Re: [微積] 求極值
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: Sent from JPTT on my Asus PadFone T004.
7.
f(w) = [1 - 1/√w][5/w]
f'(w) = (5/2)1/(w^2 √w) - [1 - 1/√w][5/w^2]
= (5/w^2)[1/(2√w) - 1 + 1/√w]
= (5/w^2)[(3 - (2√w))/(2√w)]
=> w = 9/4
8.
按題意0 <= p <= 100
t = (p/100)^2 + (p/100) + 2
t' = 2p/10000 + 1/100
=> p = -50
最大值發生在p = 100
最小值發生在p = 0
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