Open Box Production from 24 cm by 18 cm Cardboard
A packaging unit makes open rectangular boxes from rectangular cardboard sheets measuring 24 cm by 18 cm. Equal squares of side x cm are cut from each corner, and the remaining flaps are folded up. The base dimensions are therefore (24−2x) cm by (18−2x) cm, and the height is x cm. The production manager wants the box of maximum volume. A customer separately orders boxes of volume 600 cm³. The cutting machine can set any x between 0 and 9 cm, but the workshop foreman warns that not every setting in that range gives a practical box. The manager must find the cut that maximises volume, determine how many cut sizes can produce the 600 cm³ order, and explain why the larger cut is not automatically the better choice.
1. Write the volume V as a function of x, state the domain, and find the cut that maximises the volume. Give the maximum volume.
Read the worked answer
V=x(24−2x)(18−2x), with 0<x<9. Expanding gives V=4x³−84x²+432x. Then V'=12x²−168x+432=12(x²−14x+36). Setting V'=0 gives x=7±sqrt(13), so x≈3.394 or x≈10.606. Only 3.394 lies in (0,9). Since V=0 at both endpoints and V is positive inside, this critical point gives the maximum. V≈3.394(24−6.789)(18−6.789)≈3.394×17.211×11.211≈655 cm³.
2. A customer orders boxes of volume 600 cm³. Show that two cut sizes in (0,9) produce this volume, approximate them, and advise which cut to use if the box must have the larger base area.
Read the worked answer
Solve V=600: 4x³−84x²+432x−600=0, or x³−21x²+108x−150=0. Let f(x) be this cubic. f(2)=−10 and f(2.32)≈0.02, so a root is x≈2.32. Also f(4)=40 and f(4.59)≈−0.01, so another root is x≈4.59. Both lie in (0,9). The maximum volume is about 655 cm³, so the level 600 cuts the volume curve twice. Base area is (24−2x)(18−2x): at x≈2.32 it is about 258.6 cm²; at x≈4.59 it is about 130.7 cm². Choose x≈2.32 cm for the larger base.
3. The foreman says x=8 cm is 'safer because it is inside the allowed range'. Evaluate this reasoning using the volume function.
Read the worked answer
At x=8, V=8(24−16)(18−16)=8×8×2=128 cm³, far below the maximum near 655 cm³. Being inside the domain does not make a cut efficient. The endpoints x=0 and x=9 give zero volume because the box collapses, and the maximum occurs where V'=0 inside the domain. The foreman's reasoning ignores the shape of the cubic and confuses feasibility with optimality.