Question

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 \(\mathrm{y}>0\) is :

(1) 88

(2) 122

(3) 92

(4) 108

Answer 1

Correct option is (4) 108

\(A = xy\)

\(A = x(-2x^2 + 54)\)

\(A = -2x^3 + 54x\) 

\(\frac{dA}{dx} = - 6x^2 + 54 = 0\)

\(x = 3, x = -3\)

\(A_{max} = 3 (-2 \times 9 + 54)\)

\(= 3(54 - 18)\)

\(= 108\)