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F.4 AM Coordinate System (15分)

發問:

Given a triangle, which area = ((4r+2)/((r+1)^2)), where r>0 find maximum area & minimum area. 更新: 唔用微積分得唔得? 更新 2: 我都唔知-,- 期終考試最後果一題...

最佳解答:

• f(x) = ln(3x^2 + 5), f'(1)=-
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• 問BIO 2001 CE MC Q.54 and Q.56@1@

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Let A is triangle area A=(4r+2)/(r+1)^2 dA/dr=(r+1)^2(4)-(4r+2)(2)(r+1)/(r+1)^4 dA/dr=4(r^2+2r+1)-2(4r^2+4r+2r+2) dA/dr=(r+1)[(4(r+1)-2(4r+2)]/(r+1)^4 dA/dr=(r+1)(4r+4-8r-4)/(r+1)^4 dA/dr=(-4r)/(r+1)^3 dA/dr=0 (-4r)=0 r=0 又說r唔可以等於0 anyway d^2A/dr^2=[(r+1)^3(-4)-(-4r)(3)(r+1)^2]/(r+1)^6 d^2A/dr^2=(r+1)^2[(-4)(r+1)-(-4r)(3)]/(r+1)^6 d^2A/dr^2=(8r-4)/(r+1)^4 when r=0 d^2A/dr^2=[8(0)-4]/(0+1)^4 d^2A/dr^2 is less than 0 It has a maximum area [4(0)+2]/(0+1)^2 =2 2008-06-18 15:11:46 補充： What is the answer????? 2008-06-18 17:04:29 補充： sorry, the minimum area is 2

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