Open-Top Box: Domain Restrictions and Integer Cuts
A packaging technician is designing an open-top box from a single rectangular sheet of card measuring 40 cm by 30 cm. Identical squares of side x cm are cut from each corner, and the remaining card is folded up to form the sides. The technician writes the volume as V=x(40−2x)(30−2x) and differentiates to find the maximum. A colleague solves V′=0 and proposes x=17.7 cm, arguing that it is also a stationary point and gives a large value of x. The technician must reject this proposal, justify the domain of x, and classify the valid stationary point. Because the cutting machine can only make whole-centimetre cuts, the technician must also compare the best integer cut with the continuous optimum. The box must be open at the top, and the card has no overlap allowance. Volume is required to the nearest 0.1 cm³ and x to two decimal places where appropriate. The technician must also confirm that the valid stationary point gives the greatest volume on the physical domain, not merely a local maximum.
1. Expand V(x) and state the physically meaningful domain for x. Explain why the colleague's value x≈17.7 cm must be rejected.
Read the worked answer
Expand the product:
(40−2x)(30−2x)=1200−80x−60x+4x²=1200−140x+4x²,
so V=x(1200−140x+4x²)=4x³−140x²+1200x.
For a real box, both base dimensions must be positive:
40−2x>0 ⇒ x<20, and 30−2x>0 ⇒ x<15.
Together with x>0, the physical domain is 0<x<15. The value x≈17.7 cm is outside this domain and would make 30−2x negative, which is impossible for a length. It is a stationary point of the algebraic cubic but not a feasible box design.
2. Find the stationary point in the domain, classify it, and calculate the maximum volume.
Read the worked answer
Differentiate:
V′=12x²−280x+1200=4(3x²−70x+300).
Set V′=0: 3x²−70x+300=0. The discriminant is 70²−4×3×300=4900−3600=1300, so
x=(70±√1300)/6=(70±10√13)/6.
The two roots are approximately 5.657 and 17.676. Only 5.657 lies in 0<x<15, so the feasible stationary value is x≈5.66 cm.
Classify using V″=24x−280. At x≈5.657, V″≈24(5.657)−280≈−144.2<0, so the stationary point is a local maximum. Since V→0 at both endpoints x→0⁺ and x→15⁻, this local maximum is the greatest volume on the physical domain.
V(5.657)=5.657(40−11.314)(30−11.314)≈5.657×28.686×18.686≈3032.3 cm³.
3. The cutter can only make whole-centimetre cuts. Compare the best whole-centimetre choice with the continuous optimum and state the loss.
Read the worked answer
The whole-centimetre values near the maximum are x=5 and x=6; values farther away are smaller because V increases up to x≈5.66 and decreases afterwards.
V(5)=5(40−10)(30−10)=5×30×20=3000 cm³.
V(6)=6(40−12)(30−12)=6×28×18=3024 cm³.
Thus x=6 cm gives the best whole-centimetre box, with volume 3024 cm³. The continuous optimum is about 3032.3 cm³, so the loss from using an integer cut is about 3032.3−3024=8.3 cm³. Checking the adjacent integers is sufficient here because there is only one stationary maximum in the domain and the volume falls away on either side.