Answer to Question #271512 in Calculus for Emma

Question

The drawing below shows a square with side a. A straight line intersects the square and encloses

an area A. The heights x and y on the left and right side (in a distance d from the square) of

the intersecting line can be varied. Assuming that x  y and x; y  a, nd an expression for

the enclosed area A(x; y) with respect to x and y.

Answer 1

Area of the trapezoid $BCDE$

$A=\dfrac{BE+CD}{2}\cdot ED$ $\dfrac{BM}{FM}=\dfrac{CN}{FN}=\dfrac{GK}{FK}$

Substitute

$\dfrac{BE-x}{d}=\dfrac{CD-x}{a+d}=\dfrac{y-x}{a+2d}$ $d(BE-x)+a(BE-x)=d(CD-x)$ $d=\dfrac{a(BE-x)}{CD-BE}$ $a+d=\dfrac{a(CD-BE+BE-x)}{CD-BE}=\dfrac{a(CD-x)}{CD-BE}$ $a+2d=\dfrac{a(CD-BE+2BE-2x)}{CD-BE}$ $=\dfrac{a(CD+BE-2x)}{CD-BE}$ $\dfrac{BE-x}{d}=\dfrac{CD-BE}{a}$ $\dfrac{y-x}{a+2d}=\dfrac{(y-x)(CD-BE)}{a(CD+BE-2x)}$

Then

$\dfrac{CD-BE}{a}=\dfrac{(y-x)(CD-BE)}{a(CD+BE-2x)}$ $CD+BE-2x=y-x$ $CD+BE-2x=y+x$

Area of the trapezoid $BCDE$

$A(x, y)=\dfrac{x+y}{2}\cdot a \ \ ({units}^2)$