Optimising a Folded Box from a Sheet
A manufacturer 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 make an open box. The volume is V(x) = x(24 − 2x)(18 − 2x). A junior designer sets dV/dx = 0, solves, and reports that the maximum volume occurs at x = 6, giving V = 0. They are puzzled because the box has 'no volume' yet the derivative vanished, and they conclude the calculus method has failed. The senior engineer must correct the reasoning, identify the true maximum, and explain the role of the domain restriction on x, which is often ignored in textbook treatments.
1. Expand V(x), find dV/dx, and solve dV/dx = 0. Then explain the significance of each critical point in context.
Read the worked answer
Expand: (24 − 2x)(18 − 2x) = 432 − 48x − 36x + 4x² = 432 − 84x + 4x². So V = 432x − 84x² + 4x³.
Then dV/dx = 432 − 168x + 12x² = 12(x² − 14x + 36). Setting this to zero: x² − 14x + 36 = 0, so x = [14 ± √(196 − 144)]/2 = [14 ± √52]/2 = 7 ± √13.
Numerically x ≈ 3.39 or x ≈ 10.61. The physical domain requires x > 0 and 18 − 2x > 0, so 0 < x < 9. Thus x = 7 + √13 ≈ 10.61 is outside the domain and must be rejected. The only interior critical point is x = 7 − √13 ≈ 3.39.
The value x = 6 that the designer found is not a critical point of V at all; it is where one factor 24 − 2x = 12 and 18 − 2x = 6, and V(6) = 6·12·6 = 432, not 0. The designer likely set a single factor to zero or confused the corner condition. The claim V = 0 at x = 6 is simply wrong.
2. Justify that x = 7 − √13 gives the maximum, and state the maximum volume to 2 decimal places.
Read the worked answer
Use the second derivative: d²V/dx² = −168 + 24x. At x = 7 − √13 ≈ 3.39, this is −168 + 81.4 ≈ −86.6 < 0, so the critical point is a local maximum. Since the domain is 0 < x < 9 and V → 0 as x → 0⁺ or x → 9⁻ (the box flattens at both ends), the single interior critical point must be the global maximum on the open interval.
Alternatively, test signs of dV/dx: at x = 1, dV/dx = 432 − 168 + 12 = 276 > 0; at x = 5, dV/dx = 432 − 840 + 300 = −108 < 0. So V increases then decreases, confirming a maximum at x ≈ 3.39.
Volume: V = 3.39 × (24 − 6.78) × (18 − 6.78) ≈ 3.39 × 17.22 × 11.22 ≈ 654.98 cm³. More precisely, with x = 7 − √13, V ≈ 654.98 cm³. The endpoints give V = 0, so this is indeed the largest possible volume.