F4 Maths ^^－航空百大

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F4 Maths ^^

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http://img121.imageshack.us/content.php?page=done&l=img121/8574/cardboard.png&via=muploadFig. I shows a sheet of rectangular cardboard PQRS. A square of side 5 cm is cut away from each corner of the cardboard. The perimeter of the remaining part is 150 cm. Then the remaining part is... 顯示更多 http://img121.imageshack.us/content.php?page=done&l=img121/8574/cardboard.png&via=mupload Fig. I shows a sheet of rectangular cardboard PQRS. A square of side 5 cm is cut away from each corner of the cardboard. The perimeter of the remaining part is 150 cm. Then the remaining part is folded up to make an open box as shown in Fig. II. Let x cm be the length of the box. a) Express the width of the box in terms of x. Hence, express the capacity of the box in terms of x. Ans: width = 55-x cm, capacity = 275x - 5x^2 b) Find the length, width and height of the box so that its capacity is a maximum. Ans: length = 27.5 cm, width = 27.5 cm, height = 5 cm

最佳解答:

(a) The perimeter = 150 The length of the original cardboard = x + 10 The width of the original cardboard = 150/2 - (x+10) = 65 - x The width of the box = 65 - x - 10 = (55 - x) cm The capacity of the box V = x(55 - x)(5) = (275x - 5x^2) cm^3 (b) V = -5x^2 - 275x V = -5(x^2 - 55x + 756.25) + 3781.25 V = -5(x - 27.5)^2 + 3781.25 The capacity is maximum when x = 27.5 Length = 27.5 cm Width = 55 - 27.5 = 27.5 cm Height = 5 cm

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