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

A Journey Through the Parabola

Welcome, future engineer! Today, we are not just solving a problem; we are exploring the elegant dance between geometry and calculus.

Imagine you are standing on the Cartesian plane, looking at a beautiful, downward-opening parabola defined by the equation $y=-2x^2+54$. We are tasked with inscribing a triangle inside this curve, with one vertex anchored firmly at the origin $(0,0)$ and the other two vertices, $A$ and $B$, resting on the parabola itself.

Our goal is to find the maximum possible area of this triangle.

Visualizing the Symmetry

The first step in any JEE problem is to see the hidden structure. Because our parabola is symmetric about the $y$-axis, if we place one vertex at $(x,y)$, the other must be at $(-x,y)$ to keep the triangle balanced.

This symmetry is our best friend! It tells us that the base of our triangle, the segment $AB$, is a horizontal line. The length of this base is simply the distance between $x$ and $-x$, which is $2x$.

The height of our triangle is the vertical distance from the $x$-axis to the points $A$ and $B$, which is simply the $y$-coordinate.

The Algebra of Area

Now, let us translate this into the language of mathematics. We know the area of a triangle is given by the formula:

Substituting our values, we get $Area=\frac{1}{2}\times (2x)\times y$. The factor of $\frac{1}{2}$ and the $2$ cancel out beautifully, leaving us with $Area=xy$.

But we have a problem: the area depends on both $x$ and $y$. We need it in terms of a single variable. Since $y=-2x^2+54$, we substitute this into our area equation:

Expanding this, we arrive at our area function:

The Calculus of Maxima

We have a cubic function, and we want to find its peak. This is where calculus shines. To find the maximum, we calculate the derivative $\frac{dA}{dx}$ and set it to zero.

Differentiating $A(x)=54x-2x^3$ with respect to $x$, we get:

Setting this to zero, we solve $54-6x^2=0$, which leads to $6x^2=54$, or $x^2=9$. This gives us $x=3$ (we ignore $x=-3$ because we are working in the first quadrant where $x>0$).

With our optimal $x=3$, we find the maximum area by plugging it back into our area function:

This simplifies to $162-2(27)=162-54=108$.

We have arrived at our destination: the maximum area is 108 square units. Notice how the math guided us from a simple geometric sketch to a precise numerical truth. This is the power of the JEE toolkit—geometry to visualize, algebra to simplify, and calculus to conquer.