JEE Main 2024 — 30 January, Shift 1 — Question 5

The given curve is $y = -2x^2 + 54,\quad y>0$

The variable points on the curve are

$$(x,y)\ \text{and}\ (-x,y)$$

The triangle has vertices $(0,0)$, $(x,y)$, and $(-x,y)$.

$$\text{Base} = 2x,\quad \text{Height} = y$$ $$\text{Area } A = \frac{1}{2}(2x)(y) = xy$$

Substitute $y = -2x^2 + 54$:

$$A(x) = x(-2x^2 + 54) = -2x^3 + 54x$$

Differentiate:

$$\frac{dA}{dx} = -6x^2 + 54$$

Set derivative equal to zero:

$$-6x^2 + 54 = 0 \Rightarrow x^2 = 9 \Rightarrow x = 3$$

Find $y$:

$$y = -2(3)^2 + 54 = 36$$

Maximum area:

$$A_{\max} = xy = 3 \times 36 = \boxed{108}$$