Hello Khegan McLane,
I assume you meant to give the parabola as $y=36-x^2$. Let's take a look at a rectangle so inscribed:
View attachment 1679
a.) The base of the rectangle is $2x$ and the height is $y=36-x^2$, hence the area $A$ of the rectangle is:
$$A(x)=2x\left(36-x^2 \right)=72x-2x^3$$
b.) In order for the height of the rectangle to be non-negative, we require:
$$-6\le x\le6$$
and in order for the base to be non-negative, we require:
$$0\le x$$
Thus, in order to satisfy both conditions, we require:
$$0\le x\le6$$
c.) Here is a plot of the area function on the relevant domain:
View attachment 1680
d.) Looking at the plot, it appears the maximum occurs when $x$ is a little more than 3.4. Using a bit of differential calculus, we can find the exact value.
Differentiating the area function with respect to $x$ and equating the result to zero, we obtain:
$$A'(x)=72-6x^2=6\left(12-x^2 \right)=0$$
The critical value in the domain is:
$$x=\sqrt{12}=2\sqrt{3}$$
The second derivative is:
$$A''(x)=-12x$$
Since our critical value is positive, the second derivative at that value is negative, demonstrating that this critical value is at a relative maximum. Thus, we conclude that the area of the rectangle is maximized for:
$$x=2\sqrt{3}\approx3.46410161514$$
