有一正圓錐體,其體積是18πcm^3.設L,H分別是圓錐體的斜高和高度.
a)證明L^2=H^2+54/H
b)由此,求當H值變化時,L的極小值
b答案是3乘開方3cm
π=圓周率
f5 maths微分(聽日要交功課架!!!!!!!!!!
2006-11-22 2:46 am
回答 (1)
2006-11-22 2:56 am
✔ 最佳答案
a)Since L^2=H^2+R^2
1/3πR^2H=18π
R^2=54/H
SO
L^2=H^2+54/H
b)
L^2=H^2+54/H
2L(dL/dH)=2H-54/H^2
let dL/dH=0
2H-54/H^2=0
H=3
So when H=3 , L gets it minimum
the minimum value
=√(H^2+54/H)
=√(3^2+54/3)
=√27
=3√3 cm
收錄日期: 2021-04-23 18:54:29
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