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LV超3A名牌購物網
第2題仲未諗到,答住第一題先。諗到補充上去。
比best我plx!!! 等分用
1a)
q=6x-x^2 => x = 3±√(9-q) ,then sub. q = 6p-p^2
x = 3±√(9+p^2-6p)
= p or 6-p
A: (p,q) D: (6-p,q) CD=q
The Area is: (6-p-p)*q
= 2(3-p)(6p-p^2)
= 2p(3-p)(6-p)
= 2p^3-18p^2+36p
1b)
A = 2p^3-18p^2+36p
dA/dt = (6p^2-36p+36)*dp/dt [given that dp/dt is a positive constant]
= 6(p-3-√3)(p-3+√3)*dp/dt
之後你畫表(姐係書教,take dA/dt = 0 , 搵左p出黎,再代番d數落個表到,你就會見到幾時係 >1 同 <1)
參考資料
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