Find the max area or rectangle

Using similar triangle,
y/8 = (4-x)/4
y = 8 - 2x

So area of rectangle = xy
= x(8 - 2x)
= -2(x^2 - 4x)
= -2(x^2 - 4x + 4 - 4)
= -2(x - 2)^2 + 8
Thus the maximum area is 8cm^2

The key point is to find out the relation between x and y.

Since triangle ADE is similar to the triangle DCF

DE/CF=AE/CF

x/(8-x)=(4-y)/y
xy=(8-x)(4-y)
xy=32-4x-8y+xy
4x+8y=32
x+2y=8
x=8-2y

Now the Area of the rectangle is xy
=y(8-2y)
=-2y^2+8y
=-(y-4)^2+16

which attains the maximum value 16 when y=4. So, the maximum area of the rectangle is 16 cm^2.

2011-01-02 17:24:33 補充：
Some mistakes Sorry

=y(8-2y)
=-2y^2+8y
=-2(y-4)^2+8

which attains the maximum value 8 when y=4. So, the maximum area of the rectangle is 8 cm^2.

Sol
(4－x)：4=y：8
4y=8(4－x)
y=2(4－x)
y=8－2x
xy=x(8－2x)
=－2x^2+8x
=－2(x^2－4x)
=－2(x^2－4x+4)+8
=－2(x－2)^2+8
0<4
－2<2
0<=(x－2)^2<4
－4<－(x－2)^2<=0
－8<=－2(x－2)^2<=0
0<－2(x－2)^2+8<=8
０<=8
max=8