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 = ± √12
so the max area occurs when x = √12
and it is
A(√12) = 36√12 - 12√12
= 24√12 or appr. 41.57
Using a graphing calculator you are on your own, I don't have one, but 24√12 is the 'exact' answer.
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!
5 answers
"He drew the picture with the parabola at points (-6,0) (6,0) (0,36)"
The point (0,6) is not on the parabola, (0,36) is.
The parabola is given as:
y(x)=36-x²
Given the rectangle is inside the parabola and above (I think) the x-axis, we define the four corners of the rectangle as
(-x,0), (x,0), (x,y(x)), (-x, y(x))
from which the area A can be calculated as the product of the width (x-(-x) and the height (y(x)).
A(x) = 2x(y(x)
domain A(x) [-6,6]
range A(x) [0,36].
For the maximum, you can use the graphics calculator, or you can tabulate the values, refining the grid where the maximum is located. Since this is a pre-calc course, I do not assume you are allowed to find the minimum by derivatives.
All this is assuming that I interpreted the "diagram" correctly. Check my steps.
The point (0,6) is not on the parabola, (0,36) is.
The parabola is given as:
y(x)=36-x²
Given the rectangle is inside the parabola and above (I think) the x-axis, we define the four corners of the rectangle as
(-x,0), (x,0), (x,y(x)), (-x, y(x))
from which the area A can be calculated as the product of the width (x-(-x) and the height (y(x)).
A(x) = 2x(y(x)
domain A(x) [-6,6]
range A(x) [0,36].
For the maximum, you can use the graphics calculator, or you can tabulate the values, refining the grid where the maximum is located. Since this is a pre-calc course, I do not assume you are allowed to find the minimum by derivatives.
All this is assuming that I interpreted the "diagram" correctly. Check my steps.
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 = √12
and it is
A(√12) = 72√12 - 24√12
= 48√12 or appr. 166.277
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 = √12
and it is
A(√12) = 72√12 - 24√12
= 48√12 or appr. 166.277
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
pre caculates is the ansewer to above is 25545872
2 answers