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 [2024]

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Solution

Q.
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 [2024]

1 108

2 122

3 88

4 92

Ans.
(1)

Let ABC be the required triangle whose area is A.

$∴$ $A = \frac{1}{2}$ $| 0 (y - y) + x (y - 0) - x (0 - y) |$

$= \frac{1}{2}$ $| x y + x y |$ $= \frac{1}{2} (2 x y) = x y$

$= x (- 2 x^{2} + 54) = - 2 (x^{3} - 27 x)$

$\Rightarrow A = - 2 (x^{3} - 27 x) \Rightarrow \frac{d A}{d x} = - 6 x^{2} + 54$

For critical point, $\frac{d A}{d x} = 0$

$\Rightarrow 3 x^{2} - 27 = 0 \Rightarrow x^{2} = 9 \Rightarrow x = \pm 3$

Now, $\frac{d^{2} A}{d x^{2}} = - 12 x,$ $\frac{d^{2} A}{d x^{2}} |_{x = 3}$ $< 0$

$∴$ Maximum area, $A = - 2 (27 - (27 \times 3))$

$= - 2 \times (- 54) = 108 sq. units.$

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