Question

For a right angled triangle, if the sum of the length of the hypotenuse and a side is kept constant, in order to have maximum area of the triangle, the angle between the hypotensuse and the side is

• A. ${120}^o$
• B. ${60}^o$
• C. ${30}^o$
• D. ${45}^o$

Answer 1

The correct option is B ${60}^o$

Method I:



$ATQx+\sqrt{x^2+y^2}=k$ (constant) $\Rightarrow y=\sqrt{k^2-2kx}...\left(1\right)$
Now area $=\frac{1}{2}\times x\times y$
Let$u=A^2=\frac{1}{4}{x}^2{y}^2$
so $A_{max}=\sqrt{u_{max}}=?$
Hence, $u=\frac{1}{4}{x}^2(k^2-2kx)...\left(2\right)$
For maximum $u,\frac{du}{dx}=0$
$\Rightarrow \frac{2k^{2}x}{4}=\frac{3}{2}k{x}^2=0$
$\Rightarrow \frac{kx}{2}(k-3x)=0$
$\Rightarrow x=\frac{k}{3}$
Now By $\left(1\right),y=\sqrt{k^2-2kx}=\sqrt{k^2-2k\left(\frac{k}{3}\right)}=\frac{k}{\sqrt{3}}$
So $tan\theta =\frac{y}{x}=d\frac{k/\sqrt{3}}{k/3}=\sqrt{3}\Rightarrow \theta ={60}^o$

Method II:


$h+y=K,K$ is constant
Let $A$ be the area of triangle.
$A=\frac{1}{2}xy=\frac{1}{2}\left(hcos\theta \right)\left(hsin\theta \right)$
$A=\frac{h^2}{4}sin2\theta$



$\Rightarrow h+hsin\theta =K$
$\Rightarrow h=\frac{K}{1+sin\theta }$
$A=\frac{K^2}{4}\left(\frac{sin2\theta }{(1+sin\theta {)}^2}\right)$
For maximum Area $\frac{dA}{d\theta }=0$
$\Rightarrow \frac{K^2}{4}\left[\frac{(1+sin\theta {)}^2(2cos2\theta )-sin\theta .2cos\theta (1+sin\theta )}{(1+sin\theta {)}^4}\right]=0$
$\Rightarrow \theta =30\theta$ so angle between $h\&y={60}^o$