Let the length of the latus rectum of an ellipse $$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$$ (where a > b) be 30. If its eccentricity is the maximum value of the function $$f(t) = -\frac{3}{4} + 2t - t^{2}$$, then the value of $$(a^{2} + b^{2})$$ is equal to:

$$f(t) = -t^2 + 2t - \frac{3}{4}$$

Since the coefficient of $$t^2$$ is negative ($-1$), this parabola opens downward, meaning its maximum value occurs at its vertex. The $t$-coordinate of the vertex is:

$$t = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1$$

Now, we substitute t = 1 back into f(t) to find the maximum value, which is our eccentricity e:

$$e = f(1) = -(1)^2 + 2(1) - \frac{3}{4}$$ $$e = -1 + 2 - \frac{3}{4} = 1 - \frac{3}{4} = \frac{1}{4}$$

For an ellipse where a > b, the eccentricity is related to the semi-major axis a and semi-minor axis b by the formula:

$$e^2 = 1 - \frac{b^2}{a^2}$$

$$\left(\frac{1}{4}\right)^2 = 1 - \frac{b^2}{a^2}$$ $$\frac{1}{16} = 1 - \frac{b^2}{a^2}$$ $$\frac{b^2}{a^2} = 1 - \frac{1}{16} = \frac{15}{16}$$

$$b^2 = \frac{15}{16}a^2$$

The length of the latus rectum for the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ (a > b) is given by $$\frac{2b^2}{a}$$. We are given that this length is 30:

$$\frac{15}{16}a^2 = 15a$$

$$b^2 = 15a = 15(16) = 240$$

$$a^2 + b^2 = 256 + 240 = 496$$