中4既數學問題 有１０點

1)
a)
From your figure,
In triangleABC, tan A=12/8=3/2
In triangleAFE, tan A=EF/AF=EF/(8-x)

Hence, EF/(8-x)=3/2
...................EF=3(8-x)/2

b) Area of BDEF= length*width=3x(8-x)/2

c) from b), Area=3x(8-x)/2
.......................=-3(x^2-8x)/2
.......................=[-3(x^2-8x+16)+48]/2
.......................=[-3(x-4)^2+48]/2

Hence, when x=4, it attains a max.area, and max. area is 24cm^2

Dimension of area = length*width
...........................= x*3(8-x)/2
hence, dimension of max. area = 4*6

2)
Let x and y be the length and width of playground respectively.
therefor, x+2y=60

Area of playground=xy
...........................=y(60-2y)
...........................=-2(y^2-30y)
...........................=-2(y^2-30y+225)+500
...........................=-2(y-15)^2+500

hence, when y=15cm, it attains its max. of the area of playground; and max area of playround=500cm^2

c) Dimension = x*y
....................=(60-2y)*y
Hence, Dimension of max area =30*15 cm

2007-10-18 14:28:56 補充：
sorry for my typo.Area of playground should be-2(y^2-30y+225)+450=-2(y-15)^2+450Hence, max. area is 450m^2, not 500m^2

１：
a) (8-x) tan(12/8)
b) x(8-x) tan(4/3)
c)
A = x(8-x) tan(4/3)
dA/dx= (8-2x)tan(4/3)
max.Area =>>dA/dx=0
dA/dx= (8-2x)tan(4/3)=0
=>> x=4
=>> A=16 tan(4/3)
２：唔明