Lau 發問於 科學及數學數學 · 9 年 前

# Find the max area or rectangle

in the figure,a rectangle of dimensions x cm by y cm is inscribed in the right-angled triangle ABC.AB=4 cm and BC=8 cm.Find the maximum area of the rectangle.

### 3 個解答

• 蛙人
Lv 6
9 年 前
最佳解答

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

資料來源： me
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• 9 年 前

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.

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• 9 年 前

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

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