Application of Derivatives — Class 12 Maths Solution

Let $ABC$ be a triangle with $AC = h,$ $AB = x$ and $BC = y$.
Also, $\angle CAB = \theta$
Let $h + x = k$ ……..(i)

Therefore, $\cos \theta = \frac{x}{h}$

$\Rightarrow$ $x = h\cos \theta$

$\Rightarrow$ $h + h\cos \theta = k\quad$ [using Eq. (i)]

$\Rightarrow$ $h(1 + \cos \theta ) = k$

$\Rightarrow$ $h = \frac{k}{{(1 + \cos \theta )}}$ …….(ii)

Also, area of $\Delta ABC = \frac{1}{2}(AB \cdot BC)$
$A = \frac{1}{2} \cdot x \cdot y$

$= \frac{1}{2}{h^2}\sin \theta \cdot \cos \theta$

$= \frac{{2{h^2}}}{4}\sin \theta \cdot \cos \theta$
$= \frac{1}{4}{h^2}\sin 2\theta$

…….(iii)

Since, $h = \frac{k}{{1 + \cos \theta }}$

Therefore,$\quad A = \frac{1}{4}{\left( {\frac{k}{{1 + \cos \theta }}} \right)^2} \cdot \sin 2\theta$

$\Rightarrow$ $A = \frac{{{k^2}}}{4} \cdot \frac{{\sin 2\theta }}{{{{(1 + \cos \theta )}^2}}}$ …….(iv)

Therefore, $\frac{{dA}}{{d\theta }} = \frac{{{k^2}}}{4}\left[ {\frac{{{{(1 + \cos \theta )}^2} \cdot \cos 2\theta \cdot 2 - \sin 2\theta \cdot 2(1 + \cos \theta ) \cdot (0 - \sin \theta )}}{{{{(1 + \cos \theta )}^4}}}} \right]$

$= \frac{{{k^2}}}{4}\left\{ {\frac{{2(1 + \cos \theta )[(1 + \cos \theta ) \cdot \cos 2\theta + \sin 2\theta (\sin \theta )}}{{{{(1 + \cos \theta )}^4}}}} \right\}$

$= \frac{{{k^2}}}{4} \cdot \frac{2}{{{{(1 + \cos \theta )}^3}}}\left[ {(1 + \cos \theta ) \cdot \cos 2\theta + 2{{\sin }^2}\theta \cdot \cos \theta } \right]$

$= \frac{{{k^2}}}{{2{{(1 + \cos \theta )}^3}}}\left[ {(1 + \cos \theta )\left( {1 - 2{{\sin }^2}\theta } \right) + 2{{\sin }^2}\theta \cdot \cos \theta } \right]$

$= \frac{{{k^2}}}{{2{{(1 + \cos \theta )}^3}}}\left[ {1 + \cos \theta - 2{{\sin }^2}\theta - 2{{\sin }^2}\theta \cdot \cos \theta + 2{{\sin }^2}\theta \cdot \cos \theta } \right]$

$= \frac{{{k^2}}}{{2{{(1 + \cos \theta )}^3}}}\left[ {(1 + \cos \theta ) - 2{{\sin }^2}\theta } \right]$

$= \frac{{{k^2}}}{{2{{(1 + \cos \theta )}^3}}}\left[ {1 + \cos \theta - 2 + 2{{\cos }^2}\theta } \right]$

$= \frac{{{k^2}}}{{2{{(1 + \cos \theta )}^3}}}\left( {2{{\cos }^2}\theta + \cos \theta - 1} \right)$ …….(v)

For $\frac{{dA}}{{d\theta }} = 0$,

$\frac{{{k^2}}}{{2{{(1 + \cos \theta )}^3}}}\left( {2{{\cos }^2}\theta + \cos \theta - 1} \right) = 0$

$\Rightarrow$ $2{\cos ^2}\theta + \cos \theta - 1 = 0$

$\Rightarrow$ $2{\cos ^2}\theta + 2\cos \theta - \cos \theta - 1 = 0$

$\Rightarrow$ $2\cos \theta (\cos \theta + 1) - 1(\cos \theta + 1) = 0$

$\Rightarrow$ $(2\cos \theta - 1)(\cos \theta + 1) = 0$

$\Rightarrow$ $\cos \theta = \frac{1}{2}$ or $\cos \theta = - 1$

$\Rightarrow$ $\theta = \frac{\pi }{3}$ [possible]
or $\theta = 2n\pi \pm \pi$ [not possible]

Therefore,$\quad \theta = \frac{\pi }{3}$

Again, differentiating w.r.t. $\theta$ in Eq. (v), we get
$\frac{d}{{d\theta }}\left( {\frac{{dA}}{{d\theta }}} \right) = \frac{d}{{d\theta }}\left[ {\frac{{{k^2}}}{{2{{(1 + \cos \theta )}^3}}}\left( {2{{\cos }^2}\theta + \cos \theta - 1} \right)} \right]$

Therefore,$\frac{{{d^2}A}}{{d{\theta ^2}}} = \frac{d}{{d\theta }}\left[ {\frac{{{k^2}(2\cos \theta - 1)(1 + \cos \theta )}}{{2{{(1 + \cos \theta )}^3}}}} \right] = \frac{d}{{d\theta }}\left[ {\frac{{{k^2}}}{2} \cdot \frac{{(2\cos \theta - 1)}}{{{{(1 + \cos \theta )}^2}}}} \right]$

$= \frac{{{k^2}}}{2}\left[ {\frac{{{{(1 + \cos \theta )}^2} \cdot ( - 2\sin \theta ) - 2(1 + \cos \theta ) \cdot ( - \sin \theta )(2\cos \theta - 1)}}{{{{(1 + \cos \theta )}^4}}}} \right]$

$= \frac{{{k^2}}}{2}\left[ {\frac{{(1 + \cos \theta ) \cdot [1 + \cos \theta ]( - 2\sin \theta ) + 2\sin \theta (2\cos \theta - 1)}}{{{{(1 + \cos \theta )}^4}}}} \right]$

$= \frac{{{k^2}}}{2}\left[ {\frac{{ - 2\sin \theta - 2\sin \theta \cdot \cos \theta + 4\sin \theta \cdot \cos \theta - 2\sin \theta }}{{{{(1 + \cos \theta )}^3}}}} \right]$

$= \frac{{{k^2}}}{2}\left[ {\frac{{ - 4\sin \theta - \sin 2\theta + 2\sin 2\theta }}{{{{(1 + \cos \theta )}^3}}}} \right] = \frac{{{k^2}}}{2}\left[ {\frac{{\sin 2\theta - 4\sin \theta }}{{{{(1 + \cos \theta )}^3}}}} \right]$

Therefore,$\quad {\left( {\frac{{{d^2}A}}{{d{\theta ^2}}}} \right)_{{\rm{at }}\theta = \frac{\pi }{3}}} = \frac{{{k^2}}}{2}\left[ {\frac{{\sin \frac{{2\pi }}{3} - 4\sin \frac{\pi }{3}}}{{{{\left( {1 + \cos \frac{\pi }{3}} \right)}^3}}}} \right] = \frac{{{k^2}}}{2}\left[ {\frac{{\frac{{\sqrt 3 }}{2} - \frac{{4\sqrt 3 }}{2}}}{{{{\left( {1 + \frac{1}{2}} \right)}^3}}}} \right]$

$= \frac{{{k^2}}}{2}\left[ {\frac{{ - 3\sqrt 3 \cdot 8}}{{2 \cdot 27}}} \right] = - {k^2}\left( {\frac{{2\sqrt 3 }}{9}} \right)$

which is less than zero.

Hence, area of the right angled triangle is maximum, when the angle between them is $\frac{\pi }{3}$.