F.5 A.Maths

A=8z-2z^3
dA/dz = 8 - 6z^2.
Put dA/dz = 0 , we get 8=6z^2, so z= sqrt(4/3) = 2/sqrt3.
When z= 2/sqrt3, width of rectangle = 4/sqrt3.
Put z=x= 2/sqrt3 into y=4-x^2, we get y = 4- 4/3 = 8/3.
So area of rectangle = 4/sqrt3 x 8/3 = 32/3sqrt3 = 32sqrt3/9.

發問者選最佳答案的準則似乎是以先到先得, 所以我的答案還是刪了較好:
http://i238.photobucket.com/albums/ff245/chocolate328154/Maths330.jpg