Let C be the circle of minimum area enclosing the ellipse $$E : \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ with eccentricity $$\dfrac{1}{2}$$ and foci $$(\pm 2, 0)$$. Let PQR be a variable triangle, whose vertex P is on the circle C and the side QR of length 2a is parallel to the major axis of E and contains the point of intersection of E with the negative y-axis. Then the maximum area of the triangle PQR is:

To find the maximum area of the triangle $$PQR$$, we determine the equations of the ellipse, the enclosing circle, and the line containing the base $$QR$$.

Step 1: Find the Equation of the Ellipse $$E$$

The general equation of the ellipse is given by:

$$E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$

We are given the eccentricity $$e = \frac{1}{2}$$ and the foci at $$(\pm 2, 0)$$. The coordinates of the foci are defined as $$(\pm ae, 0)$$:

$$ae = 2 \implies a\left(\frac{1}{2}\right) = 2 \implies a = 4$$

Using the relationship $$b^2 = a^2(1 - e^2)$$:

$$b^2 = 16\left(1 - \frac{1}{4}\right) = 16 \times \frac{3}{4} = 12 \implies b = 2\sqrt{3}$$

Thus, the point of intersection of the ellipse with the negative y-axis is $$(0, -b) = (0, -2\sqrt{3})$$.

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Step 2: Determine the Circle $$C$$ of Minimum Area Enclosing $$E$$

For an ellipse where $$a^2 \le 2b^2$$, the circle of minimum area enclosing it is centered at the origin with a radius equal to the semi-major axis $$a$$.

Checking our values:

$$a^2 = 16 \quad \text{and} \quad 2b^2 = 2 \times 12 = 24$$

Since $$16 \le 24$$, the radius of the circle $$C$$ is $$R = a = 4$$.

The equation of the enclosing circle $$C$$ is:

$$x^2 + y^2 = 16$$

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Step 3: Identify the Line containing the Side $$QR$$

The side $$QR$$ is parallel to the major axis (x-axis) and passes through the point of intersection with the negative y-axis $$(0, -2\sqrt{3})$$. Therefore, the equation of the line containing $$QR$$ is:

$$y = -2\sqrt{3}$$

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Step 4: Calculate the Maximum Area of Triangle $$PQR$$

The area of a triangle is given by the formula:

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$

The base length is given as $$2a = 8$$.

The vertex $$P$$ lies on the circle $$x^2 + y^2 = 16$$, where the y-coordinates range from $$-4$$ to $$4$$. To maximize the height $$h$$ (the vertical distance from $$P$$ to the line $$y = -2\sqrt{3}$$), we pick the top-most point on the circle $$P(0, 4)$$:

$$h_{\text{max}} = 4 - (-2\sqrt{3}) = 4 + 2\sqrt{3}$$

Evaluating the maximum area expression:

$$\text{Maximum Area} = \frac{1}{2} \times 8 \times (4 + 2\sqrt{3}) = 4(4 + 2\sqrt{3}) = 16 + 8\sqrt{3} = 8(2 + \sqrt{3})$$

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Therefore, the maximum area of the triangle $$PQR$$ is equal to $$8(2 + \sqrt{3})$$.