I already asked this question and had some help, but I am still confused on a piece of it. I'm not sure what differentiate is.
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
I should have noticed the "Pre Calc" in your title.
You are not expected to know what "differentiate" means in that case.
So you are with me as far as the formula for the area is, right ?
It was
A(x) = 36x - x^3
Let's graph this.
since the right side factors into
A(x) = x(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?
It should be (√12,48√12)
or
appr. (3.464, 166.277)
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*------+------*
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* | | | *
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* | | | *
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- * - -+------+------+- - * -
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I found this picture online and this is the picture my teacher gave. Can we please start from the beginning. Here is the question again:
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).
Can you please explain in more detail. This all seems foreign to me. I'm sorry for seeming dense. I understand B, but A and C are confusing me.
In this question, the problem asks you to find the area of a rectangle that is inscribed between the x-axis and the parabola y=36-x^2.
a. To write the area A of the rectangle as a function of x, we need to consider that one side of the rectangle is along the x-axis and the two remaining sides are vertical with endpoints on the parabola.
Suppose we label the point of contact of the rectangle and the parabola as P(x,y). Since the rectangle is along the x-axis, the y-coordinate of P is 0. Thus, the coordinates of P can be written as P(x,0).
The length of the rectangle will be twice the distance from the origin to P(x,0), which is 2x. The width of the rectangle will be the distance from the origin to the points where the parabola intersects the rectangle, which is 2(36-x^2).
Therefore, the area A of the rectangle can be written as A(x) = 2x(36 - x^2) or A(x) = 72x - 2x^3.
b. To determine the domain of A, we need to consider the validity of the height of the rectangle, which is given by the expression 36-x^2. Since the parabola is defined for all real numbers, the height of the rectangle will also be valid for all real numbers. Hence, the domain of A is -∞ < x < ∞.
c. To determine the maximum area that the rectangle may have, we need to find the value of x that maximizes the area function A(x).
To do this, we differentiate the area function A(x) with respect to x. Differentiation is a way of finding the rate at which a function changes and is represented by the symbol d/dx, pronounced as "dee dee ex".
Differentiating A(x) = 72x - 2x^3 with respect to x, we get dA/dx = 72 - 6x^2. In this equation, dA/dx represents the derivative of the area function with respect to x.
Next, we set dA/dx equal to zero to find the critical points, which are the values of x where the maximum or minimum occurs. Solving the equation 72 - 6x^2 = 0 for x, we find x^2 = 12.
Taking the square root of both sides, we get x = ±√12, which simplifies to x = ±2√3. However, since we are dealing with the dimensions of a rectangle, we can ignore the negative value since it does not make sense in the context of the problem.
Therefore, the critical point is x = √12 = 2√3.
Substituting this value back into the area function A(x), we find A(2√3) = 72(2√3) - 2(2√3)^3 = 48√3.
Hence, the maximum area that the rectangle may have is 48√3, which is approximately equal to 83.08.