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六年制重点中学高中课本《微积分初步》第145页例2是:“用边长为60厘米的正方形铁皮做一个无盖水箱,先在四角分别截去一个小正方形,然后把四边翻转九十度,再焊接而成(如图1)。问水箱底边的长应取多少,才能使水箱容积最大,最大容积是多少?”其结果为:“当水箱底边长取40厘米时,容积最大,最大容积是16000立方厘米。”容易看出,所求水箱底面正方形的边心距与高的比等于2;侧面积与底面积相等。有趣的是,这一结论对任意圆外切n边形仍然适合。命题将圆外切n边形的各边等宽地翻转90°(各角处的多余部分截去),折成一个直棱柱形的无盖水箱。如果它的容积最大,那么其底面边心距与高
The example of the six-year key middle school high school textbook “Preliminary Calculus” on page 145 is: “Use a square iron sheet with a side length of 60 cm to make a lidless water tank. Cut off a small square in each corner and turn the four sides over. Ten degrees, then welded (Figure 1). Ask the length of the bottom of the tank should be taken in order to maximize the capacity of the tank, the maximum volume is the number? “The result is: ”When the bottom of the tank to take 40 cm long, The largest volume, the largest volume is 16000 cubic centimeters." It is easy to see that the ratio of the center-to-center distance and height of the bottom square of the tank is equal to 2; the side area is equal to the bottom area. Interestingly, this conclusion still holds true for any circle with an outer n-edge. The proposition turns the edges of the circular outer-n-lateral shape to 90° (what is removed at each corner) at equal intervals, and folds it into a straight prismatic uncovered water tank. If its volume is the largest, then the bottom side of the heart and height