Question

If the sum of lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum, when the angle between them is ${60}^{\circ }$.

Answer 1

$\text{Let AB be the hypotenuse and let}$ $AB+AC=k...\left(i\right)$

$\text{where}k\text{is a constant.}$




$LetAB=x$, then

$AC=AB\cos \theta =x\cos \theta$

$CB=AB\sin \theta =x\sin \theta$

$\Rightarrow x+x\cos \theta =kfrom\left(i\right)$.

$\Rightarrow x=\frac{k}{1+\cos \theta }$

$\text{Area of triangle}$ = $A=\frac{1}{2}Base\times Height$

$A=\frac{1}{2}\left(x\cos \theta \right)\left(x\sin \theta \right)$

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

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

$\text{Differentiating both sides w.r.t.}\theta \text{, we get:}$

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

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

$\frac{dA}{d\theta }=\frac{k^2}{2}\left(\frac{{\cos }^3\theta +{\sin }^2\theta \cos \theta -{\sin }^2\theta +{\cos }^2\theta }{(1+\cos \theta {)}^3}\right)$, $0<\theta <\pi /2$

$\frac{dA}{d\theta }=\frac{k^2}{2}\frac{\cos \theta +\cos 2\theta }{(1+\cos \theta {)}^3}$,

$[\because {\sin }^2\theta \cos \theta +{\cos }^3\theta =\cos \theta ({\sin }^2\theta +{\cos }^2\theta )=\cos \theta ]$

Now, $\frac{dA}{d\theta }=0\Rightarrow \cos \theta +\cos 2\theta =0$

$\Rightarrow \cos 2\theta =-\cos \theta =\cos (\pi -\theta )$

$\Rightarrow 2\theta =\pi -\theta \Rightarrow \theta =\pi /3$

$\frac{d^{2}A}{d{\theta }^2}=\frac{k^2}{2}\left(\frac{(1+\cos \theta {)}^6[-\sin \theta -2\sin 2\theta ]+3(1+\cos \theta {)}^2.\sin \theta [\cos \theta +\cos 2\theta ]}{(1+\cos \theta {)}^6}\right)$

$\frac{d^{2}A}{d{\theta }^2}{|}_{\frac{\pi }{3}}=\frac{k^2}{2}\left(\frac{(1+\cos \frac{\pi }{3}{)}^6[-\sin \frac{\pi }{3}-2\sin (\frac{2\pi }{3}\left)\right]+3(1+\cos \frac{\pi }{3}{)}^2.\sin \frac{\pi }{3}[\cos \frac{\pi }{3}+\cos \frac{2\pi }{3}]}{{(1+\cos \frac{\pi }{3})}^6}\right)$

$\frac{d^{2}A}{d{\theta }^2}{|}_{\frac{\pi }{3}}=\frac{k^2}{2}\left(\frac{(1+\frac{1}{2}{)}^6[-\frac{\sqrt{3}}{2}-2\times (\frac{\sqrt{3}}{2}\left)\right]+3{(1+\frac{1}{2})}^2.\frac{\sqrt{3}}{2}[\frac{1}{2}-\frac{1}{2}]}{{(1+\frac{1}{2})}^6}\right)$

$\frac{d^{2}A}{d{\theta }^2}{|}_{\frac{\pi }{3}}=\frac{k^2}{2}\left(\frac{(1+\frac{1}{2}{)}^6[-\frac{3\sqrt{3}}{2})]+0}{{(1+\frac{1}{2})}^6}\right)=-\frac{3\sqrt{3}{k}^2}{4}<0$

$\text{Therefore the area of the triangle is maximum when the}\text{angle between the hypotenuse and one of the sides is}\pi /3\text{.}$