Question

The lower corner of a leaf in a book is folded over so as to just reach the inner edge of the page. The fraction of width folded over if the area of the folded part is minimum is:

• A. $5/8$
• B. $2/3$
• C. $3/4$
• D. $4/5$

Answer 1

The correct option is B $2/3$

Hence, statement first is wrong, but second is true.

Let $AB=a$, $AP=x$

and $\angle A^{1}PB=\theta =\angle FBE$

In $△A^{1}BP$

$\cos \theta =\frac{a-x}{x}$

In $△A^{1}FE$

$A^{1}E=EF\cos \theta$

$A^{1}E=EF\times \frac{1}{sin\theta }=\frac{EF}{\sqrt{1-cos\theta }}=\frac{a}{\sqrt{1-{\left(\frac{a-x}{x}\right)}^2}}$

$A^{1}E=\frac{ax}{\sqrt{2ax-a^2}}=AE$

Area of folded part $=$ Area of $△PAE$

$=\frac{1}{2}\times AP\times AE$

$△=\frac{1}{2}\times x\times \frac{ax}{\sqrt{2ax-a^2}}$

${△}^2=\frac{1}{4}\frac{a^2{x}^4}{(2ax-a^2)}=\frac{a^2}{4(\frac{2a}{x^3}-\frac{a^2}{x^4})}$

${△}^2=\frac{a^2/4}{y}$

where, $y=\frac{2a}{x^3}-\frac{a^2}{x^4}$

For minimum area, y should be maximum.

Now, $\frac{dy}{dx}=\frac{-6a}{x^4}+\frac{{4a}^2}{x^5}$

$\frac{d^{2}y}{{dx}^2}=\frac{24a}{x^5}-\frac{{20a}^2}{x^6}$

$\frac{d^{2}y}{{dx}^2}=0$

$\frac{24a}{x^5}-\frac{20a^2}{x^6}=0\Rightarrow x=\frac{2a}{3}$

At $x=\frac{2a}{3}$, y is maximum.

$\therefore a\to$base of page

Hence 2/3 width is folded.