Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long.

Solution:

Let the base of the right angled triangle be $x$ and its height be $y$. Then,

$x^{2}+y^{2}=5^{2}$

$\Rightarrow y^{2}=25-x^{2}$

$\Rightarrow y=\sqrt{25-x^{2}}$

As, the area of the triangle, $A=\frac{1}{2} \times x \times y$

$\Rightarrow A(x)=\frac{1}{2} \times \mathrm{x} \times \sqrt{25-\mathrm{x}^{2}}$

$\Rightarrow A(x)=\frac{x \sqrt{25-x^{2}}}{2}$

$\Rightarrow A^{\prime}(x)=\frac{\sqrt{25-x^{2}}}{2}+\frac{x(-2 x)}{4 \sqrt{25-x^{2}}}$

$\Rightarrow A^{\prime}(x)=\frac{\sqrt{25-x^{2}}}{2}-\frac{x^{2}}{2 \sqrt{25-x^{2}}}$

$\Rightarrow A^{\prime}(x)=\frac{25-x^{2}-x^{2}}{2 \sqrt{25-x^{2}}}$

$\Rightarrow A^{\prime}(x)=\frac{25-2 x^{2}}{2 \sqrt{25-x^{2}}}$

For maxima or minima, we must have $f^{\prime}(x)=0$

$A^{\prime}(x)=0$

$\Rightarrow \frac{25-2 x^{2}}{2 \sqrt{25-x^{2}}}=0$

$\Rightarrow 25-2 x^{2}=0$

$\Rightarrow 2 x^{2}=25$

$\Rightarrow x=\frac{5}{\sqrt{2}}$

So, $y=\sqrt{25-\frac{25}{2}}$

$=\sqrt{\frac{50-25}{2}}$

$=\sqrt{\frac{25}{2}}$

$=\frac{5}{\sqrt{2}}$

Also, $A^{\prime \prime}(x)=\frac{\left[-4 x \sqrt{25-x^{2}}-\frac{\left(25-2 x^{2}\right)(-2 x)}{2 \sqrt{25-x^{2}}}\right]}{25-x^{2}}$

$=\frac{\left[\frac{-4 x\left(25-x^{2}\right)+\left(25 x-2 x^{3}\right)}{\sqrt{25-x^{2}}}\right]}{25-x^{2}}$

$=\frac{-100 x+4 x^{3}+25 x-2 x^{3}}{\left(25-x^{2}\right) \sqrt{25-x^{2}}}$

$=\frac{-75 x+2 x^{3}}{\left(25-x^{2}\right) \sqrt{25-x^{2}}}$

$\Rightarrow A^{\prime \prime}\left(\frac{5}{\sqrt{2}}\right)=\frac{-75\left(\frac{5}{\sqrt{2}}\right)+2\left(\frac{5}{\sqrt{2}}\right)^{3}}{\left(25-\left(\frac{5}{\sqrt{2}}\right)^{2}\right)^{\frac{3}{2}}}<0$

So, $x=\left(\frac{5}{\sqrt{2}}\right)$ is point of maxima.

$\therefore$ The largest possible area of the triangle $=\frac{1}{2} \times\left(\frac{5}{\sqrt{2}}\right) \times\left(\frac{5}{\sqrt{2}}\right)=\frac{25}{4}$ square units