Let A be the region enclosed by the parabola y² = 2x and the line x = 24. Then the maximum area of t

Q: Wait, why is the height $2y$? Shouldn't it just be $y$?

Great question! Remember, the parabola $y^2 = 2x$ exists both above and below the x-axis. If your top vertex is at $(x, y)$, the bottom vertex is at $(x, -y)$. The total vertical distance is $y - (-y) = 2y$. Always account for the full span of the shape!

Q: Why did we choose to express everything in terms of $y$ instead of $x$?

It's all about making the calculus easier. If we used $x$, the height would be $2\sqrt{2x}$. Differentiating a square root is messy and prone to errors. By using $y$, we get a simple polynomial $48y - y^3$, which is much cleaner to differentiate.

Q: Is it possible to solve this using $x$ as the variable?

Absolutely! You would define the area as $A(x) = 2\sqrt{2x}(24 - x)$. You can differentiate this using the product rule, but be careful with the chain rule. It will lead to the same answer, but the algebra is significantly more tedious.

Question

Let $A$ be the region enclosed by the parabola $y^2 = 2x$ and the line $x = 24$. Then the maximum area of the rectangle inscribed in the region $A$ is________

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Playground

Imagine you are standing on a coordinate plane, looking at the curve $y^2 = 2x$. This is a rightward-opening parabola that starts at the origin and expands gracefully.

The space trapped between the parabola and the vertical wall at $x = 24$ is our region, $A$. Our mission is to find the largest possible rectangle that can fit inside this region.

### Defining the Rectangle

To maximize the area, we place our rectangle symmetrically about the $x$-axis. Let the top-left vertex of our rectangle be at some point $(x, y)$ on the parabola.

By the law of symmetry, the bottom-left vertex must be at $(x, -y)$. Since the right side of our rectangle is pressed against the wall at $x = 24$, the right vertices are located at $(24, y)$ and $(24, -y)$.

The height of our rectangle is the vertical distance from $y$ to $-y$, which is $2y$. The width is the horizontal distance from our point $x$ to the wall at $24$, which is $24 - x$.

### The Power of Single-Variable Calculus

The area of our rectangle is given by the product of its width and height:
$$A = (24 - x)(2y)$$

We are constrained by the parabola's equation, $y^2 = 2x$, which implies $x = \frac{y^2}{2}$. Substituting this into our area equation transforms the problem into a single-variable function:
$$A(y) = \left( 24 - \frac{y^2}{2} \right)(2y)$$

Expanding this expression, we obtain:
$$A(y) = 48y - y^3$$

### Finding the Peak

To find the maximum area, we determine the critical point of the function by taking the derivative with respect to $y$:
$$\frac{dA}{dy} = 48 - 3y^2$$

Setting the derivative to zero, we solve for $y$:
$$48 - 3y^2 = 0 \Rightarrow 3y^2 = 48 \Rightarrow y^2 = 16$$

Since $y$ represents a physical dimension, we take the positive root, $y = 4$. This is the optimal value that maximizes the area.

### Final Calculation

Now that we have the optimal $y = 4$, we substitute it back into our area function to find the maximum area:
$$A(4) = 48(4) - (4)^3 = 192 - 64 = 128$$

The **maximum area** of the rectangle inscribed in this region is **$128$ square units**.

Animated Solution for Mathematics - Differentiation: Let $A$ be the region enclosed by the parabola $y^{2} = 2 x$ and the line $x = 24$ . Then the maximum area of the rectangle inscribed in the region $A$ is________

Enter Numerical Value:

Visualized Solution

Visualizing the Parabola

Bounding the Region

Inscribing the Rectangle

Defining the Vertices

Dimensions of the Rectangle

Setting up the Area Function

Single Variable Conversion

Simplifying the Area Function

The Condition for Maximum Area

Differentiating the Function

Solving for the Critical Point

Preparing for Final Calculation

Calculating the Maximum Area

Final Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Geometry of the Playground

Defining the Rectangle

The Power of Single-Variable Calculus

Finding the Peak

Final Calculation

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