Find the rectangle with the maximum area inside the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ 
Find the rectangle with the maximum area inside the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, whose edges are parallel to the axises. Hint: Find the function we need to maximize.

Well in the answers it says that the function we need to maximize is $4xy$ and the area is $2ab$. How do we solve this?

Another solution is:
the area of rectangle with one vertex $(x,y)$ (in the first quadrant) is $4xy$. From the equation of ellipse we can express $y$ by terms of $x$, as $y=b\sqrt{1-\frac{x^2}{a^2}}$, so the function to maximize is
$$A(x)=4x\cdot b\sqrt{1-\frac{x^2}{a^2}}$$
We can then apply the inequality of geometric and arithmetic means for the positive numbers $\frac{x^2}{a^2}$ and $1-\frac{x^2}{a^2}$ to get the solution.
