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Q.
The area of a right-angled triangle of given hypotenuse is maximum when the triangle is
Application of Derivatives
Solution:
Let $h$ be the hypotenuse of the right-angled triangle, and
let $x$ be its altitude. Then,
Base of the triangle $=\sqrt{h^{2} -x^{2}} $
Let $A$ be the area of the triangle.
then,
$A= \frac{1}{2} x \sqrt{h^{2} -x^{2} }$
$\Rightarrow \frac{dA}{dx} = \frac{1}{2}\left\{1\cdot\sqrt{h^{2} -x^{2}} + x \frac{1}{2} \left(h^{2} -x^{2}\right)^{-\frac{1}{2}} \frac{d}{dx}\left(h^{2}-x^{2}\right) \right\}$
$\Rightarrow \frac{dA}{dx} = \frac{1}{2}\left\{\sqrt{h^{2} -x^{2}} -\frac{x^{2}}{\sqrt{h^{2} -x^{2}}}\right\} $
$= \frac{1}{2} \left\{ \frac{h^{2} -2x^{2}}{\sqrt{h^{2} -x^{2}}}\right\}$
For maximum or minimum, we have
$ \frac{dA}{dx } = 0 \Rightarrow \frac{1}{2} \left\{\frac{h^{2} -2x}{\sqrt{h^{2} -x^{2}}}\right\} = 0$
$\Rightarrow h^{2} = 2x^{2} $
$ \Rightarrow x= \frac{h}{\sqrt{2}} $
Now,
$\Rightarrow \frac{d^{2}A}{dx^{2}}= \frac{1}{2}\left\{\left(-4x\right) \frac{1}{\sqrt{h^{2} -x^{2}}} +\left(h^{2} -2x^{2}\right) \left(-\frac{1}{2}\right)\times \left(h^{2}-x^{2}\right)^{-{3}/{2}} \frac{d}{dx}\left(h^{2}-x^{2}\right)\right\}$
$\Rightarrow \frac{d^{2}A}{ dx^{2}} = \frac{1}{2} \left\{ \frac{-4x}{\sqrt{h^{2} -x^{2}}} + \frac{x\left(h^{2} -2x\right)}{\left(h^{2} -x^{2}\right)^{{3}/{2}}}\right\}$
$ \Rightarrow \left(\frac{d^{2}A}{dx^{2}}\right)_{x=\frac{h}{\sqrt{2}}} = -2< 0$.
$\left[{\text{neglecting}}\,\, x= \frac{-h}{\sqrt{2}} \because \frac{d^{2}A}{dx^{2}}|_{x= \frac{-h}{\sqrt{2}}} >0\right] $
Thus, $A $ is maximum when $x =\frac{h}{\sqrt{2}}$.
$\therefore $ Base $= \sqrt{h^{2} -\left(\frac{h^{2}}{2}\right)} = \frac{h}{\sqrt{2}}$
Hence, $A$ is maximum when the triangle is isosceles.
