Pre-Calc - Follow-up for Reiny - MUFFY, Saturday, October 3, 2009 at 10:27pm

Question

I wasn't sure if you would notice my question since it is so much later so I re-sent this as a new question. Please let me know if I shouldn't do that.

I understand how you got a and b, but I am confused with c. What is differentiate and how did you get the equation dA/dx = 72 - 6x^2

Question:
Here I am again stuck with another geometry type question. Here goes:

A rectangle is inscribed between the x-axis and the parabola y=36-x^2
with one side along the x-axis.
He drew the picture with the parabola at points (-6,0) (6,0) (0,6)

the rectanle is drawn inside the parabola along the x axis.
Not sure if you can help me without seeing the picture.

a. Let x denote the x-coordinate of the point shown in the figure. Write the area A of the rectangle as a function of x.

b. What values of x are in the domain of A?

c. Determine the maximum area that the rectangle may have. (hint, use a graphing calculator).

I'm already confused just looking at a.

if you can guide me it would be great!


Pre-Calc - Reiny, Saturday, October 3, 2009 at 11:34am
Have seen this type of questions many many times.

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) = x(36 - x^2) or
A(x) = 36x - x^3

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

c) I would now differentiate to get
dA/dx = 36 - 3x^2
set that equal to zero for a max of A
3x^2 = 36
x^2 = 12
x = �} �ã12

so the max area occurs when x = �ã12
and it is
A(�ã12) = 36�ã12 - 12�ã12
= 24�ã12 or appr. 41.57

Using a graphing calculator you are on your own, I don't have one, but 24�ã12 is the 'exact' answer.

OOOPS - Pre-Calc - Reiny, Saturday, October 3, 2009 at 11:41am
silly me, right after telling you the base is 2x, I use only 1x in my calculation.

HERE IS THE NEW VERSION :

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) I would now differentiate to get
dA/dx = 72 - 6x^2
set that equal to zero for a max of A
6x^2 = 72
x^2 = 12

so the max area occurs when x = �ã12
and it is
A(�ã12) = 72�ã12 - 24�ã12
= 48�ã12 or appr. 166.277