If the sum of the lengths of the hypotenuse and a side of a right angled
Question:

If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is π/3.

Solution:

Let AC = x, BC = y

So, AB = √(x2 + y2)

∠ACB = θ

Let z = x + y (given)

Now, the area of ∆ABC = ½ x AB x BC

$\Rightarrow \mathrm{A}=\frac{1}{2} y \cdot \sqrt{x^{2}-y^{2}} \Rightarrow \mathrm{A}=\frac{1}{2} y \cdot \sqrt{(Z-y)^{2}-y^{2}}$

Squaring both sides, we get

$\mathrm{A}^{2}=\frac{1}{4} y^{2}\left[(Z-y)^{2}-y^{2}\right] \Rightarrow \mathrm{A}^{2}=\frac{1}{4} y^{2}\left[Z^{2}+y^{2}-2 Z y-y^{2} \mid\right.$

So, $P=\frac{1}{4} y^{2}\left[Z^{2}-2 Z y\right] \Rightarrow P=\frac{1}{4}\left[y^{2} Z^{2}-2 Z y^{3}\right]$ $\left[A^{2}=P\right]$

Differentiating both sides w.r.t. $y$ we get

$\frac{d P}{d y}=\frac{1}{4}\left[2 y Z^{2}-6 Z y^{2}\right]$ ……….(i)

For local maxima and local minima, $\frac{d \mathrm{P}}{d y}=0$

$\therefore \frac{1}{4}\left(2 y Z^{2}-6 Z y^{2}\right)=0$

$\frac{2 y Z}{4}(Z-3 y)=0 \Rightarrow y Z(Z-3 y)=0$

$y Z \neq 0$ $(\because \quad y \neq 0$ and $Z \neq 0)$

$\therefore Z-3 y=0$

$y=\frac{z}{3} \Rightarrow y=\frac{x+y}{3} \quad(\because \quad \mathrm{Z}=x+y)$

$3 y=x+y \Rightarrow 3 y-y=x \Rightarrow 2 y=x$

$\frac{y}{x}=\frac{1}{2} \Rightarrow \cos \theta=\frac{1}{2}$

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

Differentiating eq. (i) w.r.t. $y$, we have $\frac{d^{2} P}{d y^{2}}=\frac{1}{4}\left[2 Z^{2}-12 Z y\right]$

$\frac{d^{2} \mathrm{P}}{d y^{2}}$ at $y=\frac{Z}{3}=\frac{1}{4}\left[2 Z^{2}-12 Z \cdot \frac{Z}{3}\right]$

$=\frac{1}{4}\left[2 Z^{2}-4 Z^{2}\right]=\frac{-Z^{2}}{2}<0$ Maxima

$\Rightarrow \mathrm{A}=\frac{1}{2} y \cdot \sqrt{x^{2}-y^{2}} \Rightarrow \mathrm{A}=\frac{1}{2} y \cdot \sqrt{(Z-y)^{2}-y^{2}}$

Squaring both sides, we get

$A^{2}=\frac{1}{4} y^{2}\left[(Z-y)^{2}-y^{2}\right] \Rightarrow A^{2}=\frac{1}{4} y^{2}\left[Z^{2}+y^{2}-2 Z y-y^{2} \mid\right.$

So, $P=\frac{1}{4} y^{2}\left[Z^{2}-2 Z y\right] \Rightarrow P=\frac{1}{4}\left[y^{2} Z^{2}-2 Z y^{3}\right]$ $\left[A^{2}=P\right]$

Differentiating both sides w.r.t. $y$ we get

$\frac{d P}{d y}=\frac{1}{4}\left[2 y Z^{2}-6 Z y^{2}\right]$ $\ldots(i)$

For local maxima and local minima, $\frac{d \mathrm{P}}{d y}=0$

$\therefore \frac{1}{4}\left(2 y Z^{2}-6 Z y^{2}\right)=0$

$\frac{2 y Z}{4}(Z-3 y)=0 \Rightarrow y Z(Z-3 y)=0$

$y Z \neq 0$ $(\because \quad y \neq 0$ and $Z \neq 0)$

$\therefore Z-3 y=0$

$y=\frac{Z}{3} \quad \Rightarrow y=\frac{x+y}{3} \quad(\because \quad \mathrm{Z}=x+y)$

$\frac{y}{x}=\frac{1}{2} \Rightarrow \cos \theta=\frac{1}{2}$

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

Differentiating eq. (i) w.r.t. $y$, we have $\frac{d^{2} P}{d y^{2}}=\frac{1}{4}\left[2 Z^{2}-12 Z y\right]$

$\frac{d^{2} P}{d y^{2}}$ at $y=\frac{Z}{3}=\frac{1}{4}\left[2 Z^{2}-12 Z \cdot \frac{Z}{3}\right]$

$=\frac{1}{4}\left[2 Z^{2}-4 Z^{2}\right]=\frac{-Z^{2}}{2}<0$ Maxima

Therefore, the area of the given triangle is maximum when the angle between its hypotenuse and a side is π/3.