If the sum of the lengths of the hypotenuse and a side of a right-ang…

The correct answer is Option (1) → $\frac{\pi}{3}$

Let ABC be a right-angled triangle at B.

Let hypotenuse $AC = l$ and side $AB = x$, then

$BC^2 = l^2-x^2⇒ BC= \sqrt{l^2-x^2}$   ...(i)

Given $l + x = k$ (fixed)

$⇒l=k-x$   ...(ii)

Let A be area of ΔABC, then

$A=\frac{1}{2}AB×BC$

$⇒\frac{1}{2}x\sqrt{l^2-x^2}$  (using (i))

$⇒\frac{1}{2}x\sqrt{(k-x)^2-x^2}$   (using (ii))

$⇒A^2=\frac{1}{4}x^2(k^2-2kx)=\frac{1}{4}(k^2x^2-2kx^3)$.

Now A is maximum when $A^2$ is maximum. Let us write $A^2$ as $f(x)$,

so $f(x)=\frac{1}{4}(k^2x^2-2kx^3)$

$⇒f'(x)=\frac{1}{4}(2k^2x-6kx^2)$ and $f''(x)=\frac{1}{4}((2k^2-12kx)$.

Now $f'(x)=0⇒2k^2x-6kx^2=0⇒x=\frac{k}{3}$   $(∵x>0)$

$f''(\frac{k}{3})=\frac{1}{4}(2k^2-4k^2)=-\frac{1}{2}k^2<0$

⇒ f(x) is maximum when $x=\frac{k}{3}$ ⇒ A is maximum when $x=\frac{k}{3}$.

From (ii), when $x=\frac{k}{3},l=k-\frac{k}{3}=\frac{2k}{3}$

From a right angled triangle ABC, $\cos A=\frac{x}{l}=\frac{\frac{k}{3}}{\frac{2k}{3}}=\frac{1}{2}⇒A=\frac{\pi}{3}$