JEE Mathematics 2024 Question with Solution

Given: The triangle has vertices at $(0,0)$, $(x,y)$ and $(-x,y)$, where the points lie on $y = -2x^2 + 54$ and $y > 0$.

Find: The maximum possible area of the triangle and the correct option.

The base of the triangle is the distance between $(x,y)$ and $(-x,y)$, so base $= 2x$. The height is $y$.

Hence, the area is

Using $y = -2x^2 + 54$,

To maximize the area, differentiate with respect to $x$:

Set it equal to zero:

Now substitute $x = 3$ into the curve:

Therefore,

Therefore, the maximum area is $108$ square units, so the correct option is D.

The solution states "The Correct Option is B", but the working clearly gives $108$, which matches option D.

A quick verification is that the cubic $-2x^3 + 54x$ increases first and then decreases for positive $x$, so the stationary point at $x = 3$ gives the maximum area in the allowed region.

Given: The two variable vertices are symmetric about the $y$-axis, namely $(x,y)$ and $(-x,y)$, and they lie on $y = -2x^2 + 54$.

Find: The maximum area of the triangle formed with the origin.

Because the two points have the same $y$-coordinate, the segment joining them is horizontal. Its length is

The perpendicular distance from the origin to the horizontal line through these points is $y$. So

Now use the parabola equation:

Substituting,

Differentiate:

For extrema,

Since the base length is taken as $2x$ in this setup, we use the positive value $x = 3$. Now,

So the maximum area becomes

Therefore, the required maximum area is $108$.