Question

Prove that the perimeter of a right angled triangle of a given hypotenuse is maximum when triangle is isosceles.

Answer 1

Let $h$ be the hypotenuse of right angled triangle and then two sides be $x$ and $y$ respectively,

Now, $h^2=x^2+y^2$

$y=\sqrt{h^2-x^2}$

Perimeter $p=h+x+y=h+x+\sqrt{h^2-x^2}$

Now $\frac{dp}{dx}=0$ [For max or min]

$p=h+x+\sqrt{h^2-x^2}$

$\frac{dp}{dx}=0+1+\frac{1}{2\sqrt{h^2-x^2}}.(-2x)$

$0+1+\frac{(-x)}{2\sqrt{h^2-x^2}}=0$

$\Rightarrow 1-\frac{x}{\sqrt{h^2-x^2}}=0$

$\Rightarrow \frac{x}{\sqrt{h^2-x^2}}=1$

$\Rightarrow x=\sqrt{h^2-x^2}$

$\Rightarrow x^2=h^2-x^2$

$x=\frac{h}{\sqrt{2}}$

Also $\frac{d^{2}p}{{dx}^2}=0$

$=0-\frac{\sqrt{h^2-x^2-1}-x\frac{1}{2\sqrt{h^2-x^2}}.(-2x)}{h^2-x^2}=-\left(\frac{h^2-x^2+x^2}{{(h^2-x^2)}^{3/2}}\right)=\frac{-h^2}{{(h^2-x^2)}^{3/2}}$

It can be seen that $\left(\frac{d^{2}p}{{dx}^2}\right)$ is negative $p$ is maximum

$y=\sqrt{h^2-\frac{h^2}{2}}=\frac{h}{\sqrt{2}}\Rightarrow x=y$

Hence perimeter is maximum when triangle is isosceles.