Open-box optimisation with an inadmissible root
A manufacturer cuts squares of side x cm from each corner of a rectangular sheet measuring 24 cm by 18 cm, then folds the flaps to make an open box. The volume is V(x)=x(24-2x)(18-2x). A student expands this to V=4x^3-84x^2+432x, differentiates to get V'=12x^2-168x+432, sets V'=0, and solves x^2-14x+36=0, giving x=7 plus or minus sqrt(13), so x≈3.39 or x≈10.61. The student writes: 'Both roots are positive, so both are valid. I will take x=10.61 because it gives the larger box.' The workshop needs the dimensions that maximise the volume and a justification that no other cut size gives a larger volume. The student has not considered the physical restrictions on x or the endpoints of the interval.
1. Determine the physically admissible interval for x and explain why x=7+sqrt(13) must be rejected. Then find the cut size that maximises the volume and the maximum volume, showing that it is a global maximum on the admissible interval.
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The physical restrictions are x>0, 24-2x>0 and 18-2x>0, so 0<x<9. The root x=7+sqrt(13)≈10.61 lies outside this interval and would make the base dimensions negative, so it must be rejected. On (0,9), V'(x)=12(x^2-14x+36)=12(x-(7-sqrt13))(x-(7+sqrt13)). Because the second factor is negative throughout 0<x<9, the sign of V' is positive for x<7-sqrt13 and negative for x>7-sqrt13. Hence V increases then decreases, with a maximum at x=7-sqrt13≈3.39 cm. The maximum volume is V≈3.394(24-6.789)(18-6.789)≈655 cm^3. This is global on the closed interval [0,9]: V is continuous, V=0 at both endpoints, and the only interior critical point is the maximum.
2. A colleague suggests that showing V''(7-sqrt13)<0 is sufficient to guarantee the global maximum. Discuss what V'' does and does not prove, and state the extra check needed.
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V''(7-sqrt13)=24(7-sqrt13)-168≈-86.5<0, so the critical point is a local maximum. The second derivative alone does not prove a global maximum; it only describes the shape near that point. To guarantee global maximality, compare with the endpoints and check whether any other critical point lies in the admissible interval. Here V(0)=V(9)=0, and 7+sqrt13 is outside 0<x<9, so the local maximum is the only interior stationary point. Since V is continuous on the closed interval [0,9], it must attain its global maximum there; with endpoint values 0 and the interior value positive, the local maximum is global. Thus the second-derivative check must be paired with a domain and endpoint argument.