Question: Find the area of the largest rectangle contained within the region defined as $y \geq 2x^2$ and $y \leq 1-x$.
Attempts: I attempted to create a line parallel to $y = 1-x$ by taking $y = a-x$ with $a \in \left(0 , 1\right)$. These two lines form the sides of the rectangle. The other sides will be perpendicular to $y=1-x$ at $\left( \frac{1}{2},\frac{1}{2}\right)$, that is $y=x$, and through the intersection point of $y = 2x^2$ and $y = a-x$ on the opposite side of the parabola. The area can be found in terms of $a$ and maximised at $a = \frac{1}{4}$, giving the total area of the rectangle as $\frac{3}{32}(3 + 2\sqrt{3})$.
While I believe my working is fine for the method I've chosen, I'm not sure if this method ensures a maximum possible area. Is this approach justifiable? Is there another technique I can use instead? Any help or guidance would be greatly appreciated!