f.4 mathz,,,
2008-10-22 2:27 am
An open box without cover is made of wooden board 2 cm thick. The external edges of the box are all equal. If the volume of wood used is 8072 cm^3, find the length of an external edge.
回答 (2)
2008-10-30 9:14 am
✔ 最佳答案
We can consider such a box as a larger box [x cm by x cm by x cm] substract a smaller box [(x-2)cm by (x-4)cm by (x-4)cm]Hence,
(the volume of wood used) = (the vloume of the bigger box) - (the vloume of the smaaler box)
8072 = x^3 - (x-2)(x-4)^2
8072 = x^3 - (x-2)(x^2-8x+16)
8072 = x^3 - (x^3-8x^2+16x-2x^2+16x-32)
10x^2 - 32x- 8040 = 0
2(5x^2 - 16x- 4020) = 0
2(5x+134)(x-30)=0
x=30, -26.8(rejected as -ve)
So the length of the external edge is 30cm
參考: Own knowledge
2008-10-22 8:35 am
Let length = x cm,
x^3 - (x-2)(x-4)^2 = 8072
x^3 - (x-2)(x^2-8x+16) = 8072
x^3 - (x^3-8x^2+16x-2x^2+16x-32) = 8072
10x^2 - 32x- 8040 = 0
x = (32+(32^2-4*10*(-8040))^0.5)/(2*10) or (32-(32^2-4*10*(-8040))^0.5)/(2*10)
= 30 or -26.8 (rej.)
x^3 - (x-2)(x-4)^2 = 8072
x^3 - (x-2)(x^2-8x+16) = 8072
x^3 - (x^3-8x^2+16x-2x^2+16x-32) = 8072
10x^2 - 32x- 8040 = 0
x = (32+(32^2-4*10*(-8040))^0.5)/(2*10) or (32-(32^2-4*10*(-8040))^0.5)/(2*10)
= 30 or -26.8 (rej.)
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