Open-Top Box from 30 cm by 20 cm Cardboard with a Volume Promise
A technology club is making open-top storage boxes from rectangular cardboard sheets measuring 30 cm by 20 cm. From each corner they cut out a square of side x cm, then fold up the four flaps. The folding machine cannot handle squares smaller than 3 cm, and the cut must leave at least 2 cm of flap height, so 3 ≤ x ≤ 10. The club has promised boxes with a volume of at least 1000 cm³ for a science-kit fundraiser. One member proposes x = 4 cm because 'it is close to the maximum' found in a calculus exercise, while another wants x = 3 cm to save cardboard. The treasurer asks for the exact interval of allowed x values, the true maximum volume, and how sensitive the volume is to a 1 mm change in x near the lower limit.
1. Write the volume function, find its stationary points on 3 ≤ x ≤ 10, and identify the maximum volume.
Read the worked answer
After cutting squares of side x, the base is (30 − 2x) by (20 − 2x) and the height is x, so V(x) = x(30 − 2x)(20 − 2x) = 4x³ − 100x² + 600x. Differentiate: V′(x) = 12x² − 200x + 600 = 4(3x² − 50x + 150). Setting 3x² − 50x + 150 = 0 gives x = [50 ± √(2500 − 1800)]/6 = (25 ± 5√7)/3. Numerically, 5√7 ≈ 13.229, so the roots are about 3.924 and 12.743. Only 3.924 lies in [3, 10]. Check endpoints and the stationary point: V(3) = 4(27) − 100(9) + 1800 = 1008; V(10) = 0; V(3.924) ≈ 4(60.41) − 100(15.40) + 600(3.924) ≈ 241.6 − 1540 + 2354 ≈ 1056 cm³. So the maximum is about 1056 cm³ at x ≈ 3.92 cm, and the proposed x = 4 cm gives V = 256 − 1600 + 2400 = 1056 cm³, essentially the maximum.
2. Find the exact interval of x in [3, 10] for which V(x) ≥ 1000, and evaluate the two proposals.
Read the worked answer
Solve 4x³ − 100x² + 600x = 1000, i.e. x³ − 25x² + 150x − 250 = 0. Testing x = 5 gives 125 − 625 + 750 − 250 = 0, so x − 5 is a factor: x³ − 25x² + 150x − 250 = (x − 5)(x² − 20x + 50). The quadratic roots are x = 10 ± √50 = 10 ± 5√2, approximately 2.929 and 17.071. The cubic has a positive leading coefficient, so V ≥ 1000 between the two smaller roots, i.e. 2.929 ≤ x ≤ 5, and V < 1000 beyond 5 until the far root. Intersecting with 3 ≤ x ≤ 10 gives 3 ≤ x ≤ 5. Thus x = 4 is safely inside, while x = 3 is exactly at the endpoint, since V(3) = 1008 > 1000. The second member's suggestion is allowed but leaves almost no margin for cutting error.
3. Interpret V′(3) and V′(4), and use them to advise which proposal is safer for meeting the 1000 cm³ promise.
Read the worked answer
V′(x) = 12x² − 200x + 600. At x = 3: V′(3) = 108 − 600 + 600 = 108, so near x = 3 each extra 1 cm of cut side adds about 108 cm³; a 1 mm error changes the volume by roughly 10.8 cm³. Since V(3) = 1008, a cut slightly under 3 mm too deep (toward x < 3) would push the volume below 1000, and x < 3 is not even permitted by the machine. At x = 4: V′(4) = 192 − 800 + 600 = −8, so a 1 mm change alters the volume by under 1 cm³, and V(4) ≈ 1056 leaves about 56 cm³ of margin. The derivative also changes sign between 3.92 and 4, confirming x = 4 sits essentially at the maximum. Recommendation: choose x = 4 cm; it maximises volume in practice and is robust to small cutting errors, whereas x = 3 cm is a boundary value with a tiny margin.