How much will the area of rectangle ABCD increase?

Rectangle ABCD is shown in the figure above. Points A and B lie on the graph of y = 64 – x², and points C and D lie on the graph of y = x² – 36. Segments AD and BC are both parallel to the y-axis. The x-coordinates of points A and B are equal to -k and k, respectively. If the value of k changes from 2 to 4, by how much will the area of rectangle ABCD increase?

A. 352
B. 88
C. 272
D. 544
E. 176

Correct Answer: E
Explanation/Reference:
Explanation:
What we need to do is generate an expression for the area of the rectangle in terms of k. Then, all we have to do is substitute 2 and 4 in for k and see how the area changes.
Remember that the area of any rectangle is the product of the length and the width. In the case of rectangle ABCD, we can use AD to represent the length and AB to represent the height.
Let’s find the coordinates of A, B, and D in terms of k. Then we can use the distance formula to find the lengths of AB and AD.
We are told that A and B have x-coordinates equal to -k and k, respectively. In order to find the value of the y-coordinates of A and B, we need to substitute -k and k into the equation for y = 64 – x² since we know that A and B lie on this graph.
y-coordinate of A = 64 – (-k)² = 64 – k² y-coordinate of B = 64 – k² Thus, we can say that A is located at (-k, 64 – k²), and B is located at (k, 64 – k²).
Next, let’s find the coordinates of D. Because AD is parallel to the y-axis, A and D must have the same x-coordinate. Thus, the x-coordinate of point D is also -k. However, in order to find the y-coordinate of point D, we will need to use the equation y = x² – 36, since we are told that D lies on this graph. All we have to do is substitute -k in place of x.
y-coordinate of D = (-k)² – 36 = k² – 36 This means D is located at (-k, k² – 36).
We now have the coordinates of A, B, and D, which we can use to find the width and length of the rectangle.
Let’s find the distance between A and B, which will give us the rectangle’s width.
A is located at (-k, 64 – k²), and B is located at (k, 64 – k²).
The distance d between any points (x₁, y₁) and (x₂, y₂) is given below:

The length of AB is 2k. How, let’s find the length of AD. D is located at (-k, k² – 36).

The length of AD is 100 – 2k² The length of AB is 2k and the length of AD is 100 – 2k². If we multiply these two expressions together, we will have the area of rectangle ABCD.
Area of ABCD = 2k(100 – 2k²) = 200k – 4k³ Now, let’s let k = 2 and see what the area of ABCD is.
Area = 200k – 4k³ = 200(2) – 4(23) = 400 – 32 = 368 Then, we will let k = 4.
Area = 200k – 4k³ = 200(4) – 4(43) = 800 – 256 = 544 The Area changed from 368 to 544. The question asks us to find how much this increase is, so we need to find the different between 544 and 368.
Increase in area = 544 – 368 = 176The answer is 176.

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