. Along the original 30 cm direction, the two end cutouts remove a total of 2x. Therefore the base length is 30−2x. Along the original 20 cm direction, the two side cutouts likewise remove 2x. The base width is 20−2x, while the folded-up height is x. All three values must be used in the volume calculation. The repeated factor 2 arises from two opposite corner removals in each horizontal direction, not from the total count of four corners. A rectangular box has volume length×width×height. Substituting the dimensions gives V(x)=x(30−2x)(20−2x). Multiply the two base factors: (30−2x)(20−2x)=600−60x−40x+4x²=600−100x+4x². Then multiply by x to get V(x)=600x−100x²+4x³. The cubic expression is equivalent to the original factored volume formula. The factored form makes the three physical dimensions visible, while the expanded form is convenient for differentiation. A correct solution should be able to move between them without changing the model. The corner square side must be positive, so x>0. Both base dimensions must also be positive. From 30−2x>0 we obtain x<15. From 20−2x>0 we obtain x<10. Both conditions must hold, giving the permitted physical interval 0 The smaller sheet dimension controls the upper limit, because cutting ten centimetres from both sides of the 20 cm width leaves no base. A stationary root outside this interval may solve the derivative equation algebraically but cannot describe a genuine positive-volume box from the stated sheet. As x approaches zero, the folded box becomes almost flat. Its height tends to zero even though the base remains nearly 30 by 20 centimetres. As x approaches ten from below, the height approaches ten centimetres but the shorter base dimension 20−2x approaches zero. Volume again tends to zero. Therefore an intermediate cut must give more volume than either extreme. The optimisation balances decreasing base area against increasing height. This reasoning predicts an interior maximum before differentiation. It also provides a powerful check: a calculated optimum of x=0 or x=10 would contradict the purpose of maximising positive box volume. Start from V(x)=600x−100x²+4x³. Differentiate each term using the power rule. The derivative is V′(x)=600−200x+12x². A stationary volume occurs where V′(x)=0. Divide by four to simplify: 150−50x+3x²=0. Rearrange into the standard quadratic form 3x²−50x+150=0. This is an algebraic equation for candidate corner cuts, not yet the final geometry answer. Every root must still be checked against 0 Apply the quadratic formula to 3x²−50x+150=0. The roots are x=[50±√(50²−4·3·150)]/(2·3). The discriminant is 2500−1800=700=100·7. Thus x=(50±10√7)/6=(25±5√7)/3. The smaller root is x₁=(25−5√7)/3≈3.923747815. The larger root is x₂=(25+5√7)/3≈12.74291885. Both are exact stationary solutions of the algebraic cubic volume expression, but only the smaller root fits the actual rectangular sheet. The sheet’s shorter side is 20 cm. For a positive-width base, we require x<10. The larger derivative root is about 12.743 cm. A cut of that size would subtract approximately 25.486 cm from the original 20 cm width. The resulting base width would be negative, which is not physically meaningful. Therefore x₂ is inadmissible. This is not a mathematical mistake in solving the quadratic. It is an algebraic candidate outside the model’s domain. A complete optimisation solution should explicitly reject it rather than merely selecting the smaller root without explanation. The smaller value x₁≈3.924 lies strictly between zero and ten. The resulting base length is 30−2x₁≈22.153 cm and base width is 20−2x₁≈12.153 cm. Both are positive. The proposed box therefore has three valid dimensions: approximately 22.153 cm, 12.153 cm and 3.924 cm. This establishes that the stationary candidate represents a real constructible box under the stated idealised folding model. The remaining task is to prove that it gives the largest volume rather than a stationary minimum. Differentiate V′(x)=600−200x+12x² again to obtain V″(x)=−200+24x. At x≈3.924, V″≈−200+94.170=−105.830, which is negative. Thus the stationary point is a local maximum of volume. The derivative is positive before this root within the physical domain and negative afterwards, as the next root is outside the domain. Combined with volume tending to zero at both ends, this makes the admissible stationary point the global maximum for 0 At x=2, V′(2)=600−400+48=248>0, so volume is increasing there. At x=5, V′(5)=600−1000+300=−100<0, so volume is decreasing there. The derivative changes from positive to negative at the admissible stationary value near 3.924. This sign change is direct evidence of a maximum. A second-derivative check is efficient, but a first-derivative sign chart is equally valid and often more revealing to students who want to understand the trade-off rather than only memorise a test for extrema. Let x*=(25−5√7)/3. The base length is 30−2x*. Simplify: 30−2(25−5√7)/3=(90−50+10√7)/3=(40+10√7)/3. The base width is 20−2x*=(60−50+10√7)/3=(10+10√7)/3. The height remains x*=(25−5√7)/3. These exact surd expressions describe the full optimal box. They can be left in this form even though the physical measurements would later be rounded for an actual cutting plan. Substitute x* into V=x(30−2x)(20−2x), or use the stationary condition to simplify the cubic expression. The exact result is V_max=(10000+7000√7)/27 cubic centimetres. Numerically, √7≈2.64575131 gives V_max≈1056.305895 cm³. To three decimal places, that is 1056.306 cm³. The exact and approximate forms describe the same maximum. The algebraic surd preserves precision, while the decimal is easier to use in a practical volume comparison. If the question asks for an exact answer, retain the fraction involving √7. Try x=4 cm. The base length is 30−8=22 cm and the base width is 20−8=12 cm. The volume is V(4)=4·22·12=1056 cm³. This is only about 0.306 cm³ below the exact continuous maximum. That close agreement is plausible because the optimum x≈3.924 is near four. A numerical check does not prove that the continuous maximum is at x*, but it helps detect gross errors. An answer of only 105.6 or more than ten thousand cubic centimetres would deserve immediate rechecking. For x=3, the base is 24 by 14 and height 3, giving V(3)=3·24·14=1008 cm³. For x=4, volume is 1056 cm³. For x=5, the base is 20 by 10 and height 5, giving V(5)=1000 cm³. Among these three whole-centimetre choices, the four-centimetre cut is best. This comparison illustrates the trade-off: making the corner square larger can increase height but eventually removes too much base area. The continuous derivative identifies the peak, while a restricted whole-number decision requires comparing permitted cut sizes. Suppose the task requires the corner squares to have an integer side length in centimetres. The physically permitted choices are x=1,2,…,9. The continuous optimum lies between 3 and 4. Because the volume increases up to about 3.924 and decreases afterwards, only neighbouring integer candidates need comparison for the maximum. We found V(3)=1008 and V(4)=1056, so x=4 is the best permitted integer cut. The corresponding maximum under the whole-centimetre restriction is 1056 cm³, not the continuous value 1056.306 cm³. A mathematically correct answer must respect the actual decision variable’s permitted values. Let the original sheet dimensions be L and W, with L≥W>0. Cutting corner squares of side x gives base length L−2x, base width W−2x and height x. The volume is V(x)=x(L−2x)(W−2x), with domain 0 Expand to V(x)=LW x−2(L+W)x²+4x³. Differentiate: V′(x)=LW−4(L+W)x+12x². The stationary equation is 12x²−4(L+W)x+LW=0. This general form explains why the numerical problem always reduces to a quadratic stationary condition despite starting from a cubic volume expression. Solve 12x²−4(L+W)x+LW=0 using the quadratic formula. After simplification, the two roots are x=[(L+W)±√(L²−LW+W²)]/6. For a rectangular sheet with L≥W>0, the smaller root is the physically relevant maximum-volume cut. The larger root is at or beyond the geometric limit for a positive base width. Therefore x_max=[L+W−√(L²−LW+W²)]/6. Substituting L=30 and W=20 gives [50−√700]/6=(25−5√7)/3, matching the worked example. The square-root term in the general expression is positive. For L≥W>0, the smaller stationary root lies strictly inside 0 One way to see this is to inspect V′ at the endpoints of the physical interval. At x=0, V′(0)=LW>0. At x=W/2, substitution into the derivative gives V′(W/2)=W(W−L), which is nonpositive, and strictly negative when L>W. Thus at least one stationary point lies inside for a non-square rectangle. The derivative quadratic has its other stationary root outside the valid interval. This confirms that the smaller root supplies the physically meaningful maximum rather than an arbitrary choice of algebraic sign. If L=W=s, the general cut formula simplifies to x_max=[2s−√(s²−s²+s²)]/6. Thus x_max=(2s−s)/6=s/6. This is the familiar rule that each cut square should have side one-sixth of the original square sheet’s side length. The resulting base dimensions are s−2(s/6)=2s/3 in both directions, and the height is s/6. Maximum volume is (s/6)(2s/3)²=2s³/27. This special relationship should not be copied into a non-square sheet without rederiving the optimisation conditions. For a 30 by 20 cm rectangle, one-sixth of the shorter side is 20/6≈3.333 cm. But the actual optimum is approximately 3.924 cm. Using x=10/3 gives volume (10/3)(70/3)(40/3)=28000/27≈1037.037 cm³, noticeably below the true maximum near 1056.306. The mismatch arises because the two base dimensions shrink from different starting lengths. Their contributions to the volume derivative are not symmetric as in a square sheet. This is a clear example of why an optimisation formula belongs to its assumptions. A rule memorised for a square can be efficient there but inaccurate for a different rectangle. Take L=24 cm and W=18 cm. The general formula gives x_max=[42−√(24²−24·18+18²)]/6. The quantity under the square root is 576−432+324=468=36·13. Therefore x_max=[42−6√13]/6=7−√13. Numerically, x_max≈3.394 cm, which lies within 0 The optimal base dimensions are 24−2(7−√13)=10+2√13 and 18−2(7−√13)=4+2√13. Their product with height 7−√13 gives the maximum volume. This variation shows how exact radical cuts arise from different sheet proportions. Cutting four squares of area x² removes a total material area 4x². The remaining sheet material area is LW−4x². But box volume is not this remaining area multiplied by height x. The remaining material includes the base and all four vertical side flaps, whereas a box’s internal volume uses base area multiplied by height. The correct base area is (L−2x)(W−2x), not LW−4x². This distinction is especially important in modelling word problems. The mathematical quantity being maximised must match the object’s physical definition; using a nearby but different quantity can lead to a plausible-looking yet incorrect optimum. The present construction leaves the top open and uses four corner squares to create vertical sides from one sheet. If the box must have a lid made from the same sheet, the folding net and material constraints change. The base and wall dimensions might no longer follow the same simple formulas. Likewise, a box assembled from separate panels under a fixed total surface-area condition is a different optimisation problem. Do not transfer x_max=[L+W−√(L²−LW+W²)]/6 to a design that does not actually remove equal squares from the four corners and fold the remaining flaps at right angles. The idealised model treats the sheet as having negligible thickness. That makes the folded height exactly x and the internal base dimensions exactly L−2x and W−2x. Real cardboard has thickness, fold radii and possible overlap or adhesive flaps. These details can reduce internal dimensions and change material use. In Additional Mathematics, the conventional problem assumes an ideal geometric sheet unless otherwise stated. The calculus answer is exact within that mathematical model, not a manufacturing tolerance specification. This distinction helps students interpret model results accurately: mathematical validity and physical fabrication accuracy are related but not identical claims. Suppose every sheet dimension is multiplied by the same positive factor k. Then L and W become kL and kW. The general optimum cut scales to kx_max because its formula is homogeneous of degree one in L and W. The optimised base dimensions and height all scale by k, so volume scales by k³. For example, doubling both dimensions of the 30 by 20 sheet doubles the optimal corner cut and multiplies the maximum volume by eight. This is the expected behaviour of three-dimensional similarity and provides a useful units check on any general formula. Keep the shorter dimension W=20 but increase the longer dimension L. The general stationary formula still applies, yet the optimal x need not remain 3.924. A much longer rectangle can benefit from a different cut because the volume sensitivity of its length factor changes relative to the width and height. Thus the optimum depends on both sheet dimensions, not only on the shorter side. A student should not assume that fixing W fixes x_max. The formula x_max=[L+W−√(L²−LW+W²)]/6 explicitly contains L as well, which is the precise reason the shape’s aspect ratio matters. Increasing x makes the box taller, but the base length and width both decrease. At x near ten, height approaches ten centimetres, yet base width approaches zero. Volume collapses rather than increases. At x near zero, the base approaches its full 30 by 20 cm dimensions, but the height approaches zero and volume again collapses. The maximum lies between those extremes, where the marginal benefit of increasing height is exactly balanced by the marginal loss from shrinking the base. Differentiation is a way to express that balance mathematically rather than guessing a visually attractive box. Differentiate V=x(L−2x)(W−2x) directly using the product rule. The derivative is (L−2x)(W−2x)−2x(W−2x)−2x(L−2x). Expand and collect terms to obtain V′=LW−4(L+W)x+12x². This agrees with differentiating the expanded cubic. The factored derivative also gives a geometric interpretation: the first term comes from the increase in height, while the two subtracted terms reflect the shrinking base dimensions. Two derivations of the same stationary equation are useful when a student suspects a sign error in differentiating the polynomial form. For the 30 by 20 sheet, V′(x)=600−200x+12x². Near x=0, the derivative is positive because increasing the cut from almost zero creates useful height. At x=5, V′(5)=−100, so increasing the cut there reduces volume. Only one of the derivative’s two roots lies within 0 This supplies a complete global maximum argument, not merely a local second-derivative test. The value cannot be beaten near either endpoint because volume tends to zero there. A calculator can evaluate the exact cut x*=(25−5√7)/3 and return approximately 3.92375 cm. Substituting that value gives volume approximately 1056.3059 cm³. A numerical maximum tool may obtain the same answer to machine precision. But the calculator does not tell us why the interval is 0 A complete school solution should state those geometric and algebraic reasons. Use technology to check arithmetic and precision after the correct volume model has been established. One student subtracts x once from each sheet dimension rather than 2x. The repair is a labelled folding-net sketch showing a cut at both ends. Another multiplies the remaining material area LW−4x² by height instead of using the base area. That requires revisiting the geometry of volume. A third solves the derivative quadratic correctly but keeps the root x≈12.743 cm, ignoring the physical domain. A fourth reports half-width or cut size instead of the maximum volume requested. These are different first errors. Correct the earliest misunderstanding, then test a new rectangular-sheet size to see whether the student can reconstruct the model independently. Question one: a 30 by 20 cm sheet has corner squares of side x. Write its volume. Answer: V=x(30−2x)(20−2x). Question two: state the valid positive-volume range. Answer: 0 Question three: find the continuous optimal cut. Answer: x=(25−5√7)/3≈3.924 cm. Question four: find the exact maximum volume. Answer: (10000+7000√7)/27≈1056.306 cm³. Question five: if x must be a whole number of centimetres, find the best cut and volume. Answer: x=4 cm, V=1056 cm³. Question six: for a square sheet of side s, find the optimal corner cut. Answer: s/6. Question seven: for a rectangular sheet L≥W>0, give the general optimum. Answer: [(L+W)−√(L²−LW+W²)]/6. For each, justify the physical bounds before accepting stationary roots. Open-top box optimisation connects geometry, cubic volume functions, differentiation, quadratic factorisation and applied constraints. The general rectangular-sheet formula extends the classic square-sheet example and is useful for students learning to translate word problems into objective functions. Singapore families can consult the relevant SEAB syllabus pages for 2026 O-Level Additional Mathematics or 2027 SEC G3, matching the learner’s school’s assessed content. Related eduKateSG guides include How to Find the Maximum Area of a Rectangle Under a Parabola, How to Maximise a Circular Sector’s Area with Fixed Perimeter and How to Find the Maximum Area of a Rectangle Inscribed in a Semicircle. For a related square-sheet model, see The Cardboard Tray That Hides a Cubic Maximum. Browse the Additional Mathematics archive or enquire about suitable eduKate Singapore maximum three-student Mathematics tutorials near Sixth Avenue MRT. The enduring skill is defining the physical decision variable, deriving the objective exactly and rejecting algebraic solutions that do not describe a possible object. Properly taught kids shine a bright light into the future.Begin by drawing the uncut sheet
Determine which parts of the base remain
Build the correct cubic volume function
State the geometry domain before solving
Why the volume approaches zero at both extremes
Differentiate the volume function
Solve the stationary quadratic exactly
Reject the larger root for a physical reason
Verify the smaller stationary root is admissible
Use the second derivative to classify the candidate
Check the derivative sign on either side
Find the exact optimal base dimensions
Calculate the exact maximum volume
Verify the result with a nearby whole-centimetre cut
Compare cuts of three, four and five centimetres
If cuts must be whole centimetres, the answer changes
A general rectangular-sheet formula
Derive the exact general optimum cut
Why the general optimum is smaller than half the short side
A square sheet has the simpler one-sixth rule
Why a rectangular sheet is different from a square
A second rectangular-sheet example
Volume is not the same as remaining sheet area
A closed-top box would require different geometry
The thickness assumption matters in a real package
The optimum volume scales with the cube of overall size
A wider sheet can change the optimal cut
Why simply maximising height fails
An alternative derivative from the factored volume
A quick sign check on the physical interval
A numerical calculator should confirm rather than choose the model
Common errors and their first repairs
Practice questions with checked answers
Syllabus fit and connected study