Animated Solution for Mathematics - Differentiation: The area (in sq. units) of the largest rectangle $ABCD$ whose vertices $A$ and $B$ lie on the $x$-axis and vertices $C$ and $D$ lie on the parabola, $y=x^2-1$ below the $x$-axis, is :

Imagine you are standing before a coordinate plane, looking at the curve $y=x^2-1$. This is a parabola that has been pulled downward, its vertex resting at $(0,-1)$, creating a beautiful, U-shaped valley that dips below the $x$-axis.

Our mission is to fit the largest possible rectangle into this valley. This is a classic JEE Advanced challenge, and it is a perfect example of how symmetry can turn a daunting problem into a beautiful, solvable puzzle.

When you see a shape like a parabola, your first instinct should always be to look for symmetry. Because the parabola is perfectly symmetric about the $y$-axis, any rectangle inscribed within it must also be symmetric.

Let us place the right-hand vertex $B$ at some coordinate $(t,0)$ on the $x$-axis. Because of symmetry, the left-hand vertex $A$ must be at $(-t,0)$.

Now, we drop vertical lines down to the parabola to find the other two vertices, $C$ and $D$. Since $C$ is on the curve $y=x^2-1$, its coordinates are $(t,t^2-1)$. Similarly, $D$ is at $(-t,t^2-1)$. We have now reduced the entire geometry of the rectangle to a single variable, $t$.

The width of the rectangle is the horizontal distance between $A$ and $B$, which is $t-(-t)=2t$. The height is the vertical distance from the $x$-axis to the parabola.

Since the parabola is below the $x$-axis, the $y$-coordinate is negative. To get a positive height, we take the absolute value: $h=|t^2-1|=1-t^2$.

With width and height in hand, we construct our area function:

Expanding this, we get:

To find the maximum, we turn to the power of calculus. We need to find the critical point where the rate of change of the area is zero. We differentiate $A(t)$ with respect to $t$:

Setting this derivative to zero, we get $2-6t^2=0$, which simplifies to $6t^2=2$, or $t^2=\frac{1}{3}$. Taking the square root, we find $t=\frac{1}{\sqrt{3}}$. We reject the negative root because $t$ represents a physical distance.

We verify that this is a maximum using the second derivative test: $A^{\prime \prime }(t)=-12t$. Plugging in $t=\frac{1}{\sqrt{3}}$, we get $A^{\prime \prime }(\frac{1}{\sqrt{3}})=-12(\frac{1}{\sqrt{3}})<0$.

Since the second derivative is negative, the function is concave down, confirming that we have found the maximum. Finally, we calculate the area by substituting $t=\frac{1}{\sqrt{3}}$ back into our area function:

The maximum area is $\frac{4}{3\sqrt{3}}$ square units. This problem teaches us that no matter how complex a shape may seem, there is always an underlying structure—in this case, symmetry—that can simplify the path to the solution.