Let the function $y(x)$ be defined:
$y(x)=-\frac{h}{36^2}x^2+h$
This function is being restricted to the range of $y(x)\leq 0$. The important piece here is determining when $y(x)$ is zero:
$$
\begin{align}
y(x)&=0\\
\therefore x&=\pm 36
\end{align}
$$
(This was included in your photo.) We can now make specifications on $y(x)$. That is, $y(x)$ is defined on the interval $[-36,36]$. The rectangles will be bounded by two points of the form:
$$P_i=(x_i,y(x_i)):y(x_i)=2.5i$$
That is, they will be bounded by two points where the $y$ coordinate (of both points) is a multiple of $2.5$. We can thus see $x_i$ in terms of the $y$ coordinate is:
$$\begin{align}
y(x_i)&=2.5i\\
\therefore x_i&=\pm\sqrt{1296-\frac{3240i}{h}}
\end{align}$$
The length of the $i$th rectangle will be the distance between $(x_i,y(x_i))$ and $(-x_i,y(x_i))$ Since the distance is a horizontal line, the length is simply $|2x_i|$. Thus, we come to the following statement for the total area of the rectangles:
$$A(r)=5\sum_{i=1}^{r}|x_i| \quad r=\text{ the number of rectangles}$$
Determining $r$ is essentially the main crux here. I believe involving calculus is where this gets us to. Taking the definite integral with respect to $x$ would give us the total area of the parabola (dependent upon $h$):
$$\int_{-36}^{36} y(x) dx=\left[\int y(x) dx\right]^{36}_{-36}=\left[\frac{-h}{3888}x^{3}+hx+C \right]^{36}_{-36}=48h$$
To determine an upperbound on $r$, one must establish the $r$ at which
$$A(r)<48h$$ is no longer true. Manipulating this algebraically gives us:
$$\frac{|x_1|+\dots+|x_r|}{h}=\frac{1}{h^2}\sum_{i=1}^{r}\sqrt{1296h^2-hi}<9.6$$
So, for a given $h$, you can determine $1$) the upperbound on the amount of rectangles and $2$) the total area of all the rectangles (presuming you use the upperbound as the amount of rectangles).
I can't see how to take this any further, so this is the best answer I think I can provide.
Ok what you want to do here is this. Draw a rectangle with width 17 and height 11. Now circumscribe a rectangle around that rectangle of uniform width. Assign the variable $x$ as being the uniform width on the path between your inner and outer rectangle. You need to draw this.
What you should be looking at now is an 11 by 17 inner rectangle. The outer rectangle will have width length $17+x+x=17+2x$ (don't need length, but make sure your diagram is right). The height length will be $11+x+x=11+2x$. Note the lengths $x$ occur on either end (these are the corners).
Ok you are almost there. There are at least 5 ways to solve this, it just depends upon how you think. The correct diagram is essential here.
One way would be to take the outer rectangular area, subtract the inner rectangular area, set that all equal to 165, and solve for $x$. In this case your equation will be
$$(11+2x)(17+2x)-17 \cdot 11 =165.$$
You could also try an additive process in more than one way. For example, the outer rectangular height is $11+2x$, and the width is $x$, so the left piece including everything on the path from top to bottom will have area $x(11+2x)$. There are two of these, and so your left and right paths have area $2x(11+2x)$. The two remaining unaccounted for pieces have area $17x$ each, and there are two of these, so solve
$$2x(11+2x)+2 \cdot 17 =165.$$
The answer will be the same. You could also try taking the top and bottom path areas, then add the remaining unaccounted for height portions, but I will save that for you to see.
The width of your path should be $\frac{5}{2}$. Note you will get two solutions to your quadratic. You clearly take the positive solution. You will get the same solution by any of the methods.
I do not feel too bad about the plot spoiler here as you have enough work in front of yourself drawing this diagram, setting up the quadratic, and solving it. If you manage to get through this spoiler successfully I think you will have learned your homework lesson.
Best Answer
As I could not personally find a solution using only pre-calculus, this answer uses calculus:
The formula for the area of this rectangle will be $A(x)=(2x)(5-x^2)$ where we define x as the positive corner of the rectangle that intersects the parabola.
The formula is $A(x)=(2x)(5-x^2)$ because a rectangle's formula for area is $A=LW$ because the width will be the height of the function and the length will be twice x, and as the function is even, x and -x will be the same height. $0<x<\sqrt5$ because if $|x|>\sqrt5$ then $5-x^2$ will be less than 0, so it will not be above the x axis.
To approach this question, we can use calculus. So we are trying to find the maximum of $(2x)(5-x^2)$ or equivalently $-2x^3+10x$ on the interval $(0,\sqrt5)$. To do this, we must analyze the critical points of the derivative, $A'(x)$ and the endpoints of the interval.
$A'(x)=-6x^2+10$
Then to find critical points, set $A'(x)=0$
$-6x^2+10=0$
$x^2=\frac {10}6$
$ x= \pm \sqrt {\frac {5}3}$
We are looking on the interval $(0,\sqrt5)$ so we will disregard the critical point at $x=-\sqrt {\frac {5}3}$. Our largest area will either be at the critical point or at the endpoints of the interval, so we will test those three cases: $x=0, \sqrt {\frac {5}3}, \sqrt {5}$
$A(0)=(2(0))(5-(0)^2)=0$
$A(\sqrt {\frac {5}3})=(2(\sqrt {\frac {5}3}))(5-(\sqrt {\frac {5}3})^2)=(2\sqrt {\frac {5}3})({\frac {10}3})=\frac {20\sqrt {\frac {5}3}} 3≈8.60662965824$
$A(\sqrt 5)=(2(\sqrt 5))(5-(\sqrt 5)^2)=0$
Of these, $A(\sqrt {\frac {5}3})$ is greatest, so it is the maximum.
So the maximum area is about $8.607$, the base is twice x, or $2\sqrt {\frac {5}3}$, and the height is $5-{\frac {5}3}$ or $\frac {10}3$.