Key Concept: Optimization using Differential Calculus

This problem is a classic application of optimization using differential calculus. The fundamental goal is to find the maximum or minimum value of a quantity (in this case, the area of a rectangle) that depends on one or more variables. The systematic approach involves:

1. Formulating an Objective Function: Expressing the quantity to be optimized as a function of a single independent variable. This is often achieved by using given constraints to eliminate other variables.
2. Determining the Valid Domain: Identifying the range of values for the independent variable that are physically or mathematically meaningful within the context of the problem.
3. Finding Critical Points: Using the first derivative test by setting the derivative of the objective function to zero. Critical points are potential locations for local maxima or minima where the function's slope is horizontal.
4. Verifying the Nature of Critical Points: Employing the second derivative test (or analyzing the sign change of the first derivative) to confirm whether a critical point corresponds to a local maximum, local minimum, or neither.
5. Evaluating the Objective Function: Substituting the optimal value of the variable back into the original objective function to find the maximum or minimum value.

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Step-by-Step Derivation:

1. Analyze the Parabola and Visualize the Rectangle Geometry

The given equation of the parabola is $y = 12 - x^2$.

• Parabola Properties:

  • This is an inverted parabola because of the negative coefficient of $x^2$. It opens downwards.
  • Its vertex is at $(0, 12)$ (obtained by setting $x=0$).
  • Its x-intercepts are found by setting $y=0$: $0 = 12 - x^2 \implies x^2 = 12 \implies x = \pm \sqrt{12} = \pm 2\sqrt{3}$ So, the parabola intersects the x-axis at $(-2\sqrt{3}, 0)$ and $(2\sqrt{3}, 0)$.
  • The parabola is symmetric about the y-axis.

• Rectangle Placement: The problem states that the rectangle has its base on the x-axis and its other two vertices on the parabola.

  • Why symmetry is crucial: Due to the parabola's symmetry about the y-axis, for the rectangle to have the maximum possible area with its base on the x-axis, it must also be symmetric about the y-axis. This simplifies the setup significantly.
  • Let the coordinates of the two upper vertices on the parabola be $(-x, y)$ and $(x, y)$.
    • By choosing symmetric points $(-x, y)$ and $(x, y)$ (where $x > 0$), we ensure the rectangle is centered on the y-axis.
    • Since the base is on the x-axis, the bottom vertices will be $(-x, 0)$ and $(x, 0)$.

• Defining Dimensions:

  • The width of the rectangle will be the distance between the x-coordinates of its upper vertices: $x - (-x) = 2x$.
  • The height of the rectangle will be the y-coordinate of the upper vertices, which is $y$.

2. Formulate the Area Function in Terms of a Single Variable

The vertices $(x, y)$ lie on the parabola, so their coordinates must satisfy the parabola's equation: $y = 12 - x^2$

Now, we can express the area of the rectangle, $A$, using its width and height: $A = \text{width} \times \text{height}$ $A(x) = (2x) \times y$

• Why this step? To use calculus for optimization, we need the area function to be expressed in terms of a single independent variable. We achieve this by substituting the expression for $y$ from the parabola's equation into the area function. This transforms a multi-variable problem into a single-variable problem.

Substitute $y = 12 - x^2$ into the area function: $A(x) = 2x(12 - x^2)$ Expanding this, we get our objective function: $A(x) = 24x - 2x^3$

3. Determine the Valid Domain for x

For the rectangle to be physically valid and lie inside the parabola as described:

• The width $2x$ must be positive, which implies $x > 0$.

• The height $y$ must be positive (since the base is on the x-axis and the upper vertices are on the parabola above the x-axis). $y = 12 - x^2 > 0$ $12 > x^2$ $x^2 < 12$ $-\sqrt{12} < x < \sqrt{12}$ $-\sqrt{12} < x < 2\sqrt{3}$

• Why define the domain? Defining the domain is crucial because it ensures that any critical points we find are physically meaningful within the context of the problem. It restricts the possible values of $x$ and helps in identifying potential boundary maxima/minima (though not relevant in this specific problem as the maximum occurs within the open interval).

Combining the conditions $x > 0$ with $x < 2\sqrt{3}$, the valid domain for $x$ is $0 < x < 2\sqrt{3}$.

4. Find Critical Points Using the First Derivative Test

To find the value of $x$ that maximizes $A(x)$, we take the first derivative of $A(x)$ with respect to $x$ and set it to zero. $A(x) = 24x - 2x^3$ Differentiate $A(x)$ with respect to $x$: $A'(x) = \frac{d}{dx}(24x - 2x^3) = 24 - 6x^2$

• Why this step? Setting the first derivative to zero allows us to find the points where the function's slope is horizontal. These "critical points" are candidates for local maxima or minima because the function momentarily stops increasing or decreasing at these points.

Set the first derivative to zero to find the critical points: $A'(x) = 0$ $24 - 6x^2 = 0$ $6x^2 = 24$ $x^2 = \frac{24}{6}$ $x^2 = 4$ $x = \pm 2$

Since $x$ represents half the width of the rectangle and must be positive (as established in the domain $0 < x < 2\sqrt{3}$), we take $x = 2$.

5. Verify Maximum Using the Second Derivative Test

To rigorously confirm that $x=2$ corresponds to a maximum area, we use the second derivative test. Calculate the second derivative of $A(x)$: $A''(x) = \frac{d}{dx}(24 - 6x^2) = -12x$

• Why use the second derivative test? The second derivative test provides a definitive way to distinguish between a local maximum and a local minimum.
  • If $A''(x) < 0$ at a critical point, it indicates a concave-down shape, confirming a local maximum.
  • If $A''(x) > 0$, it indicates a concave-up shape, confirming a local minimum.
  • If $A''(x) = 0$, the test is inconclusive, and one would typically use the first derivative sign change test.

Now, evaluate the second derivative at the critical point $x=2$: $A''(2) = -12(2) = -24$

Since $A''(2) = -24 < 0$, the function $A(x)$ has a local maximum at $x=2$.

6. Calculate the Maximum Area

Substitute the optimal value of $x=2$ back into the original area function $A(x) = 24x - 2x^3$: $A(2) = 24(2) - 2(2)^3$ $A(2) = 48 - 2(8)$ $A(2) = 48 - 16$ $A(2) = 32$

• Why substitute back into the original area function? The derivatives help us find the value of x that optimizes the area. To find the actual maximum area, we must plug this optimal $x$ value back into the function that defines the area, not its derivatives.

Thus, the maximum area of the rectangle is 32 square units.

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Tips and Common Mistakes to Avoid:

1. Incorrect Setup of Area Function: Always double-check your algebraic formulation of the area. Ensure it correctly represents the width and height based on your chosen variables and the problem's geometry. A common error is using $x$ as the width instead of $2x$ for a symmetric figure.
2. Ignoring the Domain: Always consider the practical domain for your variable. Forgetting that $x$ must be positive or that the height $y$ must be positive can lead to extraneous solutions (like $x=-2$ in this case) or overlooking boundary conditions that might yield the maximum/minimum.
3. Not Confirming Max/Min: While setting the first derivative to zero identifies critical points, it doesn't guarantee a maximum. Always use the second derivative test or analyze the sign change of the first derivative to confirm the nature of the critical point.
4. Substituting into the Wrong Function: After finding the optimal value of the variable (e.g., $x=2$), remember to substitute it back into the original objective function (the area function, $A(x)$) to find the maximum value, not into the derivative.
5. Units: Always include the correct units in your final answer if specified (e.g., "sq. units").

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Summary and Key Takeaway:

To find the maximum area of a geometric figure constrained by another function:

1. Visualize and Parameterize: Draw the scenario and define the dimensions of the figure in terms of a variable, often leveraging symmetry to simplify.
2. Formulate Objective Function: Express the quantity to be maximized (area, volume, etc.) as a function of a single variable, using any given constraint equations (like the parabola equation here).
3. Determine Domain: Identify the valid range for your variable based on the problem's physical or geometric constraints.
4. Differentiate and Find Critical Points: Take the first derivative of the objective function and set it to zero to find potential maximum or minimum points.
5. Verify (Optional but Good Practice): Use the second derivative test (or first derivative sign change analysis) to confirm whether the critical point yields a maximum or minimum.
6. Calculate Final Value: Substitute the optimal variable value back into the original objective function to obtain the maximum (or minimum) value.

The final answer is $\boxed{\text{32}}$.