Question

One corner of a long rectangular sheet of paper of width $1$ feet is folded over, so as to reach the opposite edge of the sheet. Then the minimum length of the crease is...
A. $\dfrac{{3\sqrt 3 }}{2}$
B. $\dfrac{{\sqrt 3 }}{4}$
C. $\dfrac{{3\sqrt 3 }}{4}$
D. $\dfrac{{\sqrt 3 }}{2}$

Answer 1

Hint: Draw the picture of the given situation and then use the Pythagoras theorem to find the value of length.
* Pythagoras Theorem: In the right angled triangle hypotenuse square is equal to the sum of square of other sides.

Formula used: * Double angle formula: $\cos (2\theta ) = {\cos ^2}\theta - {\sin ^2}\theta .$
* Chain rule of derivative: If$y = f(g(x)) \Rightarrow \dfrac{{dy}}{{dx}} = f'(g(x))g'(x)$ .
* Quotient rule of derivative: If$y = \dfrac{f}{g} \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{g \cdot f' - f \cdot g'}}{{{g^2}}}$.

Complete step-by-step answer:
Draw the picture of rectangular paper $ABCD$ such that length of side $AB = 1$ as shown below:
          

Write the information from the figure.
Here $c$ is the length of the crease.
Since the rectangle has a right angle at the corner, so the folded triangle has one angle as the right angle, one angle as \[\theta \] and one angle as \[\alpha \].
Since we know sum of all angles of triangle is \[{180^ \circ }\]
 \[
   \Rightarrow \theta + \alpha + {90^ \circ } = {180^ \circ } \\
   \Rightarrow \theta + \alpha = {180^ \circ } - {90^ \circ } \\
   \Rightarrow \theta + \alpha = {90^ \circ } \\
   \Rightarrow \alpha = {90^ \circ } - \theta \\
 \]
And since \[\beta , \theta , \theta \] lie on a straight line they are supplementary,
\[
   \Rightarrow \beta + \theta + \theta = {180^ \circ } \\
   \Rightarrow \beta + 2\theta = {180^ \circ } \\
   \Rightarrow \beta = {180^ \circ } - 2\theta \\
 \]
From the figure we have,
$\alpha = 90 - \theta , \beta = 180 - 2\theta .$
Write the value of $\beta $ in terms of $\alpha .$
\[ \Rightarrow \beta = {180^ \circ } - 2\theta \]
Taking 2 common
\[ \Rightarrow \beta = 2(90 - \theta ) = 2\alpha .\]
Also, in a triangle formed in the left corner, the cos of an angle is base divided by hypotenuse.
$ \Rightarrow \cos \beta = \dfrac{{1 - x}}{x}$
Substitute value of $\beta = 2\alpha .$
$ \Rightarrow \cos (2\alpha ) = \dfrac{{1 - x}}{x}$
$ \Rightarrow {\cos ^2}\alpha - {\sin ^2}\alpha = \dfrac{{1 - x}}{x}$ {$\because $ By double angle formula}
Find the values of $\cos \alpha $ and $\sin \alpha $ in terms of $x$ and $c.$
Since in a right triangle sin of an angle is perpendicular divided by hypotenuse and cos of an angle is base divided by hypotenuse.
From the figure we have $\cos \alpha = \dfrac{{\sqrt {{c^2} - {x^2}} }}{c}$ and $\sin \alpha = \dfrac{x}{c}.$
$ \Rightarrow {\cos ^2}\alpha - {\sin ^2}\alpha = \dfrac{{1 - x}}{x}$
Substitute $\cos \alpha = \dfrac{{\sqrt {{c^2} - {x^2}} }}{c}$ and $\sin \alpha = \dfrac{x}{c}.$
$ \Rightarrow {(\dfrac{{\sqrt {{c^2} - {x^2}} }}{c})^2} - {(\dfrac{x}{c})^2} = \dfrac{{1 - x}}{x}$
Square cancels the under root from LHS.
$ \Rightarrow \dfrac{{{c^2} - {x^2}}}{{{c^2}}} - \dfrac{{{x^2}}}{{{c^2}}} = \dfrac{{1 - x}}{x}$
Take LCM in the LHS of the equation.
$ \Rightarrow \dfrac{{{c^2} - 2{x^2}}}{{{c^2}}} = \dfrac{{1 - x}}{x}$
Cross multiply both sides.
$ \Rightarrow x({c^2} - 2{x^2}) = {c^2}(1 - x)$
$ \Rightarrow {c^2}x - 2{x^3} = {c^2}(1 - x)$
Bring all values with variable c on one side of the equation.
$ \Rightarrow {c^2}x - {c^2}(1 - x) = 2{x^3}$
Take \[{c^2}\] common on the LHS of the equation.
$ \Rightarrow {c^2}[x - (1 - x)] = 2{x^3}$
$ \Rightarrow {c^2}(2x - 1) = 2{x^3}$
Divide both sides by \[(2x - 1)\]
$ \Rightarrow {c^2} = \dfrac{{2{x^3}}}{{2x - 1}}$
Take square root on both sides.
$ \Rightarrow c = \sqrt {\dfrac{{2{x^3}}}{{2x - 1}}} $ (1)
To find the minimum value of crease equate derivative of $c$ to $0.$
$c = \sqrt {\dfrac{{2{x^3}}}{{2x - 1}}} $
Differentiate $c$ with respect to $x$ and equate to $0.$
$ \Rightarrow \dfrac{{dc}}{{dx}} = 0$
$ \Rightarrow \dfrac{d}{{dx}}(\sqrt {\dfrac{{2{x^3}}}{{2x - 1}}} ) = 0$
Use chain rule to differentiate $\sqrt {\dfrac{{2{x^3}}}{{2x - 1}}} $.
$ \Rightarrow \dfrac{1}{{2\sqrt {\dfrac{{2{x^3}}}{{2x - 1}}} }}\dfrac{d}{{dx}}(\dfrac{{2{x^3}}}{{2x - 1}}) = 0$ {$\because $ If$y = f(g(x)) \Rightarrow \dfrac{{dy}}{{dx}} = f'(g(x))g'(x)$ }
Use quotient rule of derivative to find the derivative of $\dfrac{{2{x^3}}}{{2x - 1}}.$
$ \Rightarrow \dfrac{{(2x - 1)(6{x^2}) - 2{x^3}(2)}}{{(2x - 1)}} = 0$ {$\because $ If $y = \dfrac{f}{g} \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{g \cdot f' - f \cdot g'}}{{{g^2}}}$}
Multiplying both side by $2x - 1.$
$ \Rightarrow \dfrac{{6{x^2}(2x - 1) - 4{x^3}}}{{(2x - 1)}} = 0$
$
   \Rightarrow 12{x^3} - 6{x^2} - 4{x^3} = 0 \\
   \Rightarrow 8{x^3} - 6{x^2} = 0 \\
 $
Take common $2{x^2}$
$ \Rightarrow 2{x^2}(4x - 3) = 0$
$ \Rightarrow 2{x^2} = 0$ or $4x - 3 = 0$
$ \Rightarrow x = 0$ or \[4x = 3 \Rightarrow x = \dfrac{3}{4}\]
$\because x \ne 0$ because it is the length of the folded part.
$\therefore $ Length of crease is minimum when $x = \dfrac{3}{4}.$
Substitute value of $x$ in equation (1) to find value of $c$.
$c = \sqrt {\dfrac{{2{x^3}}}{{2x - 1}}} $
$ \Rightarrow c = \sqrt {\dfrac{{2{{(\dfrac{3}{4})}^3}}}{{2(\dfrac{3}{4}) - 1}}} $
$ \Rightarrow c = \sqrt {\dfrac{{2(\dfrac{{27}}{{64}})}}{{\dfrac{3}{2} - 1}}} $
Take LCM in the denominator.
$ \Rightarrow c = \sqrt {\dfrac{{2(\dfrac{{27}}{{64}})}}{{\dfrac{{3 - 2}}{2}}}} $
$ \Rightarrow c = \sqrt {\dfrac{{\dfrac{{27}}{{32}}}}{{\dfrac{1}{2}}}} $
Write fraction in simpler way.
$ \Rightarrow c = \sqrt {\dfrac{{27}}{{32}} \times \dfrac{2}{1}} $
Cancel out factors from numerator and denominator.
\[
   \Rightarrow c = \sqrt {\dfrac{{27}}{{16}}} \\
   \Rightarrow c = \sqrt {\dfrac{{{3^2} \times 3}}{{{4^2}}}} \\
 \]
Cancel square root by square power.
\[ \Rightarrow c = \dfrac{{3\sqrt 3 }}{4}\]

So, the correct answer is “Option C”.

Note: Students can make mistake of writing \[\theta , \beta , \alpha \] in the wrong form from the diagram, always keep in mind we take one angle which was the corner of the rectangle as right angle which makes our work easy, some students try to take the third angle also as a variable which will make it impossible to solve, so try to write an angle with a right angle relation or supplementary relation.