If the sum of lengths of hypotenuse and a side of a right angled triangle is given, show that area of triangle is maximum, when the angle between them is

If the sum of lengths of hypotenuse and a side of a right angled triangle is given, show that area of triangle is maximum, when the angle between them is $\frac{\pi }{3}\cdot$

Answers (1)

Let ABC be a right angled triangle with base BC = x AB = y such that $x+y= k$ (constant)
Let a be the andle between base and hypotaneous
Let A be the area of triangle then,

$A= \frac{1}{2}\times BC\times AC$
$= \frac{1}{2}\times \sqrt{y^{2}-x^{2}}$
$\Rightarrow A^{2} = \frac{x^{2}}{4}\left ( y^{2}-x^{2} \right )\left ( \because y= k-x \right )$
$\Rightarrow A^{2} = \frac{x^{2}}{4}\left [ \left ( k-x \right )^{2}-x^{2} \right ]$
$\Rightarrow A^{2} = \frac{k^{2}x^{2}-2kx^{2}}{4}---(i)$
Differentiating w.r.t x, we get
$\Rightarrow \frac{dA}{dx}= \frac{k^{2}x-3kx^{2}}{4A}$
For maximum or minimum. we have
$\frac{dA}{dx}= 0\Rightarrow \frac{k^{2}x-3kx^{2}}{4A}= 0\Rightarrow x= \frac{k}{3}$
Again differentiating (ii) w.r.t x, we get
$2\left ( \frac{dA}{dx} \right )^{2}+\frac{2Ad^{2}A}{dx^{2}}= \frac{2k^{2}-12kx}{4}---(iii)$
Putting $\frac{dA}{dx}= 0$ and $x= \frac{k}{3}$ in (iii) we get
$\frac{d^{2}A}{dx^{2}}= \frac{-k^{2}}{4A}< 0$
Thus A is maximum when $= \frac{k}{3}$ Now and $x= \frac{k}{3}$ and $y= k-x= k-\frac{k}{3}= \frac{2k}{3}$
$\therefore \cos \theta = \frac{x}{y}= \frac{\frac{k}{3}}{2\frac{k}{3}}= \frac{1}{2}$
$\theta = \cos^{-1}\left ( \frac{1}{2} \right )= \frac{\pi }{3}$ Hence proved