Key Concepts and Formulas

• Optimization using Derivatives: Finding maxima and minima of a function by finding critical points where the first derivative is zero or undefined.
• Area of a Rectangle: Area = width × height
• Second Derivative Test: If $f'(c) = 0$ and $f''(c) > 0$, then $f(c)$ is a local minimum. If $f'(c) = 0$ and $f''(c) < 0$, then $f(c)$ is a local maximum.

Step-by-Step Solution

Step 1: Understand the Region and Inscribed Rectangle

The region A is bounded by the parabola $y^2 = 2x$ and the line $x = 24$. We want to find the maximum area of a rectangle inscribed in this region. Due to the symmetry of the parabola with respect to the x-axis, we can assume the rectangle is also symmetric with respect to the x-axis.

Step 2: Define the Dimensions of the Rectangle

Let the vertices of the rectangle on the parabola be $(\frac{y^2}{2}, y)$ and $(\frac{y^2}{2}, -y)$, where $y > 0$. The other two vertices will lie on the line $x = 24$, namely $(24, y)$ and $(24, -y)$.

• The height of the rectangle is the distance between $(24, y)$ and $(24, -y)$, which is $2y$.
• The width of the rectangle is the horizontal distance between the line $x = 24$ and the parabola, which is $24 - \frac{y^2}{2}$.

Step 3: Formulate the Area Function

The area of the rectangle, $A(y)$, is given by the product of its width and height: $A(y) = \left(24 - \frac{y^2}{2}\right)(2y) = 48y - y^3$ We express the area as a function of a single variable $y$, which represents half the height of the rectangle.

Step 4: Determine the Domain of y

Since the width of the rectangle must be positive, we have: $24 - \frac{y^2}{2} > 0$ $48 > y^2$ $-\sqrt{48} < y < \sqrt{48}$ Since $y>0$, we have $0 < y < \sqrt{48} = 4\sqrt{3}$.

Step 5: Find the Critical Points

To find the maximum area, we need to find the critical points of $A(y)$ by taking its first derivative with respect to $y$ and setting it to zero: $\frac{dA}{dy} = \frac{d}{dy}(48y - y^3) = 48 - 3y^2$ Setting the derivative to zero: $48 - 3y^2 = 0$ $3y^2 = 48$ $y^2 = 16$ $y = \pm 4$ Since $y > 0$, we take $y = 4$. This value is in our domain $0 < y < 4\sqrt{3}$.

Step 6: Verify that the Critical Point Corresponds to a Maximum

We can use the second derivative test: $\frac{d^2A}{dy^2} = \frac{d}{dy}(48 - 3y^2) = -6y$ Evaluating the second derivative at $y = 4$: $\frac{d^2A}{dy^2}\Big|_{y=4} = -6(4) = -24$ Since the second derivative is negative at $y = 4$, this confirms that $y = 4$ corresponds to a local maximum.

Step 7: Calculate the Maximum Area

Substitute $y = 4$ back into the area function $A(y)$: $A(4) = 48(4) - (4)^3 = 192 - 64 = 128$

Therefore, the maximum area of the rectangle inscribed in the region A is 128 square units.

Common Mistakes & Tips

• Sign Errors: Be careful with signs when differentiating and solving equations.
• Domain Restrictions: Always consider the domain of the variable to ensure the solution makes sense geometrically. For example, width and height must be positive.
• Symmetry: Exploit symmetry to simplify the problem.

Summary

By defining the rectangle's dimensions in terms of a single parameter $y$, we transformed the geometric problem into an algebraic optimization problem solvable with calculus. We found the critical point by setting the first derivative of the area function to zero and verified that it corresponds to a maximum using the second derivative test. The maximum area is 128 square units.

Final Answer

The final answer is \boxed{128}.