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 $\frac{\pi}{3}$.

(N/A) Let $ABC$ be a right-angled triangle with hypotenuse $AC = h$,base $AB = x$,and perpendicular $BC = y$. Let $\angle CAB = \theta$.
Given that the sum of the hypotenuse and a side is constant,let $h + x = k$,where $k$ is a constant.
From the triangle,$\cos \theta = \frac{x}{h}$,so $x = h \cos \theta$.
Substituting this into the sum,$h + h \cos \theta = k$,which gives $h(1 + \cos \theta) = k$,or $h = \frac{k}{1 + \cos \theta}$.
The area of the triangle $A = \frac{1}{2} \times AB \times BC = \frac{1}{2} x y$.
Since $x = h \cos \theta$ and $y = h \sin \theta$,we have $A = \frac{1}{2} (h \cos \theta) (h \sin \theta) = \frac{1}{2} h^2 \sin \theta \cos \theta = \frac{1}{4} h^2 \sin 2\theta$.
Substituting $h = \frac{k}{1 + \cos \theta}$,we get $A = \frac{k^2}{4} \cdot \frac{\sin 2\theta}{(1 + \cos \theta)^2}$.
To find the maximum area,differentiate $A$ with respect to $\theta$:
$\frac{dA}{d\theta} = \frac{k^2}{4} \left[ \frac{(1 + \cos \theta)^2 (2 \cos 2\theta) - \sin 2\theta \cdot 2(1 + \cos \theta)(-\sin \theta)}{(1 + \cos \theta)^4} \right]$.
Simplifying this expression,we get $\frac{dA}{d\theta} = \frac{k^2}{2(1 + \cos \theta)^3} (2 \cos^2 \theta + \cos \theta - 1)$.
Setting $\frac{dA}{d\theta} = 0$ gives $2 \cos^2 \theta + \cos \theta - 1 = 0$,which factors to $(2 \cos \theta - 1)(\cos \theta + 1) = 0$.
Since $\cos \theta = -1$ is not possible for a triangle,we have $\cos \theta = \frac{1}{2}$,which implies $\theta = \frac{\pi}{3}$.
Using the second derivative test,it can be shown that $\frac{d^2A}{d\theta^2} < 0$ at $\theta = \frac{\pi}{3}$,confirming that the area is maximum.