One corner of a long rectangular sheet of paper of width 1 unit is folded over so as to reach the opposite edge of the sheet. The minimum length of the crease is

Let $ABCD$  be the rectangular sheet whose corner C is folded over along  $EF$  so as to reach the edge $AB$ at $C\text{'}$.

Let $EF=x$    $\therefore \mathrm{\angle }FEC=\theta =\mathrm{\angle }FEC$ '  $\mathrm{△}BE{C}^{\mathrm{\prime }}$ we have  $BE=C^{\mathrm{\prime }}E\cos (\pi -2\theta )$

$\Rightarrow BE=-x\cos \theta \cdot \cos 2\theta$

$\therefore BC=BE+EC$

$1=-x\cos \theta \cos 2\theta +x\cos \theta$

$\Rightarrow x=\frac{1}{\cos (1-\cos 2\theta )}$........(1)

$x$  to be minimum , Z has to be maximum i.e

$Z=\frac{1}{x}=\cos \theta (1-\cos 2\theta )$....(2)

Calculate   $\frac{\mathrm{d}Z}{\mathrm{d}\theta }=0$  and check the sign of   $\frac{{\mathrm{d}}^2\mathrm{Z}}{\mathrm{d}{\theta }^2}$

Z is maximum  $\Rightarrow x=\frac{1}{Z}$  is minimum  

$\therefore$ x is minimum

$\Rightarrow minx\Rightarrow \frac{1}{Z}=\frac{1}{(1+\sqrt{3})(1+\frac{1}{3})}=\frac{3\sqrt{3}}{4}\text{unit}$  .