ce amaths-differentation
2007-09-07 4:47 am
The sum of the height h and the base radius r of a right ciircular cone is 15.
a ) Express the volume V of the cone in term of r. Write down the range of values of r.
b)Find the value of r such that the volume of the cone is maximum.
回答 (5)
2007-09-07 4:54 am
✔ 最佳答案
h + r = 15V = (1/3)(pi)(r^2)(h)
= (1/3)(pi)(r^2)(15 - r)
V > 0, 15 - r > 0, r < 15
range 0 < r < 15
dV/dr = (1/3)(pi)(30r - 3r^2) = (pi)(10r - r^2)
set dV/dr = 0, r = 10
you can check r = 10 makes volume of the cone maximum
2007-09-06 20:55:01 補充：
慢左, 不過..請多多支持
2007-09-06 20:55:10 補充：
慢左, 不過..請多多支持
2008-07-10 9:26 pm
雞尾包再一次拉票成功....
2008-06-28 8:22 pm
點解唔check埋佢係maximum呢? 另外兩位都有但你冇喎, 咁都叫得[so far so good]?!
2007-09-07 4:59 am
參考: Myself~~~
2007-09-07 4:53 am
(a) h+r = 15, i.e. h = 15 - r
V = 1/3 pi*r^2*h = pi*r^2(15-r)/3
Range of r is 0<15
(b) dV/dr = pi/3 * 30r-3r^2 = r*pi*(10-r) = 0
r = 0 or 10
d2V/dr2 = pi(10-2r)
d2V/dr2 at (r = 0) = 10pi>0
d2V/dr2 at (r = 10) = -10pi <0
Therefore the volume of the cone is maximum when r = 10
V = 1/3 pi*r^2*h = pi*r^2(15-r)/3
Range of r is 0<15
(b) dV/dr = pi/3 * 30r-3r^2 = r*pi*(10-r) = 0
r = 0 or 10
d2V/dr2 = pi(10-2r)
d2V/dr2 at (r = 0) = 10pi>0
d2V/dr2 at (r = 10) = -10pi <0
Therefore the volume of the cone is maximum when r = 10
參考: me
收錄日期: 2021-04-13 13:25:34
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