The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two verti

Q: Wait, why is the base $2x$ and not just $x$?

Great question! Remember that the rectangle is symmetric about the y-axis. If the right vertex is at $(x, 0)$, the left vertex is at $(-x, 0)$. The distance between $x$ and $-x$ is $x - (-x) = 2x$. Always look for that symmetry!

Q: Can I use the AM-GM inequality instead of calculus?

Yes, you absolutely can! For $A(x) = 2x(12 - x^2)$, you can rewrite it as $A(x) = 4 \cdot x \cdot (6 - \frac{x^2}{2})$. However, calculus is the most robust tool for optimization in JEE, and it's safer to stick to the derivative method to avoid missing any constraints.

Q: Why do we reject the negative root when solving $\frac{dA}{dx} = 0$?

Because $x$ represents a physical distance from the origin to the right edge of the rectangle. A distance cannot be negative. Therefore, we only consider the positive root $x = 2$.

Question

The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, $y = 12-x^2$ such that the rectangle lies inside the parabola, is :-

Options:

(A) $20\sqrt{2}$

(B) $18\sqrt{3}$

(C) 32

(D) 36

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Parabola

Welcome, future engineer! Today, we are going to tackle a classic JEE Advanced problem that perfectly blends geometry and calculus. We aren't just solving for a number; we are uncovering the hidden relationship between a curve and the shapes we can carve out of it.

Imagine you are standing before the parabola $y = 12 - x^2$. It is a beautiful, downward-opening curve, symmetric about the $y$-axis, with its peak at $(0, 12)$ and roots at $\pm \sqrt{12}$. Our goal is to fit the largest possible rectangle inside this shape, with its base resting on the $x$-axis.

### Defining the Anatomy of the Rectangle

To maximize the area, we first need to define it. Let's place the right-hand vertices of our rectangle at $(x, 0)$ on the $x$-axis and $(x, 12 - x^2)$ on the parabola.

Because the parabola is symmetric, the left-hand vertices must be at $(-x, 0)$ and $(-x, 12 - x^2)$. This symmetry is our greatest ally.

The base of our rectangle is the distance between $x$ and $-x$, which is $2x$. The height is simply the $y$-coordinate of the parabola, $12 - x^2$. With these, we have our area function:

$$A(x) = 2x(12 - x^2)$$

### The Calculus Engine

Now, we expand the function to get $A(x) = 24x - 2x^3$. This is the function we need to maximize.

In the world of JEE, whenever you see "maximum" or "minimum," your mind should immediately jump to the derivative. We want to find the rate of change of the area with respect to $x$ and see where it vanishes—that is, where the slope of the area function is zero.

Differentiating $A(x)$ with respect to $x$, we get:

$$\frac{dA}{dx} = 24 - 6x^2$$

### Finding the Peak

Setting $\frac{dA}{dx} = 0$ gives us $24 - 6x^2 = 0$, which simplifies to $6x^2 = 24$, or $x^2 = 4$. This yields $x = \pm 2$.

Since $x$ represents a physical length, we discard the negative value and keep $x = 2$. This is our critical point. It is the specific width that balances the base and the height to produce the largest possible area.

### The Final Calculation

Finally, we substitute $x = 2$ back into our area function:

$$A(2) = 2(2)(12 - 2^2) = 4(12 - 4) = 4(8) = 32$$

The **maximum area is 32 square units**. It is elegant, isn't it? By using the power of calculus, we turned a geometric constraint into a simple algebraic problem.

Always remember: in optimization, visualize the symmetry, define your variables, and let the derivative guide you to the answer. You've got this!

Animated Solution for Mathematics - Differentiation: The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, $y = 12 - x^{2}$ such that the rectangle lies inside the parabola, is :-

Select Answer:

Visualized Solution

Visualizing the Parabola $y = 12 - x^{2}$

Defining Rectangle Vertices

Identifying Base and Height

Formulating the Area Function $A (x)$

Expanding the Area Function

Differentiating for Maxima

Finding the Critical Point

Solving for $x$

Substituting $x$ back into the Area formula

Final Calculation

Summary and Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Geometry of the Parabola

Defining the Anatomy of the Rectangle

The Calculus Engine

Finding the Peak

The Final Calculation

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