I am using these as references:
Problem: Find the maximum area of a rectangle inside a parabola whose equation is $f(x)=-x^2+4x+3$
- Determining the rough shape of the parabola (by determining the highest point of the graph which is $(2,7)$ and the y-intercept which is $y=3$), and placing the rough location of the rectangle inside the parabola.
- Shifting the parabola to the left so that the $Y$ axis becomes the axis of the parabola (i.e. $f(x)=f(x+(−b/2a))$). The shift results with a new equation of the parabola, which is $f(x)=-x^2+7$. Graphing it.
- From reference 2, the area's equation is $A(x)=2(x-(-b/2a))(-x^2+7)$ (I suspect this is where my mistake lies) where $-b/2a = 2$, the equation turns out to be a cubic equation $A(x)=-2x^3+4x^2+14x-28$
- $A'(x)=0$ (to obtain the maximum $A(x)$), with roots $x=-2/3$ and $x=2$
- Substitute $x=2$ to $A(x)$, $A(2)=-2(8)+4(4)+28-28$ which turns out to be $0$
Thank you for the help...