Optimising a Folded Cardboard Tray with a Material Constraint
A student cuts squares of side x cm from each corner of a rectangular sheet measuring 24 cm by 18 cm and folds up the sides to form an open tray of volume V(x) = x(24 - 2x)(18 - 2x). The student claims the maximum volume occurs at x = 4 because that is halfway between the endpoints 0 and 9. A second constraint is introduced: the tray must have a base perimeter of at least 30 cm. The student must find the true maximum volume, check whether the perimeter constraint is active, and justify that the maximum is global on the feasible domain.
1. Find the value of x that maximises V(x) on its natural domain, giving the exact maximum volume.
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Expand: V(x) = x(432 - 84x + 4x^2) = 4x^3 - 84x^2 + 432x. The natural domain requires 24 - 2x > 0 and 18 - 2x > 0 and x > 0, so 0 < x < 9.
V'(x) = 12x^2 - 168x + 432 = 12(x^2 - 14x + 36). Setting V' = 0: x = [14 ± √(196 - 144)]/2 = [14 ± √52]/2 = 7 ± √13. Numerically √13 ≈ 3.606, so x ≈ 3.394 or x ≈ 10.606. Only x = 7 - √13 ≈ 3.394 lies in (0, 9).
V''(x) = 24x - 168, so V''(7 - √13) = 24(7 - √13) - 168 = 168 - 24√13 - 168 = -24√13 < 0, confirming a local maximum. Since V(0) = 0 and V(9) = 0 and this is the only interior critical point, it is the global maximum.
Maximum volume: V = (7 - √13)(24 - 14 + 2√13)(18 - 14 + 2√13) = (7 - √13)(10 + 2√13)(4 + 2√13). Numerically x ≈ 3.394, giving V ≈ 3.394 × 17.212 × 11.212 ≈ 655 cm³.
2. Assess the student's claim that x = 4 gives the maximum, and identify the error in reasoning.
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The claim is false. Substituting x = 4: V(4) = 4(16)(10) = 640 cm³. The true maximum is about 655 cm³ at x ≈ 3.394, so x = 4 is close but not optimal.
The reasoning error is assuming the maximum of a cubic occurs at the midpoint of the domain endpoints. That would only hold for a symmetric quadratic. Here V is a cubic with V(0) = V(9) = 0, but the interior maximum need not be at the midpoint because the cubic is not symmetric about x = 4.5; the x^2 coefficient (-84/4 = -21 in monic form) shifts the turning point. Solving V'(x) = 0 is the only reliable method. Note V'(4) = 12(16 - 56 + 36) = 12(-4) = -48 < 0, so V is already decreasing at x = 4, confirming the maximum lies to the left of 4.
3. Determine whether the constraint 'base perimeter at least 30 cm' is active at the optimal x, and find the constrained maximum if it is not.
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The base dimensions are (24 - 2x) and (18 - 2x), so the base perimeter is P(x) = 2[(24 - 2x) + (18 - 2x)] = 2(42 - 4x) = 84 - 8x.
At x = 7 - √13 ≈ 3.394: P ≈ 84 - 27.15 ≈ 56.85 cm, which comfortably exceeds 30 cm. So the constraint is not active at the unconstrained optimum.
Check whether the constraint could bind anywhere: P ≥ 30 requires 84 - 8x ≥ 30, so x ≤ 6.75. Since the unconstrained maximiser 3.394 < 6.75, the constraint does not exclude it, and the constrained maximum equals the unconstrained maximum, V ≈ 655 cm³ at x ≈ 3.394. The constraint only matters for x in (6.75, 9), where V is decreasing anyway, so it does not change the answer.