The maximum area of a triangle whose one vertex is at ( 0, 0 ) and...

Let, area of triangle $ABC$ be denoted by $P$.

⇒ P = \dfrac{1}{2}\left|\begin{cases}0 & 0 & 1 \\ x & y & 1 \\ - x & y & 1\end{cases}\right|

$⇒ P = \dfrac{1}{2}\left|0 - 0 + 2xy\right|$

$⇒ P = \left|xy\right|$

$⇒ P = \left|x\left(- 2{x}^{2} + 54\right)\right|$

$⇒ P = \left|- 2{x}^{3} + 54x\right|$

$⇒ \dfrac{dP}{dx} = \left|- 6{x}^{2} + 54\right|$

For critical points, $\dfrac{dP}{dx} = 0$

$⇒ \left|- 6{x}^{2} + 54\right| = 0$

$⇒ x = \pm 3$

So, the maximum area will be,

$A = 3 ( - 2 \times 9 + 54 )$

$⇒ A = 108$