What is the biggest area of a triangle can be drawn inside the ellipse equation : $\displaystyle \left( {\frac{x}{3}} \right)^2 + \left( {\frac{y}{7}} \right)^2 = 1
$
For the circle $\displaystyle x^2+y^2=a^2$, the largest triangle that can be drawn inside it is the equilateral triangle with side $\displaystyle a\sqrt3$, whose area is $\displaystyle \tfrac{3\sqrt3}4a^2$.
The ellipse $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is obtained from the circle by the linear transformation with matrix $\displaystyle \begin{bmatrix}1&0\\0&b/a\end{bmatrix}$. (In non-technical language, you get the ellipse from the circle by leaving the x-coordinate unchanged and expanding or shrinking the y-coordinate by a factor b/a.) This transformation alters areas by a factor b/a. So the largest triangle that can be inscribed in the ellipse has area $\displaystyle \tfrac{3\sqrt3}4ab$, or $\displaystyle 63\sqrt3/4$ square units when a=3 and b=7.
If I may, here is another way using polar coordinates.
The area of the triangle can be expressed as
$\displaystyle A=\frac{1}{2}(2acos{\theta})(b+bsin{\theta})$
That is just $\displaystyle \frac{1}{2}\text{base}\cdot \text{height}$
Differentiate, set to 0 and solve for theta gives us $\displaystyle \theta=\frac{\pi}{6}$
Sub this into A and we get $\displaystyle A=\frac{3\sqrt{3}ab}{4}$
That method works fine provided that you assume that the base of the triangle is horizontal. What I found interesting (and unexpected) about this problem is that the same maximum area $\displaystyle 3\sqrt{3}ab/4$ can be achieved with a triangle whose base has any given orientation. That follows easily if you think of the ellipse as a "squashed circle", but I think it would be hard to prove by any other method.