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.
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.