Maximising an Open Tray Under a Tooling Limit
A packaging firm stamps open trays from rectangular card blanks measuring 24 cm by 18 cm. A square of side x centimetres is cut from each corner and the remaining cross is folded up to form an open box; the volume is to be maximised. The guillotine's adjustable stop is fitted with a clamp that will not allow a cut deeper than 3.0 cm, so although the mathematics may suggest a larger square, this machine physically cannot produce it. The customer's specification requires each tray to hold at least 600 cm³, and the sales team also wants a sensitivity figure: how much volume is lost if the setter is 0.05 cm away from the ideal cut. A trial batch was run at x = 3.0 cm and another at x = 3.4 cm, and both passed the 600 cm³ test, which has convinced the foreman that the setting hardly matters. The production engineer must establish the true unconstrained optimum, decide the best achievable setting on this machine, and quantify the foreman's sensitivity claim.
1. Express the tray volume V as a function of x, state its natural domain, and find and classify the stationary point.
Read the worked answer
After cutting squares of side x, the base measures (24 − 2x) by (18 − 2x) and the height is x, so
V(x) = x(24 − 2x)(18 − 2x) = 4x(12 − x)(9 − x).
Domain: x > 0 and 18 − 2x > 0, so 0 < x < 9.
Expand: 4x(108 − 21x + x²) = 432x − 84x² + 4x³.
V′(x) = 432 − 168x + 12x² = 12(x² − 14x + 36).
Setting V′(x) = 0: x = [14 ± √(196 − 144)]/2 = 7 ± √13. With √13 ≈ 3.606, the roots are x ≈ 3.394 and x ≈ 10.606. Only x = 7 − √13 ≈ 3.39 lies in (0, 9); the other root is outside the domain and corresponds to a negative base dimension.
Classify: V″(x) = 24x − 168, so V″(3.394) ≈ 81.5 − 168 = −86.5 < 0, a local maximum. Since it is the only stationary point in the domain and V → 0 at both endpoints x → 0⁺ and x → 9⁻, it is the global maximum on (0, 9).
Maximum volume: x ≈ 3.394, base 17.21 by 11.21, V ≈ 655.0 cm³.
2. Given the 3.0 cm clamp, state the best achievable setting and the resulting volume, and explain why the derivative test alone is not enough here.
Read the worked answer
The clamp restricts x to 0 < x ≤ 3.0, so the unconstrained optimum x ≈ 3.394 is not attainable. On (0, 9) the only stationary point is at 3.394, and V′(0) = 432 > 0, so V′ > 0 throughout (0, 3.394). Hence V is strictly increasing on (0, 3.0] and the best permitted value is the boundary point x = 3.0.
V(3.0) = 3(24 − 6)(18 − 6) = 3 × 18 × 12 = 648 cm³.
This satisfies the 600 cm³ specification with 48 cm³ to spare, and falls 7.0 cm³ short of the theoretical maximum of about 655.0 cm³.
Why the derivative test is insufficient: setting V′ = 0 locates only interior stationary points. Once the feasible set is truncated by the clamp, the optimum can sit on the boundary, where the derivative need not vanish (here V′(3.0) = 432 − 504 + 108 = 36 > 0). A constrained optimisation must compare interior stationary points with endpoint values; relying on V′ = 0 alone would wrongly report 3.39 cm as achievable.
3. Assess the foreman's claim that the setting hardly matters, using the second derivative to estimate the effect of a 0.05 cm error near the optimum, and contrast this with the 3.0 cm setting.
Read the worked answer
Near the stationary point the first derivative vanishes, so a first-order estimate ΔV ≈ V′(x*)Δx gives zero and the leading term is second order:
ΔV ≈ ½V″(x*)(Δx)².
With V″ = 24x − 168, at x* ≈ 3.394 we get V″ ≈ −86.5. For Δx = ±0.05:
ΔV ≈ ½(−86.5)(0.05)² = −0.108 cm³,
a loss of about 0.1 cm³ out of 655 cm³, roughly 0.017 per cent. Direct substitution confirms this: V(3.4) = 4(3.4)(8.6)(5.6) = 654.98, against 654.97 at the exact optimum.
So near the optimum the claim is correct, and the reason is geometric: the volume curve is flat at its peak, and the error is quadratic in the setting error.
But the claim does not transfer to the 3.0 cm setting. At x = 3.0 the derivative is 36, not 0, so the loss is first order: V(3.0) = 648 against 655.0, a loss of 7.0 cm³ (about 1.1 per cent) for a 0.39 cm change. The foreman's two trial batches happened to straddle the flat region and gave misleadingly similar volumes; the sensitivity statement is valid only within a few hundredths of a centimetre of the true optimum, not across the range of machine settings.