# Thread: geometrical applications of differentiation 2

1. ## 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

2. 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 $\displaystyle f\left(\frac{1}{2}\right)=\frac{5}{8}$

3. Originally Posted by chibiusagi
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.

4. Originally Posted by chibiusagi
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.