Hello Sunshine,
Because of the even symmetry of the given curve, the base $b$ of the rectangle will be $b=2x$ and the height $h$ will be $y$. And so the area of the rectangle is:
$$A=2xy$$ where $0\le x$
Substituting for $y$, we then get the area as a function of $x$:
$$A(x)=2x\left(\frac{7}{x^2+1} \right)=\frac{14x}{x^2+1}$$
Now, we want to find the critical value(s), so we equate the derivative with respect to $x$ to zero:
$$A'(x)=\frac{\left(x^2+1 \right)(14)-(14x)(2x)}{\left(x^2+1 \right)^2}=\frac{14\left(1-x^2 \right)}{\left(x^2+1 \right)^2}=0$$
And so we see that our relevant critical value is:
$$x=1$$
Using the first derivative test, we can see that on $(0,1)$ the area function is increasing and on $(1,\infty)$ the area function is decreasing, so we know our critical value is at a maximum. Hence the vertices of the rectangle of maximal area are:
$$\left(-1,0 \right),\,\left(-1,\frac{7}{2} \right),\,\left(1,0 \right),\,\left(1,\frac{7}{2} \right)$$