Consider the attached figure. The coordinates of the points are
[tex]A=(-x,0),\quad B=(-x,f(-x)),\quad C=(x,f(x)),\quad D=(x,0)[/tex]
Since f(x) is even, we have
[tex]f(x)=f(-x)=12-x^2[/tex]
So, the updated coordinates are
[tex]A=(-x,0),\quad B=(-x,12-x^2)),\quad C=(x,12-x^2),\quad D=(x,0)[/tex]
This implies that the rectangle has area
[tex]A(x)=AD\cdot AB = 2x(12-x^2) = -2x^3+24x[/tex]
And we want to maximize this function. To so do, let's compute its derivative:
[tex]A'(x)=-6x^2+24=-6(x^2-4)[/tex]
This equals 0 if
[tex]-6(x^2-4)=0 \iff x^2-4=0 \iff x=\pm 2[/tex]
Given the behaviour of the area function (cubic polynomial with negative leading coefficient), the first point is a minimum, and the second point is a maximum. So, for x=2, we have
[tex]A=(-2,0),\quad B=(-2,8)),\quad C=(2,8),\quad D=(2,0)[/tex]
Which yields an area of
[tex]A(x)=AD\cdot AB = 4\cdot 8=32[/tex]
