Open-Top Box from a 24 cm by 18 cm Sheet: True Maximum versus a Minimum Volume
A boutique wants an open-top rectangular box made from a rectangular cardboard sheet measuring 24 cm by 18 cm. Equal squares of side x cm are cut from each corner, and the flaps are folded up. The box has no lid. The volume is V(x)=x(24−2x)(18−2x), with 0<x<9 because the shorter side limits the cut. The boutique wants the box to hold as much as possible, but the production manager also insists that the finished box must hold at least 600 cm³. A worker proposes x=5 cm, arguing that 'cutting larger squares uses more of the cardboard and must give a bigger box'. A second worker proposes x=4 cm because it is a whole number and easy to measure. The manager wants a recommendation that states the true maximum and also checks whether the 600 cm³ minimum is met at the recommended cut.
1. Find the value of x in (0,9) that maximizes the volume, and compute the maximum volume to the nearest cm³.
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Expand: (V=x(24-2x)(18-2x)=x(432-84x+4x^2)=4x^3-84x^2+432x).
Differentiate: (V'(x)=12x^2-168x+432=12(x^2-14x+36)). Set (V'=0): (x=\frac{14\pm\sqrt{196-144}}{2}=7\pm\sqrt{13}). Numerically, (\sqrt{13}\approx3.606), so the roots are about 3.394 and 10.606. Only 3.394 lies in (0,9).
Second derivative: (V''=24x-168). At (x\approx3.394), (V''\approx81.5-168=-86.5<0), so this is a local maximum. The endpoints give (V(0)=0) and (V(9)=0), so the local maximum is the global maximum on the interval.
Volume: (V(3.394)=3.394(24-6.788)(18-6.788)=3.394(17.212)(11.212)\approx655\text{ cm}^3). Maximum at (x=7-\sqrt{13}\approx3.39) cm; maximum volume about 655 cm³.
2. Evaluate the two worker proposals. Does x=5 meet the 600 cm³ minimum? Does x=4 maximize volume? If the cut must be a whole number of centimetres, what should the manager choose?
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x=5: (V=5(24-10)(18-10)=5(14)(8)=560\text{ cm}^3). This is below 600, so it fails the manager's minimum. The worker's reasoning is also wrong: a larger cut removes more from the base, reducing the base area; volume does not keep increasing.
x=4: (V=4(24-8)(18-8)=4(16)(10)=640\text{ cm}^3). It meets the 600 minimum, but it is not the maximum; the true maximum is about 655 cm³ at 3.39 cm.
If the cut must be an integer, check nearby values: (x=2): (2(20)(14)=560); (x=3): (3(18)(12)=648); (x=4): 640; (x=5): 560. The best integer is x=3 cm, giving 648 cm³, not x=4. If exact measurement is possible, use 3.39 cm for 655 cm³.
3. The manager asks whether a cut of x=4.6 cm is acceptable under the 600 cm³ minimum. Compute V(4.6) and interpret.
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(V(4.6)=4.6(24-9.2)(18-9.2)=4.6(14.8)(8.8)). First, (14.8\times8.8=130.24). Then (4.6\times130.24=599.104), about 599 cm³.
This is just below 600, so if 600 cm³ is a strict minimum, x=4.6 cm is not acceptable. It illustrates that near the upper boundary of the feasible region, a small change in x can push the volume below the requirement. Do not round 599.104 up to 600 and declare it compliant; either reduce x slightly, for example around 4.55 cm, and recompute, or choose a safer value such as x=4 cm, which gives 640 cm³.