The maximum area of a triangle whose one vertex is at ( 0 , 0 ) and the other two vertices lie on the curve y = - 2 x 2 + 54 at points ( x , y ) and ( - x , y ) where y > 0 is :

Step-by-step solution

Let, area of triangle A B C be denoted by P . ⇒ P = 1 2 0 0 1 x y 1 - x y 1 ⇒ P = 1 2 0 - 0 + 2 x y ⇒ P = x y ⇒ P = x - 2 x 2 + 54 ⇒ P = - 2 x 3 + 54 x ⇒ d P d x = - 6 x 2 + 54 For critical points, d P d x = 0 ⇒ - 6 x 2 + 54 = 0 ⇒ x = ± 3 So, the maximum area will be, A = 3 ( - 2 × 9 + 54 ) ⇒ A = 108