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To solve the problem of finding the maximum area of a rectangle under a curve, we'll use a common example (a rectangle inscribed under the parabola (y = 16 - x^2)) to demonstrate the step-by-step process.

Problem Statement

Find the maximum area of a rectangle with its base on the x-axis and upper vertices on the parabola (y = 16 - x^2).

Step 1: Model the Area Function

The parabola (y = 16 - x^2) is symmetric about the y-axis. Let the rectangle’s base span from (-x) to (x) (so half the base is (x)).

• Base length: (2x)
• Height: (y = 16 - x^2)
• Area function: (A(x) = \text{base} \times \text{height} = 2x(16 - x^2) = 32x - 2x^3)

Step 2: Find Critical Points

Take the first derivative of (A(x)) and set it to zero to find critical points:
[A'(x) = 32 - 6x^2]
Set (A'(x) = 0):
[32 - 6x^2 = 0 \implies x^2 = \frac{32}{6} = \frac{16}{3} \implies x = \frac{4}{\sqrt{3}} \quad (\text{since } x > 0)]

Step 3: Confirm Maximum

Take the second derivative to check if the critical point is a maximum:
[A''(x) = -12x]
For (x = \frac{4}{\sqrt{3}}), (A''(x) < 0), so this is a maximum.

Step 4: Calculate Maximum Area

Substitute (x = \frac{4}{\sqrt{3}}) into (A(x)):
[A\left(\frac{4}{\sqrt{3}}\right) = 2\left(\frac{4}{\sqrt{3}}\right)\left(16 - \frac{16}{3}\right) = \frac{8}{\sqrt{3}} \times \frac{32}{3} = \frac{256\sqrt{3}}{9} \approx 50.4]

General Approach

For any curve (y = f(x)):

1. Model the area as a function of one variable (use symmetry if possible).
2. Compute the first derivative and find critical points.
3. Verify maximum using the second derivative test.
4. Substitute critical value to get the maximum area.

Answer: (\boxed{\dfrac{256\sqrt{3}}{9}}) (for the parabola example). Adjust based on the actual curve in the image.
If the curve was different (e.g., semicircle (x^2 + y^2 = r^2)), the maximum area would be (r^2).

Let me know if you need to adjust for the specific curve in your image!
(\boxed{50.4}) (approximate value for the example) or the exact form as above.

But for the exact answer in the example: (\boxed{\dfrac{256\sqrt{3}}{9}}) is the precise form.

Final Answer
(\boxed{\dfrac{256\sqrt{3}}{9}})

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