S = (x-8)(432/x-6) = 480 - 6x - 3456/x
The only stagnation point of s' =-6+3456/x 2 = 6 (576-x 2)/x 2 is x = 24.
Because of practical problems, the only stagnation point is the maximum point, which is the maximum point.
At this time, the width is 24 and the height is 18, so the paper is the least.
2. If the height is h, the bottom area of the cylinder is s = π r 2 = π (r 2-h 2/4).
V = hs = π (hr 2-h 3/4), V v' = π (r 2-3h 2/4/4), and the unique stagnation point h = 2R/√3.
Because of practical problems, the only stagnation point is the maximum point, which is the maximum point.
At this time, the height is 2R/√3, and the maximum volume VMAX = 4 π r 3/(3 √ 3).