Find the maximum possible area of a rectangle in quadrant 1 under the curve y=(x-4)^2 (with one corner at the origin and one corner on the curve y=(x-4)^2)

To find the maximum possible area of a rectangle in quadrant 1 under the curve y=(x-4)^2, we need to find the coordinates of the corner point of the rectangle on the curve.

1. Let's assume the coordinates of the corner point of the rectangle on the curve are (a, (a-4)^2).
2. The area of the rectangle is given by length * width.
3. The length of the rectangle is a, and the width is given by ((a-4)^2 - 0) = (a-4)^2.
4. Therefore, the area of the rectangle is A = a * (a-4)^2.

To find the maximum area, we need to maximize the function A. We can do this by finding the derivative of A and setting it equal to 0.

5. Differentiating A with respect to a, we get:
dA/da = (a-4)^2 + a * 2(a-4) * 1

6. Setting dA/da = 0, we have:
(a-4)^2 + a * 2(a-4) * 1 = 0

7. Simplifying the equation, we get:
(a-4)^2 + 2(a-4) * a = 0
(a-4)^2 + 2a(a-4) = 0
(a-4)^2 + 2a^2 - 8a = 0
a^2 - 8a + 16 + 2a^2 - 8a = 0
3a^2 - 16a + 16 = 0

8. Solving this quadratic equation, we get:
a = (16 ± sqrt((-16)^2 - 4*3*16))/ (2*3)
a = (16 ± 4) / 6
a = 10/3 or a = 2

9. The value of a should be in the range (0, 4) since we are considering quadrant 1.
Therefore, a = 2.

10. Now, substitute the value of a into the area formula A = a * (a-4)^2:
A = 2 * (2-4)^2
A = 2 * (-2)^2
A = 2 * 4
A = 8

Therefore, the maximum possible area of a rectangle in quadrant 1 under the curve y=(x-4)^2 is 8 square units.

To find the maximum possible area of a rectangle in quadrant 1 under the curve y = (x-4)^2, we need to determine the dimensions of the rectangle that will yield the largest area.

Let's assume the rectangle has its bottom left corner at the origin (0, 0) and its top right corner on the curve y = (x-4)^2.

To find the top right corner of the rectangle, we need to find the x-coordinate that corresponds to the maximum value of y = (x-4)^2.

The maximum value of y = (x-4)^2 occurs at the vertex of the parabola, so we need to find the x-coordinate of the vertex.

The x-coordinate of the vertex can be determined using the formula x = -b/(2a), where the equation of the parabola is in the form y = ax^2 + bx + c.

In this case, the equation y = (x-4)^2 is already in vertex form, with a = 1, b = -8, and c = 16. Therefore, the x-coordinate of the vertex is x = -(-8)/(2*1) = 4.

So, the top right corner of the rectangle will have the coordinates (4, (4-4)^2) = (4, 0), since the y-coordinate is 0 at the vertex.

Now, we can calculate the dimensions of the rectangle. The width will be the x-coordinate of the top right corner, which is 4 units, and the height will be the y-coordinate of the top right corner, which is 0 units.

The area of the rectangle is given by length times width, so in this case, it is 4 units times 0 units, which is 0 square units.

Therefore, the maximum possible area of the rectangle is 0 square units.