You stated on Saturday that you understood my solution as far as part a) and b) were, so let me just recap what we did.

Question

(I will not repeat the actual question)

My Solution:

Suppose we label the point of contact P(x,y). I bet P is in the first quadrant.

but we have the equation for y, so we could call the point P(x,36-x^2)

Isn't the contact point on the x-axis (x,0) ?
And isn't the base of the rectangle 2x (The distance from the origin to the right is the same as the distance to the left)

a) so the area is
A(x) = 2x(36 - x^2) or
A(x) = 72x - 2x^3

b) wouldn't the domain be -6 < x < +6 or else the height 36-x^2 wouldn't make any sense.

c) So you are with me as far as the formula for the area is, right ?
It was
A(x) = 72x - 2x^3

Let's graph this.
since the right side factors into
A(x) = 2x(x-6)(x+6) you should have learned that x = 0, x = 6, and x = -6 are the x -intercepts of the graph.

so the domain of the graph that would make sense for "area" is the part of the graph from x = 0 to x = 6, which looks like a downwards parabola.

I think this is where your graphing calculator will come in handy.
Can you use your calculator to find the coordinates of the vertex of that part of the graph?

appr. (3.464, 166.277)

Since A(3.464) = 166.277

the maximum area is 166.277 when the x value is 3.464