(1)解 ∵PN // BC, AD⊥BC
∴AD⊥PN于F
∵PN // BC
∴△ABC∽△APN,PQ = FD
∴AF/AD = PN/BC,AF = AD - FD = 8 - x
∴(8 - x)/8 = PN/12
∴PN=(3/2)· (8 - x)
(2)y =PN·PQ=(3/2)· (8 - x)x
(3) y = (-3/2)(x² - 8x)=(-3/2)(x - 4)² + 24
∴当x =4时 面积有最大值24,此时F为AD中点,∴P为AB中点
(1)解 ∵PN // BC, AD⊥BC
∴AD⊥PN于F
∵PN // BC
∴△ABC∽△APN,PQ = FD
∴AF/AD = PN/BC,AF = AD - FD = 8 - x
∴(8 - x)/8 = PN/12
∴PN=(3/2)· (8 - x)
(2)y =PN·PQ=(3/2)· (8 - x)x
(3) y = (-3/2)(x² - 8x)=(-3/2)(x - 4)² + 24
∴当x =4时 面积有最大值24,此时F为AD中点,∴P为AB中点