Question

If the sum of length 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 $\frac{\pi }{3}.$

Answer 1

Let the hypotenuse of the right triangle be $x$, and the height be $y.$

Hence, its base is $\sqrt{x^2-y^2}$

Hence the area

$=\frac{1}{2}\times Base\times Height$

$=\frac{1}{2}\times \sqrt{x^2-y^2}\times y$

But it is given that,

$x+y=a\left(say\right)$

$x=a-y$

Such that,

Area

$=\frac{1}{2}y\sqrt{{(p-y)}^2-y^2}$

$=\frac{1}{2}y\sqrt{p^2+y^2-2py-y^2}$

$=\frac{1}{2}y\sqrt{p^2-2py}$

On squaring both side and we get,

${\left(Area\right)}^2=\frac{1}{4}{y}^2(p^2-2py)$

$=\frac{1}{4}{y}^2{p}^2-\frac{1}{2}p{y}^3$

For maximum and minimum

$\frac{dy}{dA}=0$

Here the area of the triangle is maximum when

$x=\frac{2p}{3}$ and $y=\frac{p}{3}$

$\cos \theta =\frac{y}{x}$

$=\frac{\frac{p}{3}}{2\frac{p}{3}}$

$=\frac{1}{2}$

$\cos \theta =\frac{1}{2}$

$\cos \theta =\cos \frac{\pi }{3}$

$\theta =\frac{\pi }{3}$

Hence, this is the answer.