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Math Help - geometrical applications of differentiation 2

  1. #1
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    geometrical applications of differentiation 2

    hello again everyone ><>< sorry but i need a bit more help

    ABCD is a square of unit length and points Eand F are taken on the sides AB and AD respectively such that AE = AF = x

    a) Express the area of the Quadrilateral CDFE as a function of x
    b) Find the greatest area the quadrilateral can have

    and if its not too much trouble can you please draw a diagram it would be most helpful
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let EG\perp CD, \ G\in CD
    Then A_{CDFE}=A_{GDFE}+A_{CGE}.
    \displaystyle A_{GDFE}=A_{GDAE}-A_{AEF}=x-\frac{x^2}{2}.
    \displaystyle A_{CGE}=\frac{CG\cdot EG}{2}=\frac{1-x}{2}.
    Then \displaystyle A_{CDFE}=\frac{-x^2+x+1}{2}=f(x)

    \displaystyle f'(x)=\frac{-2x+1}{2}
    \displaystyle f'(x)=0\Rightarrow x=\frac{1}{2}, which is the point of maximum.
    The greatest area is \displaystyle f\left(\frac{1}{2}\right)=\frac{5}{8}
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  3. #3
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    Quote Originally Posted by chibiusagi View Post
    hello again everyone ><>< sorry but i need a bit more help

    ABCD is a square of unit length and points Eand F are taken on the sides AB and AD respectively such that AE = AF = x

    a) Express the area of the Quadrilateral CDFE as a function of x
    b) Find the greatest area the quadrilateral can have

    and if its not too much trouble can you please draw a diagram it would be most helpful
    A square of unit length. Meaning, ABCD is 1 by 1.

    Area of CDFE, K = (area of ABCD) minus (area of rigtht triangle EAF) minus (area of right triangle EBC)

    K = 1*1 -[(1/2)(x)(x)] -[(1/2)(1-x)(1)]
    K = 1 -(1/2)(x^2) -1/2 +(1/2)(x)
    K = 1/2 +(1/2)(x) -(1/2)(x^2)
    K = (1/2)(1 +x -x^2) ---------------the area of CDFE.

    Greatest K is when dK/dx = 0, so,
    dK/dx = (1/2)(1 -2x)
    Set that to zero,
    0 = 1 -2x
    2x = 1
    x = 1/2
    Hence, greatest K = (1/2)[1 +1/2 -(1/2)^2] = (1/2)[1 +1/4] = 1/2 +1/8 = 5/8 sq.units ----answer.
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  4. #4
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    Quote Originally Posted by chibiusagi View Post
    hello again everyone ><>< sorry but i need a bit more help

    ABCD is a square of unit length and points Eand F are taken on the sides AB and AD respectively such that AE = AF = x

    a) Express the area of the Quadrilateral CDFE as a function of x
    b) Find the greatest area the quadrilateral can have

    and if its not too much trouble can you please draw a diagram it would be most helpful
    Hello,

    to a)

    The area of the quadrilateral is calculated by:
    A_{square} - A_{triangleI} - A_{triangleII}

    Therefore:

    A(x) = 1^2 - \frac12 \cdot x^2 - \frac12 \cdot 1 \cdot (1-x)=-\frac12 \cdot x^2+\frac12 \cdot x + \frac12

    to b)
    You'll get the extreme value of this function (minimum or maximum) if the 1st drivative equals zero:

    A'(x) = -x+\frac12

    I'll leave the rest for you.
    Attached Thumbnails Attached Thumbnails geometrical applications of differentiation 2-area_quadlat.gif  
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