Optimising an Open-Top Box When the Constraint Bites
A packaging workshop starts with flat card sheets measuring 20 cm by 32 cm. Equal squares of side x cm are cut from the four corners, the flaps are folded up and taped, producing an open-top box of height x cm and volume V cm³. Two constraints apply. First, geometry alone limits x to 0 < x < 10. Second, the product to be packed is 5.5 cm deep, so the production manager insists the finished box must be at least 6 cm tall. A customer orders boxes of volume 1000 cm³. The supervisor says this is no problem: 'Cut 5 cm squares - those give the biggest box possible, so 1000 cm³ must fit easily inside the maximum.' Meanwhile a trainee who solved dV/dx = 0 on a calculator has obtained two values, x = 4 and x = 13.33, and asks which one to use. The workshop can only cut squares whose side is measured to the nearest 0.1 cm.
1. Show that V = 4x³ − 104x² + 640x, locate the stationary values of V, and decide which of the trainee's two values is meaningful. Hence test the supervisor's claim that 5 cm squares give the biggest box.
Read the worked answer
Forming the model
After cutting, the base is (20 − 2x) by (32 − 2x), so
V = x(20 − 2x)(32 − 2x) = x(640 − 104x + 4x²) = 4x³ − 104x² + 640x.
Stationary points
dV/dx = 12x² − 208x + 640 = 4(3x² − 52x + 160) = 0
x = [52 ± √(52² − 4·3·160)]/(2·3) = (52 ± √784)/6 = (52 ± 28)/6
x = 4 or x = 80/6 ≈ 13.33.
Judgement on the roots
The problem requires 0 < x < 10 (the 20 cm side runs out first). So x = 13.33 is rejected: at that value 20 − 2x = −6.67 cm, a negative length. It is a root of the algebra, not a solution of the problem.
Nature of x = 4
d²V/dx² = 24x − 208; at x = 4 this is 96 − 208 = −112 < 0, so x = 4 gives a maximum.
V(4) = 4(12)(24) = 1152 cm³.
The supervisor's claim
At x = 5: V = 5(10)(22) = 1100 cm³, which is less than 1152 cm³. Five-centimetre squares give a box that is close to, but not at, the maximum - the claim is false.
2. With the 6 cm minimum height enforced, find the greatest volume actually available, and determine whether the 1000 cm³ order can be met. If it cannot, state the smallest change that would make it possible.
Read the worked answer
The 6 cm rule restricts x to the interval [6, 10).
V'(x) = 4(3x² − 52x + 160) is a upward-opening quadratic with roots x = 4 and x = 13.33, so V' < 0 for 4 < x < 13.33. Hence on [6, 10) the volume is strictly decreasing, and the greatest usable volume is at the endpoint x = 6:
V(6) = 6(20 − 12)(32 − 12) = 6(8)(20) = 960 cm³.
960 cm³ < 1000 cm³, so the order cannot be met under the height rule. Note that the unconstrained maximum of 1152 cm³ occurs at x = 4, a height of only 4 cm, which the product depth rules out - this is why the calculus answer alone is not the answer to the customer's problem.
How close can we get to 1000 cm³?
Solve V = 1000 with x < 6:
V(5.7) = 5.7(8.6)(20.6) ≈ 1009.7 cm³
V(5.8) = 5.8(8.4)(20.4) ≈ 993.9 cm³
So the root is x ≈ 5.8 cm (to the nearest 0.1 cm), i.e. a box only about 5.8 cm tall. Two honest options: relax the 6 cm height rule, or use larger card. On 22 cm by 34 cm card, x = 6 gives V = 6(10)(22) = 1320 cm³, comfortably above 1000 cm³.
3. Explain, in terms of the physical situation, why x = 13.33 must be discarded even though it satisfies dV/dx = 0, and why the stationary maximum at x = 4 does not by itself answer the customer's order.
Read the worked answer
Why x = 13.33 is discarded
A stationary point is only a candidate; it must also be a physically possible configuration. The sheet is 20 cm wide, so cutting squares wider than 10 cm leaves a negative base dimension (20 − 26.67 = −6.67 cm). No box exists at that value, so the equation dV/dx = 0 has produced a spurious solution in the context of this problem. The valid domain 0 < x < 10 is part of the model, not an afterthought.
Why x = 4 is not the final answer
The stationary value gives the largest volume over the whole interval 0 < x < 10, but the manager's 6 cm height requirement changes the feasible set to [6, 10). On that smaller set the maximum sits at the endpoint x = 6, giving 960 cm³. Judgement required: identify the feasible interval first, then optimise on it, checking endpoints as well as stationary points. Quoting 1152 cm³ to the customer would promise a box that the product cannot use.