The lower corner of a leaf in a book is folded over so as to reach the inner edge of the page. Show that the fraction of the width folded over when the area of the folded part is minimum is `2//3.`

Explanation

Let the corner A of the leaf ABCD be folded over to A Which is one the inner edge BC of the page.
Let AP=x and AB=a. Therefore,BP=a-x

`If angle APB==theta then angleEA'F=theta.EF` is parallel to AB, then EF=a.
In `triangleA'BP,cos theta = BP/PA'=(a-x)/(x)`
In `triangleA'FE,A'E=EF cosec theta= a //sqrt(1-cos^(2)theta`
`=a//sqrt(1-({a-x})/({x}^(2))`
`=(ax)sqrt(x^(2))-(a-x)^(2)=(ax)/sqrt(2ax-a)^(2)=aE`
Triangle APe is folded to triangle A'PE .Therefore,
Area of folded part=Area of `triangle PAE=(1)/(2) AP.AE`
`=(1)/(2)x(ax)/sqrt(2ax-a^(2))`=A (say)
or `A^(2)=(a^(2)x^(4))/(4(2ax-a^(2)))=(a^(2)//4)/(2a//x^(3)-a^(2)//x^(4)=(a^(2)//4)/(y)`
where y=`(2a//x^(3))-a^(2)//x^(4),0ltxlta`
Obviously`A^(2)` (i.e,A) is minimum when y is maximum .Now,
`y=2a//x^(3)-a^(2)//x^(4)`
or `(dy)/(dx)=-(6a)/(x^(4))+(4a^(2))/(x^(5))and(d^(2)y)/(dx^(2))and (d^(2)y)/(dx^(2))=(24a)/(x^(5))-(20a^(2))/(x^(6))`
For maximum or mnimum of y,
`(d)/(dx)=-(6a)/(x^(4))+(4a^(2))/(x^(5))=0 or x=2a//3`
When `x= 2a//3,-(6a)/(x^(45))+(4a^(2))/(x^(5))=0 or x=2a//3`
when `x=2a//3,(d^(2)y)/(dx^(2))=(4a)/(x^(5))(6-(5a)/(x))`
`=4a(3)/(2a)^(5)6-5a(3)/(2a)=-4a(3)/(2a)^(5)(3)/(2)lt0`
Thus y is maximum i.e A is minimum when `x=2a//3` which is the only criticla point (least).Hence the folded area is minimum when `2//3` of the width of the page is folded over.