Open-Top Box Optimisation with Machine Increments
A packaging workshop makes open-top boxes from rectangular sheets of card measuring 24 cm by 18 cm. From each corner a square of side x cm is cut out, and the remaining flaps are folded up. The workshop wants the box with the largest possible volume, but the cutting machine can remove squares only with 0.5 cm increments and cannot cut a square smaller than 1 cm or larger than 8 cm. A client asks for a box with volume at least 500 cm³. The supervisor sketches the volume as a cubic and claims the maximum occurs at x = 4 because “the derivative is zero there.” A trainee notices that the graph may have another stationary point and that the machine’s 0.5 cm restriction could change the answer. The workshop needs a defensible recommendation, including whether the 500 cm³ target is achievable under the machine’s constraints.
1. Derive V(x), state its domain under the machine’s constraints, and find all stationary points. Check the supervisor’s claim that x = 4 is a stationary point.
Read the worked answer
After cutting squares of side x, the base dimensions are (24-2x) and (18-2x), and the height is x. Thus [V(x)=x(24-2x)(18-2x).]
Expanding: [V(x)=x(432-84x+4x^2)=4x^3-84x^2+432x.]
The machine allows (x\in[1,8]) in 0.5 cm steps.
Differentiate: [V'(x)=12x^2-168x+432=12(x^2-14x+36).]
Set (V'(x)=0): [x^2-14x+36=0 \Rightarrow x=7\pm\sqrt{13}.]
Since (\sqrt{13}\approx3.606), the roots are about 3.394 and 10.606. Only (x=7-\sqrt{13}\approx3.394) lies in [1,8].
At x = 4, [V'(4)=12(16-56+36)=12(-4)=-48\ne0.] So the supervisor’s claim is false: x = 4 is not a stationary point.
2. Find the volume-maximising cut size under the 0.5 cm machine increment and give the maximum volume.
Read the worked answer
The continuous maximum is near (x=3.394), so check the allowed 0.5 cm values around it:
[V(3)=3(18)(12)=648,] [V(3.5)=3.5(17)(11)=654.5,] [V(4)=4(16)(10)=640.]
The endpoints are much smaller: (V(1)=352) and (V(8)=128). Therefore the best permitted cut is [x=3.5\ \text{cm},] with maximum volume [V(3.5)=654.5\ \text{cm}^3.]
The 0.5 cm restriction means the workshop cannot use the exact calculus maximiser 3.394 cm, but 3.5 cm is the closest allowed value and gives the largest permitted volume.
3. Determine whether the client’s 500 cm³ target is achievable under the machine’s constraints, and give a cut size that meets it.
Read the worked answer
Check allowed values near where the volume crosses 500 cm³. On the increasing side: [V(1.5)=1.5(21)(15)=472.5<500,] [V(2)=2(20)(14)=560>500.] So x = 2 cm already meets the target.
On the decreasing side, check: [V(5.5)=5.5(13)(7)=500.5\ge500,] [V(6)=6(12)(6)=432<500.] So x = 5.5 cm also meets the target.
Therefore the 500 cm³ target is achievable. A simple choice is (x=2) cm, giving 560 cm³. If the client wants the largest possible box, the workshop should still recommend (x=3.5) cm with 654.5 cm³.