Maximising Tray Volume on a Constrained Blank
A sheet-metal worker makes open-topped trays from rectangular blanks measuring 30 cm by 20 cm. A square of side x cm is cut from each corner and the flaps are folded up, giving a tray with base (30 − 2x) by (20 − 2x) and height x. The workshop wants the largest possible volume. The guillotine advances at a steady 0.4 cm per second along the corner cuts, so x increases with time during a run. A trainee has already cut one trial tray with x = 5 cm, producing a volume of 1000 cm³, and argues that this must be the maximum because 'the base is then 20 by 10, a tidy proportion'. The supervisor wants three things before the jig is set: the true optimum, evidence that it is a maximum rather than a minimum or an endpoint, and an estimate of how fast the volume is growing at the moment x = 3 cm.
1. Form the volume function, state its valid domain, and find all stationary points.
Read the worked answer
V(x) = x(30 − 2x)(20 − 2x).
Expand the quadratic first: (30 − 2x)(20 − 2x) = 600 − 100x + 4x². Hence
V = 4x³ − 100x² + 600x.
Domain: the base dimensions must be positive, so 30 − 2x > 0 and 20 − 2x > 0, giving 0 < x < 10. The stricter condition is the 20 cm side; x = 10 is outside the domain because the base would have zero width.
dV/dx = 12x² − 200x + 600 = 4(3x² − 50x + 150).
Setting 3x² − 50x + 150 = 0: discriminant = 2500 − 1800 = 700, √700 = 26.458.
x = (50 ± 26.458)/6 → x = 3.924 or x = 12.74.
Only x = 3.924 lies in (0, 10) - note it also satisfies x < 10, whereas 12.74 does not. So there is exactly one stationary point in the physical domain.
2. Show that x = 3.924 cm gives the largest volume, and compare it with the trainee's 1000 cm³ tray.
Read the worked answer
Second derivative: d²V/dx² = 24x − 200.
At x = 3.924: 24(3.924) − 200 = 94.2 − 200 = −105.8 < 0, so the stationary point is a local maximum.
A local test is not sufficient on its own, so close the argument with the domain. On 0 < x < 10, V is continuous and differentiable, and the only stationary point is x = 3.924. At the endpoints V → 0 (x → 0 flattens the box, x → 10 collapses the base). A continuous function on a closed interval attains its extremes either at stationary points or endpoints; since the endpoints give 0 and the interior stationary point gives a positive value, it is the global maximum.
V(3.924) = 4(3.924)³ − 100(3.924)² + 600(3.924) = 241.6 − 1539.5 + 2354.4 = 1056 cm³ (3 s.f.).
The trainee's tray is 1056 − 1000 = 56 cm³ smaller, about 5.3% below the optimum. The 'tidy proportion' argument is not a mathematical reason: the base 20 by 10 comes from x = 10/2, which treats the 20 cm side in isolation and ignores the 30 cm side and the height penalty. The optimum happens to sit near a quarter of the 20 cm side (x ≈ 3.92 against 5), which is why the trial was not disastrous - but coincidence of scale is not a maximisation argument.
3. Given that x increases at 0.4 cm/s, find the rate at which the volume is increasing when x = 3 cm, and comment on the sign of your answer relative to the maximum.
Read the worked answer
Use the chain rule: dV/dt = (dV/dx)(dx/dt).
At x = 3: dV/dx = 12(9) − 200(3) + 600 = 108 − 600 + 600 = 108 cm².
dx/dt = 0.4 cm/s, so dV/dt = 108 × 0.4 = 43.2 cm³/s.
The positive sign is consistent with x = 3 lying below the optimal 3.924 cm: volume is still rising as the cut deepens. The rate is also instructive in size - dV/dx is only 108 near x = 3, whereas at the optimum dV/dx = 0; the volume gains are large early and vanish at the turning point, so overshooting the cut is far more costly than undershooting it. That asymmetry is the practical reason to advance the guillotine slowly near x = 3.9, not at 0.4 cm/s.