Find the maximum area and its corresponding value of k.
Show workings,thz a lot
更新1:
By using derivative please
Maximization
2011-03-23 1:09 am
Area=(4/3)(k)(144-16k^2)^1/2
Find the maximum area and its corresponding value of k.
Show workings,thz a lot
Find the maximum area and its corresponding value of k.
Show workings,thz a lot
回答 (4)
2011-03-23 9:32 am
✔ 最佳答案
A=(4/3)(k)(144-16k^2)^1/2(A>=0,∴ k >=0 )A=(4/3)(144k^2-16k^4)^1/2令 k^2 = t(4/3)(144t-16t^2)^1/2 (t>=0)-16t^2 + 144t= -16(t^2 -9t +81/4) +324when t=3 ,the max of ( -16t^2 + 144t )=324t=3, k=√3A=(4/3) √324=(4*18/3)=24
2011-03-23 2:29 am
A^2 = (16k^2/9)(144 - 16k^2)
dA^2/dk = (16k^2/9)(-32k) + (144-16k^2)(32k/9)
When dA^2/dk = 0 => k = 3/sqrt2
By first derivative test, max A^2 is attained when k = 3/sqrt 2 (this part done by urself)
Thus max A^2 = [16(3/sqrt2)^2/9][144 - 16(3/sqrt2)^2] = 8*72 = 576
max A = sqrt 576 = 24
dA^2/dk = (16k^2/9)(-32k) + (144-16k^2)(32k/9)
When dA^2/dk = 0 => k = 3/sqrt2
By first derivative test, max A^2 is attained when k = 3/sqrt 2 (this part done by urself)
Thus max A^2 = [16(3/sqrt2)^2/9][144 - 16(3/sqrt2)^2] = 8*72 = 576
max A = sqrt 576 = 24
2011-03-23 1:55 am
(4/3)(k)(144-16k^2)^1/2
2011-03-23 1:30 am
(4/3)(k)(144-16k^2)^1/2
or
[(4/3)(k)(144-16k^2)]^1/2
or
[(4/3)(k)(144-16k^2)]^1/2
收錄日期: 2021-04-25 17:07:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110322000051KK00648