Global Optimization of an Open-Top Box
In a calculus class, students are optimizing the volume of an open-top box made from a 20 cm by 30 cm rectangular sheet by cutting congruent squares of side x from each corner and folding up the sides. One student writes V = x(20 − 2x)(30 − 2x), tests x = 5 because it is halfway between 0 and 10, obtains V = 1000 cm³, and concludes this is the maximum. The teacher knows the true maximum occurs at an irrational critical point near 3.92 cm. The class has 40 minutes and graphing calculators. The teacher wants students to justify global optimality on a closed interval and to confront the 'midpoint' misconception without simply giving the formula. The student did not check endpoints or other critical points, and the physical interval is 0 < x < 10.
1. Find the exact critical point(s) of V on the physical interval and determine the maximum volume. Show the derivative and the interval check.
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V = x(20 − 2x)(30 − 2x). Expand: V = 600x − 100x² + 4x³. Differentiate: V′ = 600 − 200x + 12x². Set V′ = 0: 12x² − 200x + 600 = 0; divide by 4: 3x² − 50x + 150 = 0. Quadratic formula: x = (50 ± √(2500 − 1800))/6 = (50 ± √700)/6 = (25 ± 5√7)/3. Numerically, (25 − 13.229)/3 ≈ 3.924 and (25 + 13.229)/3 ≈ 12.743. Only 3.924 lies in (0, 10). Check endpoints: V(0) = 0 and V(10) = 10(0)(10) = 0. Since V is continuous on [0, 10], its maximum is at a critical point or endpoint; thus the global maximum is at x ≈ 3.924 cm. V ≈ 3.924 × 12.153 × 22.153 ≈ 1056 cm³, which is greater than the student's 1000 cm³ at x = 5.
2. Diagnose the student's reasoning error and explain why checking the midpoint is not a valid optimization strategy.
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The student assumes the midpoint of the interval is automatically optimal. That is not a theorem: the maximum of a differentiable function on a closed interval can occur at an endpoint or at a critical point, and the midpoint need not be either. Here V(5) = 1000, but V(3.924) ≈ 1056 and V(0) = V(10) = 0. A sign chart of V′ shows V′ > 0 before the critical point and V′ < 0 after it, so V increases then decreases; x = 5 is on the decreasing side, not at the peak. The valid strategy is to solve V′ = 0, discard roots outside the physical interval, and compare the remaining critical values with endpoint values.
3. Design a one-question exit ticket that would reveal whether students understand global versus local extrema in this context, and give the expected response.
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Exit ticket: 'A classmate claims that for this box the maximum must occur at x = 5 because 5 is halfway between 0 and 10. Without graphing, give one reason this reasoning is invalid and state the complete set of points that must be checked to justify a global maximum on the physical interval.' Expected response: the midpoint is not a guaranteed extremum; the maximum can occur at an endpoint or at a critical point where V′ = 0. The complete check is to find all critical points in (0, 10), evaluate V there, evaluate V at endpoints 0 and 10, and choose the largest value. A strong response may add that here V(0) = V(10) = 0 and the only interior critical point is x = (25 − 5√7)/3 ≈ 3.92, so the midpoint claim is false.