標題:
Maximization
發問:
Area=(4/3)(k)(144-16k^2)^1/2 Find the maximum area and its corresponding value of k. Show workings,thz a lot 更新: By using derivative please
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最佳解答:
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
其他解答:
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|||||(4/3)(k)(144-16k^2)^1/2|||||(4/3)(k)(144-16k^2)^1/2 or [(4/3)(k)(144-16k^2)]^1/234CDDE348BEB263C
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