What is the maximum area of a triangle that can be inscribed in an ellipse with semi-axes $a$ and $b$?

$\text{(A) }ab\frac{3\sqrt{3}}{4}\qquad\text{(B) }ab\qquad\text{(C) }ab\sqrt{2}\qquad\text{(D) }\left(a+b\right)\frac{3\sqrt{3}}{4}\qquad\text{(E) }\left(a+b\right)\sqrt{2}$
1. Consider an ellipse with semi-major axis \(a\) and semi-minor axis \(b\). We need to find the maximum area of a triangle that can be inscribed in this ellipse.
2. To simplify the problem, we can use an affine transformation to map the ellipse to a circle. This transformation will preserve the ratios of areas.
3. Suppose the ellipse is transformed into a circle with radius \(r\). For an ellipse, the transformation scales the \(x\)-coordinates by \(a\) and the \(y\)-coordinates by \(b\). Thus, the radius of the circle is \(r = \sqrt{ab}\).
4. The maximum area of a triangle inscribed in a circle is achieved by an equilateral triangle. The area \(A\) of an equilateral triangle with side length \(s\) is given by:
   \[
   A = \frac{\sqrt{3}}{4} s^2
   \]
5. The side length \(s\) of the equilateral triangle inscribed in a circle of radius \(r\) is:
   \[
   s = r \sqrt{3}
   \]
6. Substituting \(r = \sqrt{ab}\) into the side length formula, we get:
   \[
   s = \sqrt{ab} \sqrt{3} = \sqrt{3ab}
   \]
7. Now, substituting \(s = \sqrt{3ab}\) into the area formula for the equilateral triangle, we get:
   \[
   A = \frac{\sqrt{3}}{4} (\sqrt{3ab})^2 = \frac{\sqrt{3}}{4} \cdot 3ab = \frac{3\sqrt{3}}{4} ab
   \]
8. Therefore, the maximum area of a triangle that can be inscribed in the ellipse is:
   \[
   \boxed{ab \frac{3\sqrt{3}}{4}}
   \]