Lau 發問於 科學及數學數學 · 9 年 前

Find the max area or rectangle

圖片參考:http://imgcld.yimg.com/8/n/HA00837720/o/7011010200...

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

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  • 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
  • 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.

  • 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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