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?
Consider the first quadrant for simplicity.
Let $x = a\cos \theta$, $y =b\sin \theta$, then the area of the rectangle in the first quadrant is $$ A = xy = \frac12ab\sin(2\theta) $$ As $0\le\sin(2\theta)\le 1$, we can see the maximum area of $A_{\max}$ is $\frac12ab$. Because of the axial symmetry of the ellipse about both $X$ and $Y$ axis, the maximum area of the rectangle inside the ellise would be $4A_{\max}$, that is, $2ab$.