Constrained Box Optimization
An open-top box is made from a 60 cm by 40 cm rectangular sheet by cutting a square of side x cm from each corner and folding up the sides. The finished box must have height between 8 cm and 15 cm inclusive, and the base must remain rectangular with positive dimensions. The sheet thickness and fold allowance are negligible, and no lid is added. A student writes the volume as V(x)=x(60−2x)(40−2x), differentiates, solves V′(x)=0, and obtains x≈7.85 cm as the only critical point in 0<x<20. The student then reports that the maximum volume occurs at x≈7.85 cm. The workshop supervisor points out that this value is outside the required height range and asks for the correct maximum under the stated constraints. The supervisor also wants a written justification that distinguishes the unconstrained critical point from the constrained optimum.
1. Determine the admissible interval for x and decide whether x≈7.85 cm can be the constrained maximum. Show the inequality that eliminates it.
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Admissible x: 0<60−2x and 0<40−2x give x<20; the height constraint gives 8≤x≤15. Intersection: 8≤x≤15. The value 7.85 is below 8, so it is not admissible; a constraint cannot be ignored after differentiation. V′(x)=2400−400x+12x². On [8,15], V′(8)=−32<0, and the other root is 25.5, so V′ is negative throughout the admissible interval. Hence V decreases on [8,15], and the constrained maximum is at x=8. Base: 60−16=44 cm by 40−16=24 cm; V=8×44×24=8448 cm³.
2. A classmate argues that because V′(7.85)=0, the volume at x=8 must be lower than the unconstrained maximum but could still be the constrained maximum. Explain the logic and compute the volume at x=8.
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V′(7.85)=0 means 7.85 is the unconstrained stationary point; it says nothing about feasibility. Since V decreases for x>7.85, moving from 7.85 to 8 lowers volume only slightly, but x=8 is the smallest admissible height. V(7.85)≈7.847×44.306×24.306≈8450 cm³; V(8)=8448 cm³. The constrained maximum is therefore the boundary value 8448 cm³ at x=8. The difference is small, but the answer x≈7.85 is outside the stated height range.
3. If the height constraint were removed but the base perimeter had to be at most 130 cm, find whether the critical point remains feasible and state the new maximum.
Read the worked answer
Base perimeter P=2[(60−2x)+(40−2x)]=200−8x. Requiring P≤130 gives 200−8x≤130, so x≥70/8=8.75. With x<20 and positive base, the admissible interval is [8.75,20). The critical point 7.85 is again infeasible, and V is decreasing on this interval, so the new maximum is at x=8.75. Base: 60−17.5=42.5 cm, 40−17.5=22.5 cm; V=8.75×42.5×22.5=8367.1875≈8.37×10³ cm³.